3.3: Trigonometric Equations

In the previous section on trigonometric identities we worked with equations that would be true for all values of a particular angle ( heta .) These are sort of like the algebraic equations whose solution set is "all real numbers," like (2 x+10=) (2(x+1)+8 .) In this section, we will solve trigonometric equations whose solution set involves only certain values for the angle in question. Because of the cyclical nature of the angles we're working with, there will often be an infinite number of solutions although not "all real numbers."

Example 1
Here's an example. Suppose that we consider the equation (sin x=0.5 .) Whether we use technology, a table or reasoning to solve this equation, it's clear that one solution is (30^{circ} .) However, remember from the beginning of Chapter 2 that the sine function is positive in Quadrant (Pi). That means that a second quadrant angle with a reference angle of (30^{circ}) also has a sine equal to (0.5 .) Recall the ASTC diagram from Chapter 2 :

So, the sine function is positive in Quadrants I and (Pi). This means that in addition to a solution of (30^{circ},) there is another solution in Quadrant (Pi). As mentioned above, this second quadrant solution has a reference angle of (30^{circ})

To find this angle, we simply subtract (180^{circ}-30^{circ}=150^{circ})
In Quadrant (Pi,) we subtract the reference angle from (180^{circ})

In Quadrant (Pi V), we subtract the reference angle from (360^{circ})
So, the solutions to the equation (sin x=0.5) between (0^{circ}) and (360^{circ}) are (x=30^{circ}, 150^{circ}) In this chapter we will consider mainly solutions with this restriction:
[0^{circ} leq x<360^{circ}
]The infinite solutions to this equation can be expressed as:
[30^{circ}+n cdot 360^{circ} ext { and } 150^{circ}+n cdot 360^{circ}
]

Let's look at another example:
Example 2
Find all solutions of the given equation for (0^{circ} leq x<360^{circ})
( an x=4)
Using a calculator to find ( an ^{-1}(4),) we find that it returns an answer of (x approx 75.96^{circ}) So this is the solution to the equation that lies in QuadrantI. The tangent function is also positive in Quadrant (Pi), so we should also consider the third quadrant angle with a reference angle of (75.96^{circ})

(180^{circ}+75.96^{circ}=255.96^{circ},) so our solutions for this equation are (x approx 75.96^{circ}, 255.96^{circ})
Often, calculators are programmed to return an angle value that is not between (0^{circ} leq x<360^{circ})

Example 3
Find all solutions of the given equation for (0^{circ} leq x<360^{circ})
(sin x=-0.25)
Solving this on a TI calculator would generally return a value of (-14.5^{circ} .) However, (-14.5^{circ}) is clearly not between (0^{circ}) and (360^{circ}), so we need to use this information to find the solutions that are between (0^{circ}) and (360^{circ})

With the calculator returning a vlue of (-14.5^{circ},) we know that the reference angle for all answers will be (14.5^{circ} .) Knowing this, we can say that the sine is negative in Quadrants III and IV, so we'll need angles in those quadrants with reference angles of (14.5^{circ})

In Quadrant (Pi) we'll subtract the reference angle from (360^{circ}: 360^{circ}-14.5^{circ}=345.5^{circ})
So, (x approx 194.5^{circ}, 345.5^{circ})

Some trigonometric equations have no real number solutions. The equation (sin x=2) has no real number solutions. Recall that the sine ratio was originally defined as the ratio of the side opposite an angle to the hypotenuse. The hypotenuse is always the longest side in a right triangle so there is no way the sine function could be greater than 1 if we're working with real-valued angles. However, in the same way that complex numbers are used to solve equations like (x^{2}=-7,) complex-valued angles can be used to solve equations such as (sin x=2) We won't go into this here, however, there is a relatively straightforward way to solve these equations.
If you encounter an equation like (cos x=3) and are solving for values of (x) (0^{circ} leq x<360^{circ},) then the proper response is "no solution" or "no real solution." However, remember that the tangent function can take any value between (-infty) and (infty)

Example 4
Solving an equation that includes a reciprocal trigonometric function simply involves the extra step of finding the reciprocal:
Find all solutions of the given equation for (0^{circ} leq x<360^{circ})
(sec x=12)
The trick here is to restate the equation so that we can use the preprogrammed values from a calculator to find the solution.
If (sec x=12) then (cos x=frac{1}{12} .) Finding (cos ^{-1}left(frac{1}{12} ight)) gives a solution of (x approx 85.2^{circ})
The cosine and the secant are both positive in Quadrant (Pi mathrm{V}), so we'll also want a fourth quadrant angle whose reference angle is (85.2^{circ})

In Quadrant (Pi), we'll subtract the reference angle from (360^{circ})
[egin{array}{c}
360^{circ}-85.2^{circ} approx 274.8^{circ}
x approx 85.2^{circ}, 274.8^{circ}
end{array}
]

Example 5
Solving a quadratic trigonometric equation often involves the use of the quadratic formula:
Find all solutions of the given equation for (0^{circ} leq x<360^{circ})
(2 sin ^{2} x-sin x-2=0)
Using the quadratic formula we arrive at approximate values for (sin x) of (sin x approx-0.7808,1.2808)
The solution (sin x approx 1.2808) yields no real solutions, so we will focus on solving (sin x approx-0.7808)
Finding (sin ^{-1}(-0.7808)) gives us an answer of (approx-51.3^{circ} .) This means our answers will lie in Quadrants III and IV with reference angles of (51.3^{circ} .) In Quadrant III we'll say (180^{circ}+51.3^{circ} approx 231.3^{circ} .) In Quadrant (mathrm{IV}), we'll subtract the reference angle from (360^{circ}: 360^{circ}-51.3^{circ} approx 308.7^{circ})
(mathrm{So}, x approx 231.3^{circ}, 308.7^{circ})

Exercises 3.3
Find all solutions for (0^{circ} leq x<360^{circ}) Round all angle measures to the nearest (10^{ ext {th }}) of a degree.
1. (cos x-0.75=0)
2. (sin x+0.432=0)
3. ( 3 sin x-5=0)
4. (sin x-4=0)
5. ( 3 sec x+8=0)
6. ( 4 csc x+9=0)
7. ( 3-5 sin x=4 sin x+1)
8. ( 4 cos x-5=cos x-3)
9. ( 3 an ^{2} x+2 an x=0)
10. ( 4 cos ^{2} x-cos x=0)
11. ( 3 cos ^{2} x+5 cos x-2=0)
12. ( 2 cot ^{2} x-7 cot x+3=0)
13. ( 2 an ^{2} x- an x-10=0)
14. ( 2 sin ^{2} x+5 sin x+3=0)
15. ( 2 cos ^{2} x-5 cos x-5=0)
16. ( 3 sin ^{2} x-sin x-1=0)

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Learning trigonometry equations table:

The above indicate the trigonometric table. Here we are having the value for the trigonometric functions value 0 o , 30 o , 45 o , 60 o o . and 90

Learning trigonometric equations problems:

Solve the trigonometric equation 2sin x – 1 =0

Here we are going to learn the simple trigonometric equation 2sin x – 1 = 0

From the above table we get x = 30 o

We know Sec 2 x = 1 + tan 2 x

Plug it into the above trigonometric equation

1 + tan 2 x – 2 tan x -4 = 4 – 4

x = tan -1 (3) and x = tan -1 (-1)

From the above problem we learn how to solve the twisted trigonometric equations.

I am planning to write more post on sine graphs with example, Name Four Types of Quadrilaterals. Keep checking my blog.

Trigonometry

The three basic trigonometric functions occur so often as the denominator of a fraction that it is convenient to give names to their reciprocals. We define three new trigonometric functions as follows.

Definition 8.47 . Three More Functions.

If ( heta) is an angle in standard position, and (P(x,y)) is a point on the terminal side, then we define the following functions.

We can find exact values for all six trig functions at a given angle if we know the value of any one of them.

Example 8.48 .

If (sec heta = 3 ext<,>) and (-dfrac <2>le heta le 0 ext<,>) find exact values for the other five trig functions.

Because (-dfrac <2>le heta le 0 ext<,>) we draw a reference triangle in the fourth quadrant, as shown at right. Because (sec heta = 3 = dfrac<3><1> ext<,>) we label the horizontal leg with (x=1) and the hypotenuse with (r=3 ext<.>)

From the Pythagorean theorem, we find (y=-sqrt<8>=-2sqrt<2> ext<.>) We can now compute the values of the six trigonometric ratios.

Checkpoint 8.49 .

If (csc heta =4 ext<,>) and (90degree le heta le 180degree ext<,>) find exact values for the other five trig functions.

By comparing the definitions of secant, cosecant, and cotangent to the three basic trigonometric functions, we find the following relationships.

Reciprocal Trigonometric Functions.

Calculators do not have keys for the secant, cosecant, and cotangent functions instead, we calculate their values as reciprocals.

Example 8.50 .

Use a calculator to approximate (sec 47degree) to three decimal places.

With the calculator in degree mode, enter

to obtain (sec 47degree approx 1.466 ext<.>) Or we can calculate (cos 47degree) first, and then use the reciprocal key:

Checkpoint 8.51 .

Use a calculator to approximate (csc 132degree) to three decimal places.

Of course, we can also evaluate the reciprocal trig functions for angles in radians, or for real numbers. Thus for example,

In particular, the exact values for the reciprocal trig functions of the special angles are easily obtained.

 Exact Values for the Special Angles ( heta) (sec heta) (sec heta) (cot heta) (0) (1) undefined undefined (dfrac<6>) (dfrac<2sqrt<3>><3>) (2) (sqrt<3>) (dfrac<4>) (sqrt<2>) (sqrt<2>) (1) (dfrac<3>) (2) (dfrac<2sqrt<3>><3>) (dfrac<1>>) (dfrac<2>) undefined (1) (0)
Caution 8.52 .

The reciprocal functions are not the same as the inverse trig functions!

For example, (sec 0.8) is not equal to (cos^<-1>(0.8) ext<.>) Remember that (

) is an angle, namely the angle whose cosine is 0.8, while (

) is the reciprocal of the cosine of 0.8 radians, or (dfrac<1> ext<.>) You can check on your calculator that

Each of the reciprocal functions is undefined when its denominator is equal to zero. For example, the secant is undefined when (cos heta = 0 ext<,>) or when ( heta) is an odd multiple of (90degree ext<.>)

Example 8.53 .

For which angles is the cosecant undefined?

The cosecant is undefined when its denominator, (sin heta ext<,>) equals zero, and (sin heta = 0) when ( heta) is a multiple of (180degree ext<.>) In radians, (csc heta) is undefined if ( heta) is a multiple of (pi ext<.>)

Checkpoint 8.54 .

Multiples of (180degree ext<,>) or multiples of (pi ext<.>)

Note 8.55 .

Although ( an dfrac<2>) is undefined, (cot dfrac<2>=0 ext<.>)

Subsection Application to Right Triangles

In Chapter 2 we defined three trigonometric ratios for an acute angle namely, sine, cosine, and tangent. When we take the reciprocals of those ratios, we obtain expressions for the secant, cosecant, and cotangent.

Reciprocal Trigonometric Ratios.

If ( heta) is one of the acute angles in a right triangle,

Although we can express any relationship between the sides of a right triangle using sine, cosine, and tangent, sometimes it is more convenient to use one of the reciprocal functions.

Example 8.56 .

The length, (L ext<,>) of the shadow cast by a flagpole on a sunny day depends on the height, (h ext<,>) of the flagpole and the angle, ( heta ext<,>) that the sun's rays make with ground.

1. Write an expression for the length, (L ext<,>) of the shadow cast by a flagpole of height (h) when the sun makes an angle of ( heta) from the ground.
2. Find the length (to the nearest 0.01 meter) of the shadow cast by a 3-meter flagpole when the sun makes an angle of (20degree) from the ground.
1. From the figure, we see that (dfrac=cot heta ext<,>) or (L=hcot heta ext<.>)
2. Substiting (alert<3>) for (h) and (alert<20degree>) for ( heta ext<,>) we find
Checkpoint 8.57 .

The area (A) of a regular polygon with (n) sides having perimeter (L) satisfies

Refer to the figure at right showing (n=6) to prove this formula in the following steps.

1. Find an expression for the angle ( heta) in terms of (n ext<.>)
2. Find an expression for the base of the triangle shown.
3. Find an expression for the height of the triangle.
4. Write an expression for the area of the triangle, and then for the area of the entire polygon.
1. (displaystyle heta = dfrac)
2. (displaystyle b = dfrac)
3. (displaystyle h=dfrac<2n>cot dfrac)
4. (displaystyle A_T=dfrac<4n^2>cot dfrac,

Subsection Graphs of the Reciprocal Functions

We can obtain graphs of the reciprocal trig functions by plotting points, as we did for the sine, cosine and tangent functions. However, it is more enlightening to construct these graphs as the reciprocals of the three basic functions.

Example 8.58 .

Use the graph of (y=cos x) to construct a graph of (f(x)=sec x ext<.>)

Consider the graph of (y=cos x) shown at left below.

cos x=0 ext<,>) so (sec x) is undefined at these (x)-values, and we insert vertical asymptotes at those (x)-values to start our graph of (y=sec x ext<,>) as shown at right below.

To find some points on the graph, we look at points on the graph of (y=cos x ext<.>) At each (x)-value, the (y)-coordinate of the point on the graph of (y=sec x) is the reciprocal of (cos x ext<.>)

For example, at (x=0) and (x=2pi ext<,>) we have (cos x = 1 ext<,>) so (sec x = frac<1> <1>= 1 ext<.>) Thus, we plot the points ((0,1)) and ((2pi,1)) on the graph of (f(x)=sec x ext<.>) Similarly, at (x=-pi) and (x=pi ext<,>) (cos x = -1 ext<,>) so the value of (sec x) is (frac<1> <-1>= -1 ext<,>) and we plot the points ((-pi,-1)) and ((pi,-1)) on the graph of (f(x)=sec x ext<.>)

Finally, we notice that the values of (cos x) are decreasing toward (0) as (x) increases from (0) to (dfrac<2> ext<,>) so the graph of (f(x)=sec x) increases toward (infty) on the same interval.

By similar arguments, we fill in the graph of (f(x)=sec x) between each of the vertical asymptotes, to produce the graph below.

Checkpoint 8.59 .

Use the graph of (y= an x) to sketch a graph of (g(x)=cot x ext<.>)

The graphs of the three new functions are shown below, with (x) in radians. Note that the secant function is undefined at odd multiples of (dfrac<2> ext<,>) the values at which (cos x=0 ext<.>) The cosecant is undefined where (sin x=0 ext<,>) namely at multiples of (pi ext<.>) The cotangent is also undefined at multiples of (pi ext<,>) because ( an x=0) at those values.

Example 8.60 .

State the domain and range of the secant function.

Because the cosine is defined for all real numbers, the domain of the secant includes all real numbers except for values where the cosine is zero. These values are the odd multiples of (dfrac<2> ext<,>) that is, (dfrac<2>,

ldots ext<,>) and their opposites.

Because the range of the cosine consists of all (y)-values with (-1 le y le 1 ext<,>) the range of the secant includes the reciprocals of those values, namely (y ge 1) and (y le -1 ext<.>)

Checkpoint 8.61 .

State the domain and range of the cosecant and cotangent functions.

Domain of cosecant: all real numbers except integer multiples of (pi ext<>) Range of cosecant: ((-infty,-1] cup [1, infty))

Domain of cotangent: all real numbers except integer multiples of (pi ext<>) Range of cotangent: all real numbers

Subsection Solving Equations

From the graph of the secant function, we can see that the equation (sec heta =k) has two solutions between (0) and (2pi) if (k ge 1) or (k le -1 ext<,>) but no solution for (-1 lt k lt 1 ext<.>) The same is true of the cosecant function: the equation (csc heta = k) has no solution for (-1 lt k lt 1 ext<.>)

Example 8.62 .

) for ( heta) between (0) and (2pi ext<.>)

We take the reciprocal of each side of the equation to obtain

Because (dfrac><2>) is one of the special values, we recognize that one of the solutions is ( heta =dfrac<3> ext<.>) The sine and the cosecant are also positive in the second quadrant, so the second solution is (pi - dfrac <3>= dfrac<2pi><3> ext<.>)

Example Dish Example

It is not always apparent that the three angles to specify a rotation are not independent of each other and must be applied in a certain order. For example imagine that we are aiming a dish at a satellite. The azimuth and elevation are independent of each other, for example we can aim south and then elevate up by the required inclination, or we can set the elevation and then turn and point toward the south. However there is a third angle, we can rotate about the line to the satellite, to correctly align with the horizontal and vertically polarised signal from the satellite, this third angle is dependant on the others so we cant escape from this issue.

When the angles are small, then they are nearly independent of each other, for example if we are aiming at a small area of sky.

Solved Examples on Trigonometric Equations

For ACT Students
The ACT is a timed exam. $60$ questions for $60$ minutes
This implies that you have to solve each question in one minute.
Some questions will typically take less than a minute a solve.
Some questions will typically take more than a minute to solve.
The goal is to maximize your time. You use the time saved on those questions you solved in less than a minute, to solve the questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no "negative" penalty for any wrong answer.

For JAMB and CMAT Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For WASSCE Students
Any question labeled WASCCE is a question for the WASCCE General Mathematics
Any question labeled WASSCE:FM is a question for the WASSCE Further Mathematics/Elective Mathematics

For GCSE Students
All work is shown to satisfy (and actually exceed) the minimum for awarding method marks.
Calculators are allowed for some questions. Calculators are not allowed for some questions.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits. separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve the trigonometric equations
Determine the specific solutions of the trigonometric equations in the specified intervals
Write the general solution of the trigonometric equations
State the reasons for every step
Show all work
Unless otherwise specified or implied, only the exact values are allowed.

(1.) ACT For all real values of $x$, which of the following equations is true?

$F.:: sin(7x) + cos(7x) = 7 [3ex] G.:: sin(7x) + cos(7x) = 1 [3ex] H.:: 7sin(7x) + 7cos(7x) = 14 [3ex] J.:: sin^2(7x) + cos^2(7x) = 7 [3ex] K.:: sin^2(7x) + cos^2(7x) = 1$

This question should take you at most $5$ seconds.

$F.:: sin(7x) + cos(7x) = 7. NOT [3ex] G.:: sin(7x) + cos(7x) = 1. NOT [3ex] H.:: 7sin(7x) + 7cos(7x) = 14. NOT [3ex] J.:: sin^2(7x) + cos^2(7x) = 7. NOT [3ex] K.:: sin^2(7x) + cos^2(7x) = 1. Pythagorean:: Identity. YES$

(2.) Determine all the solutions of $4sin heta + 5 = 3$ in the interval $[0, 360^circ)$

$4sin heta + 5 = 3 [3ex] 4sin heta = 3 - 5 [3ex] 4sin heta = -2 [3ex] sin heta = -dfrac<2> <4>[5ex] sin heta = -dfrac<1> <2>[5ex] Argument = heta [3ex] heta = sin^ <-1>left(-dfrac<1><2> ight) [5ex] heta = 210^circ, 330^circ . :Unit:: Circle:: Trig [3ex] 210 = 210 * dfrac <180>= dfrac<7pi> <6>[5ex] 330 = 330 * dfrac <180>= dfrac<11pi> <6>[5ex] heta = dfrac<7pi><6>, dfrac<11pi> <6>[5ex] underline [3ex] heta = 210 + 360k . coterminal:: angle s [3ex] heta = 330 + 360k . coterminal:: angle s [3ex]$ Check
 $underline [3ex] heta = 210^circ [3ex] [3ex] 4sin heta + 5 [3ex] = 4 * sin210 + 5 [5ex] = 4 * -dfrac<1> <2>+ 5 [5ex] = -2 + 5 [3ex] = 3 [3ex]$ $heta = 330^circ [3ex] 4sin heta + 5 [3ex] = 4 * sin 330 + 5 [3ex] = 4 * -dfrac<1> <2>+ 5 [5ex] = -2 + 5 [3ex] = 3$ $underline [3ex] 3$

(3.) Determine all the solutions of $2sineta - sqrt <3>= 0$ in the interval $[0, 2pi)$

$2sineta - sqrt <3>= 0 [3ex] 2sineta = sqrt <3>[3ex] sineta = dfrac> <2>[5ex] Argument = eta [3ex] eta = sin^<-1>left(dfrac><2> ight) [5ex] eta = 60^circ, 120^circ . Unit:: Circle:: Trig [3ex] 60 = 60 * dfrac <180>= dfrac <3>[5ex] 120 = 120 * dfrac <180>= dfrac<2pi> <3>[5ex] underline [3ex] eta = dfrac <3>+ 2pi k . coterminal:: angle s [5ex] eta = dfrac<2pi> <3>+ 2pi k . coterminal:: angle s [5ex]$ Check
 $underline [3ex] eta = dfrac <3>[5ex] 2sineta - sqrt <3>[3ex] = 2 * sindfrac <3>- sqrt <3>[5ex] = 2 * dfrac> <2>- sqrt <3>[5ex] = sqrt <3>- sqrt <3>[3ex] = 0 [3ex]$ $eta = dfrac<2pi> <3>[5ex] 2sineta - sqrt <3>[3ex] = 2 * sindfrac<2pi> <3>- sqrt <3>[5ex] = 2 * dfrac> <2>- sqrt <3>[5ex] = sqrt <3>- sqrt <3>[3ex] = 0$ $underline [3ex] 0$

(4.) ACT What are the values of $heta$, between $heta$ and $2pi$, when $an heta = -1$?

$an heta = -1 [3ex] Argument = heta [3ex] Assume:: an heta = 1 [3ex] heta = an^<-1>(1) = 45^circ = 45 * dfrac <180>= dfrac <4>[5ex] But:: an heta = -1 [3ex] an ::is:: negative:: in:: 2nd ::and:: 4th ::quadrants [3ex] - an left(dfrac<4> ight) = an left(pi - dfrac<4> ight). 2nd:: Quadrant:: Identity [5ex] = an left(dfrac<4pi> <4>- dfrac<4> ight) [5ex] = an left(dfrac<4pi - pi><4> ight) [5ex] = an left(dfrac<3pi><4> ight) [5ex] Also, [3ex] - an left(dfrac<4> ight) = an left(2pi - dfrac<4> ight). 4th:: Quadrant:: Identity [5ex] = an left(dfrac<8pi> <4>- dfrac<4> ight) [5ex] = an left(dfrac<8pi - pi><4> ight) [5ex] = an left(dfrac<7pi><4> ight) [5ex] heta = dfrac<3pi><4>,: dfrac<7pi> <4>[5ex]$ Check
 $underline [3ex] an heta [3ex] heta = dfrac<3pi> <4>[5ex] an left(dfrac<3pi><4> ight) = -1 [5ex]$ $heta = dfrac<7pi> <4>[5ex] an left(dfrac<7pi><4> ight) = -1$ $underline [3ex] -1$

(5.) ACT If ^circ le x^circ le 90^circ$, and$2sin^2x^circ - 1 = 0$, then$x^circ = $?$ A.:: 0^circ [3ex] B.:: 30^circ [3ex] C.:: 45^circ [3ex] D.:: 60^circ [3ex] E.:: 90^circ  2sin^2x^circ - 1 = 0 [3ex] 2sin^2x^circ = 1 [3ex] sin^2x^circ = dfrac<1> <2>[3ex] sin x = pm sqrt<2>> [5ex] Argument = x [3ex] sqrt<2>> = dfrac>> [5ex] = dfrac<1>> [5ex] = dfrac<1>> * dfrac>> [5ex] = dfrac> <2>[5ex] ightarrow sin x = pm dfrac> <2>[5ex] x = sin^<-1>left(-dfrac><2> ight) ::OR:: x = sin^<-1>left(dfrac><2> ight) [5ex] x = -45^circ ::OR:: x = 45^circ [3ex] Because:: 0^circ le x^circ le 90^circ [3ex] x = 45^circ [3ex] $Check $ underline [3ex] 2sin^2x^circ - 1 [3ex] x = 45^circ [3ex] 2sin^2(45) - 1 [3ex] 2 * (sin 45)^2 - 1 [3ex] sin 45 = dfrac> <2>[5ex] (sin 45)^2 = left(dfrac><2> ight)^2 [5ex] (sin 45)^2 = dfrac^2> <2^2>[5ex] (sin 45)^2 = dfrac<2> <4>= dfrac<1> <2>[5ex] = 2 * dfrac<1> <2>- 1 [5ex] = 1 - 1 [3ex] = 0  underline [3ex] 0 $(6.) ACT What are the values of$ heta$, between$ and $360^circ$, when $an heta = -1$?

$A.:: 225^circ ::and:: 315^circ ::only [3ex] B.:: 135^circ ::and:: 315^circ ::only [3ex] C.:: 135^circ ::and:: 225^circ ::only [3ex] D.:: 45^circ ::and:: 135^circ ::only [3ex] E.:: 45^circ, :135^circ, :225^circ, ::and:: 315^circ$

$an heta = -1 [3ex] Argument = heta [3ex] Assume:: an heta = 1 [3ex] heta = an^<-1>(1) = 45^circ = 45 * dfrac <180>= dfrac <4>[5ex] But:: an heta = -1 [3ex] an ::is:: negative:: in:: 2nd ::and:: 4th ::quadrants [3ex] - an 45 = an (180 - 45). 2nd:: Quadrant:: Identity [3ex] = an 135 [3ex] Also, [3ex] - an 45 = an (360 - 45). 4th:: Quadrant:: Identity [3ex] = an 315 [3ex] heta = 135^circ,: 315^circ [5ex]$ Check
 $underline [3ex] an heta [3ex] heta = 135^circ [3ex] an 135 = -1 [3ex]$ $heta = 315^circ [3ex] an 315 = -1$ $underline [3ex] -1$

(7.) Determine all the solutions of $sinalpha - cosalpha = 0$ in the interval $[0, 2pi)$

This is not straightforward. It has two functions - the sine function and the cosine function.
What do we do to express it in terms of only one function?

$sinalpha - cosalpha = 0 [3ex] Square:: both:: sides [3ex] (sinalpha - cosalpha)^2 = 0^2 [3ex] (sinalpha - cosalpha)(sinalpha - cosalpha) = 0 [3ex] sin^2alpha - sinalphacosalpha - sinalphacosalpha + cos^2alpha = 0 [3ex] sin^2alpha + cos^2alpha - 2sinalphacosalpha = 0 [3ex] sin^2alpha + cos^2alpha = 1 . Pythagorean:: Identity [3ex] 1 - 2sinalphacosalpha = 0 [3ex] 1 = 2sinalphacosalpha [3ex] 2sinalphacosalpha = 1 [3ex] 2sinalphacosalpha = sin2alpha . Double:: Angle:: Formula [3ex] sin2alpha = 1 [3ex] Argument = 2alpha [3ex] herefore interval = [0, 4pi) [3ex] 2alpha = sin^<-1>(1) [3ex] sin^<-1>(1) = 90^circ . Unit:: Circle:: Trig [5ex] Also,:: based:: on:: [0, 4pi) = [0, 4 * 180^circ) = [0, 720^circ) [3ex] sin^<-1>(1) = 90 + 360(1) . coterminal angle s [3ex] sin^<-1>(1) = 90 + 360 [3ex] sin^<-1>(1) = 450^circ [3ex] herefore 2alpha = 90, ::2alpha = 450 [5ex] alpha = dfrac<90><2>, ::alpha = dfrac<450> <2>[5ex] alpha = 45^circ, 225^circ [3ex] 45^circ = 45 * dfrac <180>= dfrac <4>[5ex] 225^circ = 225 * dfrac <180>= dfrac<5pi> <4>[5ex] alpha = dfrac<4>, dfrac<5pi> <4>[5ex]$ Check

 $underline [3ex] sinalpha - cosalpha [3ex] alpha = dfrac <4>[5ex] sinalpha = sindfrac <4>= dfrac> <2>[5ex] cosalpha = cosdfrac <4>= dfrac> <2>[5ex] = dfrac> <2>- dfrac> <2>[5ex] = 0 [5ex]$ $alpha = dfrac<5pi> <4>[5ex] sinalpha = sindfrac<5pi> <4>= -dfrac> <2>[5ex] cosalpha = cosdfrac<5pi> <4>= -dfrac> <2>[5ex] = -dfrac> <2>- -dfrac> <2>[5ex] = -dfrac> <2>+ dfrac> <2>[5ex] = 0$ $underline [3ex] 0$

$underline [3ex] alpha = dfrac <4>+ 2pi k . coterminal:: angle s [5ex] alpha = dfrac<5pi> <4>+ 2pi k . coterminal:: angle s$

(8.) Determine all the solutions of $cosalpha + sqrt<3>sinalpha = 1$ in the interval $[0, 2pi]$

This is not straightforward. It has two functions - the cosine function and the sine function.
What do we do to express it in terms of only one function?

$cosalpha + sqrt<3>sinalpha = 1 [3ex] sqrt<3>sinalpha = 1 - cosalpha [3ex] Square:: both:: sides [3ex] (sqrt<3>sinalpha)^2 = (1 - cosalpha)^2 [3ex] (sqrt<3>)^2 * sin^2alpha = (1 - cosalpha)(1 - cosalpha) [3ex] 3sin^2alpha = 1 - cosalpha - cosalpha + cos^2alpha [3ex] 3sin^2alpha = 1 - 2cosalpha + cos^2alpha [3ex] sin^2alpha + cos^2alpha = 1 . Pythagorean:: Identity [3ex] sin^2alpha = 1 - cos^2alpha [3ex] 3(1 - cos^2alpha) = 1 - 2cosalpha + cos^2alpha [3ex] 3 - 3cos^2alpha = 1 - 2cosalpha + cos^2alpha [3ex] 0 = 1 - 2cosalpha + cos^2alpha - 3 + 3cos^2alpha [3ex] 1 - 2cosalpha + cos^2alpha - 3 + 3cos^2alpha = 0 [3ex] 4cos^2alpha - 2cosalpha - 2 = 0 [3ex] Divide:: all:: terms:: by:: 2 [3ex] 2cos^2alpha - cosalpha - 1 = 0 [3ex] Let:: cosalpha = p [3ex] 2p^2 - p - 1 = 0 [3ex] 2p^2 + p - 2p - 1 = 0 [3ex] p(2p + 1) - 1(2p + 1) = 0 [3ex] (2p + 1)(p - 1) = 0 [3ex] 2p + 1 = 0 :::OR::: p - 1 = 0 [3ex] 2p = -1 :::OR::: p = 1 [3ex] p = -dfrac<1> <2>:::OR::: p = 1 [3ex] Substitute:: back [3ex] ightarrow cosalpha = -dfrac<1> <2>[3ex] alpha = cos^<-1>left(-dfrac<1><2> ight) [5ex] alpha = 120^circ, 240^circ . Unit:: Circle:: Trig [3ex] ightarrow cosalpha = 1 [3ex] alpha = cos^<-1>(1) [3ex] alpha = 0^circ, 360^circ . Unit:: Circle:: Trig [3ex]$ Check

 $underline [3ex] cosalpha + sqrt<3>sinalpha [3ex] alpha = 120 [5ex] cosalpha = cos 120 = -dfrac<1> <2>[5ex] sinalpha = sin 120 = dfrac> <2>[5ex] sqrt<3>sinalpha = sqrt <3>* dfrac> <2>= dfrac<3> <2>[5ex] = -dfrac<1> <2>+ dfrac<3> <2>[5ex] = -dfrac<1> <2>+ dfrac<3> <2>[5ex] = dfrac<-1 + 3> <2>[5ex] = -dfrac<2> <2>[5ex] = 1 [3ex] YES [3ex]$ $alpha = 240 [5ex] cosalpha = cos 240 = -dfrac<1> <2>[5ex] sinalpha = sin 240 = -dfrac> <2>[5ex] sqrt<3>sinalpha = sqrt <3>* -dfrac> <2>= -dfrac<3> <2>[5ex] = -dfrac<1> <2>+ - dfrac<3> <2>[5ex] = -dfrac<1> <2>- dfrac<3> <2>[5ex] = dfrac<-1 - 3> <2>[5ex] = -dfrac<4> <2>[5ex] = -2 [3ex] Extraneous: root [3ex] NO [3ex]$ $alpha = 0 [3ex] cosalpha = cos 0 = 1 [3ex] sinalpha = sin 0 = 0 [3ex] sqrt<3>sinalpha = sqrt <3>* 0 = 0[3ex] = 1 + 0 [5ex] = 1 [3ex] YES [3ex]$ $alpha = 360 [3ex] cosalpha = cos 360 = 1 [3ex] sinalpha = sin 360 = 0 [3ex] sqrt<3>sinalpha = sqrt <3>* 0 = 0[3ex] = 1 + 0 [5ex] = 1 [3ex] YES$ $underline [3ex] 1$

$underline [3ex] alpha = 120^circ = 120 * dfrac <180>= dfrac<2pi> <3>[5ex] alpha = 0^circ = 120 * dfrac <180>= 0 [5ex] alpha = 360^circ = 360 * 120 * dfrac <180>= 2pi [5ex] underline [3ex] alpha = 0 + 2pi k = 2pi k . coterminal:: angle s [3ex] alpha = dfrac<2pi> <3>+ 2pi k . coterminal:: angle s$

(9.) Determine all the solutions of $6cos^2 heta = 3$ in the interval $[0, 2pi)$

$6cos^2 heta = 3 [3ex] cos^2 heta = dfrac<3> <6>= dfrac<1> <2>[5ex] cos heta = pm sqrt<2>> [5ex] Argument = heta [3ex] sqrt<2>> = dfrac>> [5ex] = dfrac<1>> [5ex] = dfrac<1>> * dfrac>> [5ex] = dfrac> <2>[5ex] ightarrow cos heta = pm dfrac> <2>[5ex] heta = cos^<-1>left(dfrac><2> ight) [5ex] heta = 45^circ, 315^circ . Unit:: Circle:: Trig [3ex] 45 = 45 * dfrac <180>= dfrac <4>[5ex] 315 = 315 * dfrac <180>= dfrac<7pi> <4>[5ex] heta = cos^<-1>left(-dfrac><2> ight) [5ex] heta = 135^circ, 225^circ . Unit:: Circle:: Trig [3ex] 135 = 135 * dfrac <180>= dfrac<3pi> <4>[5ex] 225 = 225 * dfrac <180>= dfrac<5pi> <4>[5ex] heta = dfrac<4>, dfrac<3pi><4>, dfrac<5pi><4>, dfrac<7pi> <4>[5ex] underline [3ex] eta = dfrac <4>+ 2pi k . coterminal:: angle s [5ex] eta = dfrac<3pi> <4>+ 2pi k . coterminal:: angle s [5ex] eta = dfrac<5pi> <4>+ 2pi k . coterminal:: angle s [5ex] eta = dfrac<7pi> <4>+ 2pi k . coterminal:: angle s [5ex]$ Check
 $underline [3ex] 6cos^2 heta [3ex] heta = dfrac <4>[5ex] cos heta = cosdfrac <4>= dfrac> <2>[5ex] cos^2 heta = left(dfrac><2> ight)^2 = dfrac<(sqrt<2>)^2> <2^2>= dfrac<2> <4>= dfrac<1> <2>[5ex] = 6 * dfrac<1> <2>[5ex] = 3 [3ex]$ $heta = dfrac<3pi> <4>[5ex] cos heta = cosdfrac<3pi> <4>= -dfrac> <2>[5ex] cos^2 heta = left(-dfrac><2> ight)^2 = dfrac<(-sqrt<2>)^2> <2^2>= dfrac<2> <4>= dfrac<1> <2>[5ex] = 6 * dfrac<1> <2>[5ex] = 3 [3ex]$ $heta = dfrac<5pi> <4>[5ex] cos heta = cosdfrac<5pi> <4>= dfrac> <2>[5ex] cos^2 heta = left(dfrac><2> ight)^2 = dfrac<(sqrt<2>)^2> <2^2>= dfrac<2> <4>= dfrac<1> <2>[5ex] = 6 * dfrac<1> <2>[5ex] = 3 [3ex]$ $heta = dfrac<7pi> <4>[5ex] cos heta = cosdfrac<7pi> <4>= -dfrac> <2>[5ex] cos^2 heta = left(-dfrac><2> ight)^2 = dfrac<(-sqrt<2>)^2> <2^2>= dfrac<2> <4>= dfrac<1> <2>[5ex] = 6 * dfrac<1> <2>[5ex] = 3$ $underline [3ex] 3$

(10.) Determine all the solutions of $cosalpha - sqrt<3>sinalpha = 1$ in the interval $[0, 2pi]$

This is not straightforward. It has two functions - the cosine function and the sine function.
What do we do to express it in terms of only one function?

$cosalpha - sqrt<3>sinalpha = 1 [3ex] -sqrt<3>sinalpha = 1 - cosalpha [3ex] Square:: both:: sides [3ex] (-1 * sqrt<3>sinalpha)^2 = (1 - cosalpha)^2 [3ex] (-1)^2 * (sqrt<3>)^2 * sin^2alpha = (1 - cosalpha)(1 - cosalpha) [3ex] 1 * 3 * sin^2alpha = 1 - cosalpha - cosalpha + cos^2alpha [3ex] 3sin^2alpha = 1 - 2cosalpha + cos^2alpha [3ex] sin^2alpha + cos^2alpha = 1 . Pythagorean:: Identity [3ex] sin^2alpha = 1 - cos^2alpha [3ex] 3(1 - cos^2alpha) = 1 - 2cosalpha + cos^2alpha [3ex] 3 - 3cos^2alpha = 1 - 2cosalpha + cos^2alpha [3ex] 0 = 1 - 2cosalpha + cos^2alpha - 3 + 3cos^2alpha [3ex] 1 - 2cosalpha + cos^2alpha - 3 + 3cos^2alpha = 0 [3ex] 4cos^2alpha - 2cosalpha - 2 = 0 [3ex] Divide:: all:: terms:: by:: 2 [3ex] 2cos^2alpha - cosalpha - 1 = 0 [3ex] Let:: cosalpha = p [3ex] 2p^2 - p - 1 = 0 [3ex] 2p^2 + p - 2p - 1 = 0 [3ex] p(2p + 1) - 1(2p + 1) = 0 [3ex] (2p + 1)(p - 1) = 0 [3ex] 2p + 1 = 0 :::OR::: p - 1 = 0 [3ex] 2p = -1 :::OR::: p = 1 [3ex] p = -dfrac<1> <2>:::OR::: p = 1 [3ex] Substitute:: back [3ex] ightarrow cosalpha = -dfrac<1> <2>[3ex] alpha = cos^<-1>left(-dfrac<1><2> ight) [5ex] alpha = 120^circ, 240^circ . Unit:: Circle:: Trig [3ex] ightarrow cosalpha = 1 [3ex] alpha = cos^<-1>(1) [3ex] alpha = 0^circ, 360^circ . Unit:: Circle:: Trig [3ex]$ Check

 $underline [3ex] cosalpha - sqrt<3>sinalpha [3ex] alpha = 120 [5ex] cosalpha = cos 120 = -dfrac<1> <2>[5ex] sinalpha = sin 120 = dfrac> <2>[5ex] sqrt<3>sinalpha = sqrt <3>* dfrac> <2>= dfrac<3> <2>[5ex] = -dfrac<1> <2>- dfrac<3> <2>[5ex] = -dfrac<1> <2>- dfrac<3> <2>[5ex] = dfrac<-1 - 3> <2>[5ex] = -dfrac<4> <2>[5ex] = -2 [3ex] Extraneous: root [3ex] NO [3ex]$ $alpha = 240 [5ex] cosalpha = cos 240 = -dfrac<1> <2>[5ex] sinalpha = sin 240 = -dfrac> <2>[5ex] sqrt<3>sinalpha = sqrt <3>* -dfrac> <2>= -dfrac<3> <2>[5ex] = -dfrac<1> <2>- - dfrac<3> <2>[5ex] = -dfrac<1> <2>+ dfrac<3> <2>[5ex] = dfrac<-1 + 3> <2>[5ex] = -dfrac<2> <2>[5ex] = 1 [3ex] YES [3ex]$ $alpha = 0 [3ex] cosalpha = cos 0 = 1 [3ex] sinalpha = sin 0 = 0 [3ex] sqrt<3>sinalpha = sqrt <3>* 0 = 0[3ex] = 1 - 0 [5ex] = 1 [3ex] YES [3ex]$ $alpha = 360 [3ex] cosalpha = cos 360 = 1 [3ex] sinalpha = sin 360 = 0 [3ex] sqrt<3>sinalpha = sqrt <3>* 0 = 0[3ex] = 1 - 0 [5ex] = 1 [3ex] YES$ $underline [3ex] 1$

$underline [3ex] alpha = 240^circ = 240 * dfrac <180>= dfrac<4pi> <3>[5ex] alpha = 0^circ = 120 * dfrac <180>= 0 [5ex] alpha = 360^circ = 360 * 120 * dfrac <180>= 2pi [5ex] underline [3ex] alpha = 0 + 2pi k = 2pi k . coterminal:: angle s [3ex] alpha = dfrac<4pi> <3>+ 2pi k . coterminal:: angle s$

(11.) Determine all the solutions of $2cos^2 heta + 3cos heta = -1$ in the interval $[0, 2pi]$

$2cos^2 heta + 3cos heta = -1 [3ex] Let:: cos heta = p [3ex] 2p^2 + 3p = -1 [3ex] 2p^2 + 3p + 1 = 0 [3ex] 2p^2 + 2p + p + 1 = 0 [3ex] 2p(p + 1) + 1(p + 1) = 0 [3ex] p + 1 = 0 :::OR::: 2p + 1 = 0 [3ex] p = -1 :::OR::: 2p = -1 [3ex] p = -1 :::OR::: p = -dfrac<1> <2>[5ex] Substitute:: back [3ex] ightarrow cos heta = -1 heta = cos^<-1>(-1) [3ex] heta = 180^circ . Unit:: Circle:: Trig [3ex] ightarrow cos heta = -dfrac<1> <2>[5ex] heta = cos^<-1>left(-dfrac<1><2> ight) [5ex] heta = 120^circ, 240^circ . Unit:: Circle:: Trig [3ex]$ Check

 $underline [3ex] 2cos^2 heta + 3cos heta [3ex] heta = 180 [3ex] cos heta = cos 180 = -1 [3ex] cos^2 heta = (cos 180)^2 = (-1)^2 = 1 [3ex] = 2(1) + 3(-1) [3ex] = 2 - 3 [3ex] = -1 [3ex] YES [3ex]$ $heta = 120 [3ex] cos heta = cos 120 = -dfrac<1> <2>[3ex] cos^2 heta = (cos 120)^2 = left(-dfrac<1><2> ight)^2 = dfrac<1> <4>[3ex] = 2left(dfrac<1><4> ight) + 3left(-dfrac<1><2> ight) [3ex] = dfrac<1> <2>- dfrac<3> <2>[3ex] = dfrac<1 - 3> <2>[5ex] = -dfrac<2> <2>[5ex] = -1 [3ex] YES [3ex]$ $heta = 240 [3ex] cos heta = cos 240 = -dfrac<1> <2>[3ex] cos^2 heta = (cos 240)^2 = left(-dfrac<1><2> ight)^2 = dfrac<1> <4>[3ex] = 2left(dfrac<1><4> ight) + 3left(-dfrac<1><2> ight) [3ex] = dfrac<1> <2>- dfrac<3> <2>[3ex] = dfrac<1 - 3> <2>[5ex] = -dfrac<2> <2>[5ex] = -1 [3ex] YES$ $underline [3ex] -1$

$underline [3ex] alpha = 180^circ = pi [5ex] alpha = 120^circ = 120 * dfrac <180>= dfrac<2pi> <3>[5ex] alpha = 240^circ = 240 * dfrac <180>= dfrac<4pi> <3>[5ex] underline [3ex] alpha = pi + 2pi k . coterminal:: angle s [3ex] alpha = dfrac<2pi> <3>+ 2pi k . coterminal:: angle s [5ex] alpha = dfrac<4pi> <3>+ 2pi k . coterminal:: angle s$

(12.) Determine all the solutions of $sqrt<2>sin heta = 2sin^2 heta$ in the interval $[0, 2pi)$

$sqrt<2>sin heta = 2sin^2 heta [3ex] 2sin^2 heta = sqrt<2>sin heta [3ex] 2sin^2 heta - sqrt<2>sin heta = 0 [3ex] Let:: sin heta = p [3ex] 2p^2 - psqrt <2>= 0 [3ex] p(2p - sqrt<2>) = 0 [3ex] p = 0 :::OR::: 2p - sqrt <2>= 0 [3ex] p = 0 :::OR::: 2p = sqrt <2>[3ex] p = 0 :::OR::: p = dfrac> <2>[5ex] Substitute:: back [3ex] ightarrow sin heta = 0 heta = sin^<-1>(0) [3ex] heta = 0^circ, 180^circ . Unit:: Circle:: Trig [3ex] ightarrow sin heta = dfrac> <2>[5ex] heta = sin^<-1>left(dfrac><2> ight) [5ex] heta = 45^circ, 135^circ . Unit:: Circle:: Trig [3ex]$ Check

 $underline [3ex] sqrt<2>sin heta [3ex] heta = 0 [5ex] sin heta = sin 0 = 0 [3ex] = sqrt <2>* 0 [5ex] = 0 [3ex] YES [3ex]$ $heta = 180 [5ex] sin heta = sin 180 = 0 [3ex] = sqrt <2>* 0 [5ex] = 0 [3ex] YES [3ex]$ $heta = 45 [5ex] sin heta = sin 45 = dfrac> <2>[5ex] = sqrt <2>* dfrac> <2>[5ex] = dfrac<2> <2>[5ex] = 1 [3ex] YES [3ex]$ $heta = 135 [5ex] sin heta = sin 135 = dfrac> <2>[5ex] = sqrt <2>* dfrac> <2>[5ex] = dfrac<2> <2>[5ex] = 1 [3ex] YES$ $underline [3ex] 2sin^2 heta [3ex] heta = 0 [5ex] sin heta = sin 0 = 0 [3ex] sin^2 heta = (sin heta)^2 = 0^2 = 0 [3ex] = 2 * 0 [3ex] = 0 [3ex] YES [3ex]$ $heta = 180 [5ex] sin heta = sin 180 = 0 [3ex] sin^2 heta = (sin heta)^2 = 0^2 = 0 [3ex] = 2 * 0 [3ex] = 0 [3ex] YES [3ex]$ $heta = 45 [5ex] sin heta = sin 45 = dfrac> <2>[3ex] sin^2 heta = (sin heta)^2 [3ex] sin^2 heta = left(dfrac> <2> ight) [5ex] sin^2 heta = dfrac<(sqrt<2>)^2> <2^2>[5ex] sin^2 heta = dfrac<2> <4>= dfrac<1> <2>[5ex] = 2 * dfrac<1> <2>[3ex] = 1 [3ex] YES [3ex]$ $heta = 135 [5ex] sin heta = sin 135 = dfrac> <2>[3ex] sin^2 heta = (sin heta)^2 [3ex] sin^2 heta = left(dfrac> <2> ight) [5ex] sin^2 heta = dfrac<(sqrt<2>)^2> <2^2>[5ex] sin^2 heta = dfrac<2> <4>= dfrac<1> <2>[5ex] = 2 * dfrac<1> <2>[3ex] = 1 [3ex] YES$

$underline [3ex] alpha = 0^circ = 0 * dfrac <180>= 0 [5ex] alpha = 45^circ = 45 * dfrac <180>= dfrac <4>[5ex] alpha = 135^circ = 135 * dfrac <180>= dfrac<3pi> <4>[5ex] alpha = 180^circ = pi [5ex] underline [3ex] alpha = 0 + 2pi k = 2pi k . coterminal:: angle s [3ex] alpha = dfrac <4>+ 2pi k . coterminal:: angle s [5ex] alpha = dfrac<3pi> <4>+ 2pi k . coterminal:: angle s [5ex] alpha = pi + 2pi k . coterminal:: angle s$

(13.) Determine all the solutions of $6coseta - 6sineta = 3sqrt<6>$ in the interval $[0, 2pi)$

This is not straightforward. It has two functions - the cosine function and the sine function.
What do we do to express it in terms of only one function?

$6coseta - 6sineta = 3sqrt <6>[3ex] Divide:: each:: term:: by:: 3 . Simplify [3ex] 2coseta - 2sineta = sqrt <6>[3ex] 2(coseta - sineta) = sqrt <6>[3ex] Square:: both:: sides [3ex] 2^2 * (coseta - sineta)^2 = (sqrt<6>)^2 [3ex] 4[(coseta - sineta)(coseta - sineta)] = 6 [3ex] 4(cos^2eta - sinetacoseta - sinetacoseta + sin^2eta) = 6 [3ex] 4(cos^2eta + sin^2eta - 2sinetacoseta) = 6 [3ex] cos^2eta + sin^2eta = 1 . Pythagorean:: Identity [3ex] 2sinetacoseta = sin2eta . Double:: Angle:: Formula [3ex] 4(1 - 2sin 2eta) = 6 [3ex] 4 - 4sin 2eta = 6 [3ex] 4 - 6 = 4sin 2eta [3ex] -2 = 4sin 2eta [3ex] 4sin 2eta = -2 [3ex] sin2eta = -dfrac<2> <4>[5ex] sin2eta = -dfrac<1> <2>[5ex] Argument = 2eta [3ex] herefore interval = [0, 4pi) [3ex] 2eta = sin^<-1>left(-dfrac<1><2> ight) [5ex] sin^<-1>left(-dfrac<1><2> ight) = 210^circ, 330^circ . Unit:: Circle:: Trig [3ex] Based:: on:: [0, 4pi) = [0, 4 * 180^circ) = [0, 720^circ) [3ex] sin^<-1>(1) = 210 + 360(1) . coterminal angle s [3ex] sin^<-1>(1) = 210 + 360 [3ex] sin^<-1>(1) = 570^circ [3ex] Also,:: sin^<-1>(1) = 330 + 360(1) . coterminal angle s [3ex] sin^<-1>(1) = 330 + 360 [3ex] sin^<-1>(1) = 690^circ [3ex] herefore 2eta = 210, ::2eta = 330, ::2eta = 570, ::2eta = 690 [3ex] eta = dfrac<210><2>, ::eta = dfrac<330><2>, ::eta = dfrac<570><2>, ::eta = dfrac<690> <2>[5ex] eta = 105^circ, 165^circ, 285^circ, 345^circ [3ex] 105^circ = 105 * dfrac <180>= dfrac<7pi> <12>[5ex] 165^circ = 165 * dfrac <180>= dfrac<11pi> <12>[5ex] 285^circ = 285 * dfrac <180>= dfrac<19pi> <12>[5ex] 345^circ = 345 * dfrac <180>= dfrac<23pi> <12>[5ex] eta = dfrac<7pi><12>, dfrac<11pi><12>, dfrac<19pi><12>, dfrac<23pi> <12>[5ex]$ Check

 $underline [3ex] 6coseta - 6sineta = 6(coseta - sineta) [3ex] eta = 105^circ [3ex] coseta = cos105 = dfrac - sqrt<6>> <4>. Special:: Angles:: Kids [5ex] sineta = sin105 = dfrac + sqrt<6>> <4>. Special:: Angles:: Kids [5ex] coseta - sineta = dfrac - sqrt<6>> <4>- left(dfrac + sqrt<6>><4> ight) [5ex] coseta - sineta = dfrac - sqrt <6>- (sqrt <2>+ sqrt<6>)> <4>[5ex] coseta - sineta = dfrac - sqrt <6>- sqrt <2>- sqrt<6>> <4>[5ex] coseta - sineta = dfrac<-2sqrt<6>> <4>= dfrac<-sqrt<6>> <2>[5ex] = 6left(-dfrac><2> ight) [5ex] = 3 * -sqrt <6>[3ex] = -3sqrt <6>[3ex] Extraneous:: root [3ex] NO [3ex]$ $eta = 165^circ [3ex] coseta = cos165 = dfrac <-sqrt<6>- sqrt<2>> <4>. Special:: Angles:: Kids [5ex] sineta = sin165 = dfrac - sqrt<2>> <4>. Special:: Angles:: Kids [5ex] coseta - sineta = dfrac <-sqrt<6>- sqrt<2>> <4>- left(dfrac - sqrt<2>><4> ight) [5ex] coseta - sineta = dfrac <-sqrt<6>- sqrt <2>- (sqrt <6>- sqrt<2>)> <4>[5ex] coseta - sineta = dfrac <-sqrt<6>- sqrt <2>- sqrt <6>+ sqrt<2>> <4>[5ex] coseta - sineta = dfrac<-2sqrt<6>> <4>= dfrac<-sqrt<6>> <2>[5ex] = 6left(-dfrac><2> ight) [5ex] = 3 * -sqrt <6>[3ex] = -3sqrt <6>[3ex] Extraneous:: root [3ex] NO [3ex]$ $eta = 285^circ [3ex] coseta = cos285 = dfrac - sqrt<2>> <4>. Special:: Angles:: Kids [5ex] sineta = sin285 = dfrac <-sqrt<2>- sqrt<6>> <4>. Special:: Angles:: Kids [5ex] coseta - sineta = dfrac - sqrt<2>> <4>- left(dfrac <-sqrt<2>- sqrt<6>><4> ight) [5ex] coseta - sineta = dfrac - sqrt <2>- (-sqrt <2>- sqrt<6>)> <4>[5ex] coseta - sineta = dfrac - sqrt <2>+ sqrt <2>+ sqrt<6>> <4>[5ex] coseta - sineta = dfrac<2sqrt<6>> <4>= dfrac> <2>[5ex] = 6left(dfrac><2> ight) [5ex] = 3 * sqrt <6>[3ex] = 3sqrt <6>[3ex] YES [3ex]$ $eta = 345^circ [3ex] coseta = cos345 = dfrac + sqrt<6>> <4>. Special:: Angles:: Kids [5ex] sineta = sin345 = dfrac - sqrt<6>> <4>. Special:: Angles:: Kids [5ex] coseta - sineta = dfrac + sqrt<6>> <4>- left(dfrac - sqrt<6>><4> ight) [5ex] coseta - sineta = dfrac + sqrt <6>- (sqrt <2>- sqrt<6>)> <4>[5ex] coseta - sineta = dfrac + sqrt <6>- sqrt <2>+ sqrt<6>> <4>[5ex] coseta - sineta = dfrac<2sqrt<6>> <4>= dfrac> <2>[5ex] = 6left(dfrac><2> ight) [5ex] = 3 * sqrt <6>[3ex] = 3sqrt <6>[3ex] YES$ $underline [3ex] 3sqrt <6>$

$underline [3ex] alpha = 285^circ = dfrac<19pi> <12>[5ex] alpha = 345^circ = dfrac<23pi> <12>[5ex] underline [3ex] alpha = dfrac<19pi> <12>+ 2pi k . coterminal:: angle s [5ex] alpha = dfrac<23pi> <12>+ 2pi k . coterminal:: angle s$

(14.) Determine all the solutions of $cosalpha + sqrt<3>sinalpha = 1$ in the interval $[0, 2pi]$

This is not straightforward. It has two functions - the cosine function and the sine function.
What do we do to express it in terms of only one function?

$cosalpha + sqrt<3>sinalpha = 1 [3ex] sqrt<3>sinalpha = 1 - cosalpha [3ex] Square:: both:: sides [3ex] (sqrt<3>sinalpha)^2 = (1 - cosalpha)^2 [3ex] (sqrt<3>)^2 * sin^2alpha = (1 - cosalpha)(1 - cosalpha) [3ex] 3sin^2alpha = 1 - cosalpha - cosalpha + cos^2alpha [3ex] 3sin^2alpha = 1 - 2cosalpha + cos^2alpha [3ex] sin^2alpha + cos^2alpha = 1 . Pythagorean:: Identity [3ex] sin^2alpha = 1 - cos^2alpha [3ex] 3(1 - cos^2alpha) = 1 - 2cosalpha + cos^2alpha [3ex] 3 - 3cos^2alpha = 1 - 2cosalpha + cos^2alpha [3ex] 0 = 1 - 2cosalpha + cos^2alpha - 3 + 3cos^2alpha [3ex] 1 - 2cosalpha + cos^2alpha - 3 + 3cos^2alpha = 0 [3ex] 4cos^2alpha - 2cosalpha - 2 = 0 [3ex] Divide:: all:: terms:: by:: 2 [3ex] 2cos^2alpha - cosalpha - 1 = 0 [3ex] Let:: cosalpha = p [3ex] 2p^2 - p - 1 = 0 [3ex] 2p^2 + p - 2p - 1 = 0 [3ex] p(2p + 1) - 1(2p + 1) = 0 [3ex] (2p + 1)(p - 1) = 0 [3ex] 2p + 1 = 0 :::OR::: p - 1 = 0 [3ex] 2p = -1 :::OR::: p = 1 [3ex] p = -dfrac<1> <2>:::OR::: p = 1 [3ex] Substitute:: back [3ex] ightarrow cosalpha = -dfrac<1> <2>[3ex] alpha = cos^<-1>left(-dfrac<1><2> ight) [5ex] alpha = 120^circ, 240^circ . Unit:: Circle:: Trig [3ex] ightarrow cosalpha = 1 [3ex] alpha = cos^<-1>(1) [3ex] alpha = 0^circ, 360^circ . Unit:: Circle:: Trig [3ex]$ Check

 $underline [3ex] cosalpha + sqrt<3>sinalpha [3ex] alpha = 120 [5ex] cosalpha = cos 120 = -dfrac<1> <2>[5ex] sinalpha = sin 120 = dfrac> <2>[5ex] sqrt<3>sinalpha = sqrt <3>* dfrac> <2>= dfrac<3> <2>[5ex] = -dfrac<1> <2>+ dfrac<3> <2>[5ex] = -dfrac<1> <2>+ dfrac<3> <2>[5ex] = dfrac<-1 + 3> <2>[5ex] = -dfrac<2> <2>[5ex] = 1 [3ex] YES [3ex]$ $alpha = 240 [5ex] cosalpha = cos 240 = -dfrac<1> <2>[5ex] sinalpha = sin 240 = -dfrac> <2>[5ex] sqrt<3>sinalpha = sqrt <3>* -dfrac> <2>= -dfrac<3> <2>[5ex] = -dfrac<1> <2>+ - dfrac<3> <2>[5ex] = -dfrac<1> <2>- dfrac<3> <2>[5ex] = dfrac<-1 - 3> <2>[5ex] = -dfrac<4> <2>[5ex] = -2 [3ex] Extraneous:: root [3ex] NO [3ex]$ $alpha = 0 [3ex] cosalpha = cos 0 = 1 [3ex] sinalpha = sin 0 = 0 [3ex] sqrt<3>sinalpha = sqrt <3>* 0 = 0[3ex] = 1 + 0 [5ex] = 1 [3ex] YES [3ex]$ $alpha = 360 [3ex] cosalpha = cos 360 = 1 [3ex] sinalpha = sin 360 = 0 [3ex] sqrt<3>sinalpha = sqrt <3>* 0 = 0[3ex] = 1 + 0 [5ex] = 1 [3ex] YES$ $underline [3ex] 1$

$underline [3ex] alpha = 120^circ = 120 * dfrac <180>= dfrac<2pi> <3>[5ex] alpha = 0^circ = 120 * dfrac <180>= 0 [5ex] alpha = 360^circ = 360 * 120 * dfrac <180>= 2pi [5ex] underline [3ex] alpha = 0 + 2pi k = 2pi k . coterminal:: angle s [3ex] alpha = dfrac<2pi> <3>+ 2pi k . coterminal:: angle s$

(15.) Determine all the solutions of $cos(pi + heta) + sinleft( heta - dfrac<2> ight) = 1$ in the interval $[0, 2pi)$

This is not straightforward. It has two functions - the cosine function and the sine function.
How do we express it in terms of only one function?
Let us apply the Sum and Difference Formulas first

$cos(pi + heta) + sinleft( heta - dfrac<2> ight) = 1 [5ex] cos(pi + heta) = cospicos heta - sinpisin heta . Addition:: Formula [3ex] cospi = -1 ::and:: sinpi = 0 . Unit:: Circle:: Trig [3ex] ightarrow cos(pi + heta) = -1 * cos heta - 0 * sin heta = -cos heta - 0 = -cos heta [3ex] sinleft( heta - dfrac<2> ight) = sin hetacosdfrac <2>- cos hetasindfrac <2>. Difference:: Formula [5ex] cosdfrac <2>= 0 ::and:: sindfrac <2>= 1 . Unit:: Circle:: Trig [5ex] ightarrow sinleft( heta - dfrac<2> ight) = sin heta * 0 - cos heta * 1 = 0 - cos heta = -cos heta [5ex] cos(pi + heta) + sinleft( heta - dfrac<2> ight) = 1 [5ex] ightarrow -cos heta + -cos heta = 1 [3ex] -cos heta - cos heta = 1 [3ex] -2cos heta = 1 [5ex] cos heta = -dfrac<1> <2>[5ex] heta = cos^<-1><-dfrac<1><2>> [5ex] heta = 120^circ, 240^circ [3ex]$ Check

 $underline [3ex] cos(pi + heta) + sinleft( heta - dfrac<2> ight) [5ex] pi = 180^circ [3ex] dfrac <2>= dfrac<180> <2>= 90^circ [5ex] heta = 120^circ [3ex] = cos(180 + 120) + sin(120 - 90) [3ex] = cos 300 + sin 30 [3ex] = dfrac<1> <2>+ dfrac<1> <2>. Unit:: Circle:: Trig [5ex] = 1 [3ex] YES [3ex]$ $heta = 240^circ [3ex] = cos(180 + 240) + sin(240 - 90) [3ex] = cos 420 + sin 150 [3ex] coterminal:: angle ::of:: 420 = 420 - 360 = 60 [3ex] = cos 60 + sin 150 [3ex] = dfrac<1> <2>+ dfrac<1> <2>. Unit:: Circle:: Trig [5ex] = 1 [3ex] YES$ $underline [3ex] 1$

$underline [3ex] eta = 120^circ = 120 * dfrac <180>= dfrac<2pi> <3>[5ex] eta = 240^circ = 240 * dfrac <180>= dfrac<4pi> <3>[5ex] underline [3ex] eta = dfrac<2pi> <3>+ 2pi k . coterminal:: angle s [5ex] eta = dfrac<4pi> <3>+ 2pi k . coterminal:: angle s$

(16.) Determine all the solutions of $coteta + sqrt <3>= csceta$ in the interval le eta lt 2pi$Interval of le eta lt 2pi$ means $[0, 2pi)$
This is not straightforward. It has two functions - the cotangent function and the cosecant function.
How do we express it in terms of only one function?

$coteta + sqrt <3>= csceta [3ex] Square:: both:: sides [3ex] (coteta + sqrt<3>)^2 = (csceta)^2 [3ex] (coteta + sqrt<3>)(coteta + sqrt<3>) = csc^2eta [3ex] cot^2eta + sqrt<3>coteta + sqrt<3>coteta + (sqrt<3>)^2 = csc^2eta [3ex] cot^2eta + 2sqrt<3>coteta + 3 = csc^2eta [3ex] csc^2eta = 1 + cot^2eta . Pythagorean:: Identity [3ex] cot^2eta + 2sqrt<3>coteta + 3 = 1 + cot^2eta [3ex] cot^2eta - cot^2eta + 2srt<3>coteta + 3 - 1 = 0 [3ex] 2sqrt<3>coteta + 2 = 0 [3ex] 2sqrt<3>coteta = -2 [3ex] coteta = -dfrac<2><2sqrt<3>> [5ex] coteta = -dfrac<1>> [5ex] coteta = -dfrac<1>> * dfrac>> [5ex] coteta = -dfrac> <3>[5ex] eta = cos^<-1>left(<-dfrac><3>> ight) [5ex] eta = 120^circ, 300^circ . Unit:: Circle:: Trig [3ex]$ Check

 $underline [3ex] coteta + sqrt <3>[3ex] eta = 120^circ [5ex] coteta = cot 120 = -dfrac> <3>[5ex] = -dfrac> <3>+ sqrt <3>[5ex] = -dfrac> <3>+ dfrac<3sqrt<3>> <3>[5ex] = dfrac <-sqrt<3>+ 3sqrt<3>> <3>[5ex] = dfrac<2sqrt<3>> <3>[3ex] YES [3ex]$ $eta = 300^circ [5ex] coteta = cot 300 = -dfrac> <3>[5ex] = -dfrac> <3>+ sqrt <3>[5ex] = -dfrac> <3>+ dfrac<3sqrt<3>> <3>[5ex] = dfrac <-sqrt<3>+ 3sqrt<3>> <3>[5ex] = dfrac<2sqrt<3>> <3>[3ex] LHS e RHS [3ex] NO$ $underline [3ex] csceta [3ex] eta = 120^circ [3ex] = csc 120 [3ex] = dfrac<2sqrt<3>> <3>[3ex] YES [3ex]$ $eta = 300^circ [3ex] = csc 300 [3ex] = -dfrac<2sqrt<3>> <3>[5ex] RHS e LHS [3ex] NO$

$underline [3ex] eta = 120^circ = 120 * dfrac <180>= dfrac<2pi> <3>[5ex] underline [3ex] eta = dfrac<2pi> <3>+ 2pi k . coterminal:: angle s$

(17.) Determine all the solutions of $cos(2alpha) + 14sin^2alpha = 10$ in the interval $[0, 2pi)$

This is not straightforward. It has two functions - the cosine function and the sine function.
How do we express it in terms of only one function?
Let us apply the Sum and Difference Formulas first

$cos(2alpha) + 14sin^2alpha = 10 [3ex] cos(2alpha) = 1 - 2sin^2alpha . Double-Angle:: Formula [3ex] ightarrow 1 - 2sin^2alpha + 14sin^2alpha = 10 [3ex] 1 + 12sin^2alpha = 10 [3ex] 12sin^2alpha = 10 - 1 [3ex] 12sin^2alpha = 9 [3ex] sin^2alpha = dfrac<9> <12>= dfrac<3> <4>[5ex] sinalpha = pm sqrt<4>> = pm dfrac> <2>[5ex] alpha = sin^<-1>left(dfrac><2> ight) ::OR:: alpha = sin^<-1>left(-dfrac><2> ight) [5ex] alpha = 60, 120 ::OR:: alpha = 240, 300 [3ex] alpha = 60^circ, 120^circ, 240^circ, 300^circ . Unit:: Circle:: Trig [3ex]$ Check

 $underline [3ex] cos(2alpha) + 14sin^2alpha [3ex] alpha = 60^circ [3ex] 2alpha = 2 * 60 = 120 [3ex] cos2alpha = cos 120 = -dfrac<1> <2>[5ex] sin^2alpha = (sin 60)^2 = left(dfrac><2> ight)^2 = dfrac<(sqrt<3>)^2> <2^2>= dfrac<3> <4>[5ex] 14sin^2alpha = 14left(dfrac<3><4> ight) = 7left(dfrac<3><2> ight) = dfrac<21> <2>[5ex] = -dfrac<1> <2>+ dfrac<21> <2>[5ex] = dfrac<-1 + 21> <2>[5ex] = dfrac<20> <2>[5ex] = 10 [3ex] YES [3ex]$ $alpha = 120^circ [3ex] 2alpha = 2 * 120 = 240 [3ex] cos2alpha = cos 240 = -dfrac<1> <2>[5ex] sin^2alpha = (sin 120)^2 = left(dfrac><2> ight)^2 = dfrac<(sqrt<3>)^2> <2^2>= dfrac<3> <4>[5ex] 14sin^2alpha = 14left(dfrac<3><4> ight) = 7left(dfrac<3><2> ight) = dfrac<21> <2>[5ex] = -dfrac<1> <2>+ dfrac<21> <2>[5ex] = dfrac<-1 + 21> <2>[5ex] = dfrac<20> <2>[5ex] = 10 [3ex] YES [3ex]$ $alpha = 240^circ [3ex] 2alpha = 2 * 240 = 480 [3ex] cos2alpha = cos 480 [3ex] coterminal:: angle ::of:: 480 = 480 - 360 = 120 [3ex] cos2alpha = cos 480 = cos 120 = -dfrac<1> <2>[5ex] sin^2alpha = (sin 480)^2 = (sin 120)^2 = left(dfrac><2> ight)^2 = dfrac<(sqrt<3>)^2> <2^2>= dfrac<3> <4>[5ex] 14sin^2alpha = 14left(dfrac<3><4> ight) = 7left(dfrac<3><2> ight) = dfrac<21> <2>[5ex] = -dfrac<1> <2>+ dfrac<21> <2>[5ex] = dfrac<-1 + 21> <2>[5ex] = dfrac<20> <2>[5ex] = 10 [3ex] YES [3ex]$ $alpha = 300^circ [3ex] 2alpha = 2 * 300 = 600 [3ex] cos2alpha = cos 600 [3ex] coterminal:: angle ::of:: 600 = 600 - 360 = 240 [3ex] cos2alpha = cos 600 = cos 240 = -dfrac<1> <2>[5ex] sin^2alpha = (sin 600)^2 = (sin 240)^2 = left(dfrac><2> ight)^2 = dfrac<(sqrt<3>)^2> <2^2>= dfrac<3> <4>[5ex] 14sin^2alpha = 14left(dfrac<3><4> ight) = 7left(dfrac<3><2> ight) = dfrac<21> <2>[5ex] = -dfrac<1> <2>+ dfrac<21> <2>[5ex] = dfrac<-1 + 21> <2>[5ex] = dfrac<20> <2>[5ex] = 10 [3ex] YES$ $underline [3ex] 10$

$underline [3ex] alpha = 60^circ = 60 * dfrac <180>= dfrac <3>[5ex] alpha = 120^circ = 120 * dfrac <180>= dfrac<2pi> <3>[5ex] alpha = 240^circ = 240 * dfrac <180>= dfrac<4pi> <3>[5ex] alpha = 300^circ = 300 * dfrac <180>= dfrac<5pi> <3>[5ex] underline [3ex] alpha = dfrac <3>+ 2pi k . coterminal:: angle s [5ex] alpha = dfrac<2pi> <3>+ 2pi k . coterminal:: angle s [5ex] alpha = dfrac<4pi> <3>+ 2pi k . coterminal:: angle s [5ex] alpha = dfrac<5pi> <3>+ 2pi k . coterminal:: angle s$

(18.) Determine all the solutions of $3 analphasinalpha - 3 analpha = 0$ in the interval $[0, 2pi)$

This is not straightforward. It has two functions: the tangent function and the cosecant function.
How do we express it in terms of only one function?
If we cannot express it in terms of only one function, can we factor by GCF?

$3 analphasinalpha - 3 analpha = 0 [3ex] GCF = 3 analpha [3ex] Factor:: by:: GCF [3ex] 3 analpha(sinalpha - 1) = 0 [3ex] 3 analpha = 0 ::OR:: sinalpha - 1 = 0 [3ex] analpha = dfrac<0> <3>::OR:: sinalpha = 1 [3ex] analpha = 0 ::OR:: sinalpha = 1 [3ex] alpha = an^<-1>(0) ::OR:: alpha = sin^<-1>(1) [3ex] alpha = 0, 180 ::OR:: alpha = 90 [3ex] alpha = 0^circ, 90^circ, 180^circ . Unit:: Circle:: Trig [3ex]$ Check

 $underline [3ex] 3 analphasinalpha - 3 analpha [3ex] alpha = 0^circ [5ex] analpha = an 0 = 0 [3ex] sinalpha = sin 0 = 0 [3ex] = 3 * 0 * 0 - 3 * 0 [3ex] = 0 - 0 [3ex] = 0 [3ex] YES [3ex]$ $alpha = 90^circ [5ex] analpha = an 90 ::is:: undefined [3ex] STOP [3ex] NO [3ex]$ $alpha = 180^circ [5ex] analpha = an 180 = 0 [3ex] sinalpha = sin 180 = 0 [3ex] = 3 * 0 * 0 - 3 * 0 [3ex] = 0 - 0 [3ex] = 0 [3ex] YES$ $underline [3ex] 0$

$underline [3ex] alpha = 0^circ = 0 * dfrac <180>= 0 [5ex] alpha = 180^circ = 180 * dfrac <180>= pi [5ex] underline [3ex] alpha = 0 + 2pi k = 2pi k . coterminal:: angle s [3ex] alpha = pi + 2pi k . coterminal:: angle s$

NCERT Solutions for Math Class 11 Chapter-3 Trigonometric Functions

Q3 :A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?
Number of revolutions made by the wheel in 1 minute = 360
∴Number of revolutions made by the wheel in 1 second =
In one complete revolution, the wheel turns an angle of 2π radian.
Hence, in 6 complete revolutions, it will turn an angle of 6 × 2π radian, i.e.,
Thus, in one second, the wheel turns an angle of 12π radian.

Q4 :Find the degree measure of the angle subtended at the centre of a circle of radius 100 cm by an arc of length 22 cm.
We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then
Therefore, forr = 100 cm, l = 22 cm, we have

Thus, the required angle is 12°36“².

Q5 :In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor arc of the chord.
Diameter of the circle = 40 cm
∴Radius (r) of the circle =
Let AB be a chord (length = 20 cm) of the circle.

In ΔOAB, OA = OB = Radius of circle = 20 cm
Also, AB = 20 cm
Thus, ΔOAB is an equilateral triangle.
∴θ = 60° =
We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then
Thus, the length of the minor arc of the chord is

Q6 :If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.
Let the radii of the two circles be and. Let an arc of length l subtend an angle of 60° at the centre of the circle of radius r1, while let an arc of length l subtend an angle of 75° at the centre of the circle of radius r2.
Now, 60° = and 75° =
We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then

Thus, the ratio of the radii is 5:4.

Q7 :Find the angle in radian though which a pendulum swings if its length is 75 cm and the tip describes an arc of length
(i) 10 cm (ii) 15 cm (iii) 21 cm
We know that in a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then
It is given that r = 75 cm
(i) Here, l = 10 cm

(ii) Here, l = 15 cm

(iii) Here, l = 21 cm

Exercise 3.2 : Solutions of Questions on Page Number : 63
Q1 :Find the values of other five trigonometric functions if , x lies in third quadrant.

Since x lies in the 3rd quadrant, the value of sin x will be negative.

Q2 :Find the values of other five trigonometric functions if , x lies in second quadrant.

Since x lies in the 2nd quadrant, the value of cos x will be negative

Q3 :Find the values of other five trigonometric functions if , x lies in third quadrant.

Since x lies in the 3rd quadrant, the value of sec x will be negative.

Q4 :Find the values of other five trigonometric functions if , x lies in fourth quadrant.

Since x lies in the 4th quadrant, the value of sin x will be negative.

Q5 :Find the values of other five trigonometric functions if , x lies in second quadrant.

Since x lies in the 2nd quadrant, the value of sec x will be negative.
∴sec x =

Q6 :Find the value of the trigonometric function sin 765°
It is known that the values of sin x repeat after an interval of 2π or 360°.

Q7 :Find the value of the trigonometric function cosec (-1410°)
It is known that the values of cosec x repeat after an interval of 2π or 360°.

Q8 :Find the value of the trigonometric function
It is known that the values of tan x repeat after an interval of π or 180°.

Q9 :Find the value of the trigonometric function
It is known that the values of sin x repeat after an interval of 2π or 360°.

Q10 :Find the value of the trigonometric function
It is known that the values of cot x repeat after an interval of π or 180°.

Exercise 3.3 : Solutions of Questions on Page Number : 73
Q1 :
L.H.S. =

Q2 :Prove that
L.H.S. =

Q3 :Prove that
L.H.S. =

Q4 :Prove that
L.H.S =

Q5 :Find the value of:
(i) sin 75°
(ii) tan 15°
(i) sin 75° = sin (45° + 30°)
= sin 45° cos 30° + cos 45° sin 30°
[sin (x + y) = sin x cos y + cos x sin y]

(ii) tan 15° = tan (45° – 30°)

Q6 :Prove that:

Q7 :Prove that:
It is known that
∴L.H.S. =

Q8 :Prove that

Q9 :
L.H.S. =

Q10 :Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
L.H.S. = sin (n + 1)x sin(n + 2)x + cos (n + 1)x cos(n + 2)x
4

Q11 :Prove that
It is known that
∴L.H.S. =

Q12 :Prove that sin 2 6x – sin2 4x = sin 2x sin 10x
It is known that
∴L.H.S. = sin 2 6x – sin 2 4x
= (sin 6x + sin 4x) (sin 6x – sin 4x)
= (2 sin 5x cos x) (2 cos 5x sin x)
= (2 sin 5x cos 5x) (2 sin x cos x)
= sin 10x sin 2x
= R.H.S.

Q13 :Prove that cos 2 2x – cos 2 6x = sin 4x sin 8x
It is known that
∴L.H.S. = cos 2 2x – cos 2 6x
= (cos 2x + cos 6x) (cos 2x – 6x)

= [2 cos 4x cos 2x] [-2 sin 4x (-sin 2x)]
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x
= R.H.S.

Q14 :Prove that sin 2x + 2sin 4x + sin 6x = 4cos 2 x sin 4x
L.H.S. = sin 2x + 2 sin 4x + sin 6x
= [sin 2x + sin 6x] + 2 sin 4x

= 2 sin 4x cos (- 2x) + 2 sin 4x
= 2 sin 4x cos 2x + 2 sin 4x
= 2 sin 4x (cos 2x + 1)
= 2 sin 4x (2 cos 2 x – 1 + 1)
= 2 sin 4x (2 cos 2 x)
= 4cos 2 x sin 4x
= R.H.S.

Q15 :Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)
L.H.S = cot 4x (sin 5x + sin 3x)

= 2 cos 4x cos x
R.H.S. = cot x (sin 5x – sin 3x)

Q16 :Prove that
It is known that
∴L.H.S =

Q17 :Prove that
It is known that
∴L.H.S. =

Q18 :Prove that
It is known that
∴L.H.S. =

Q19 :Prove that
It is known that
∴L.H.S. =

Q20 :Prove that
It is known that
∴L.H.S. =

Q21 :Prove that
L.H.S. =

Q22 :Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1
L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x
= cot x cot 2x – cot 3x (cot 2x + cot x)
= cot x cot 2x – cot (2x + x) (cot 2x + cot x)

= cot x cot 2x – (cot 2x cot x – 1)
= 1 = R.H.S.

Q23 :Prove that
It is known that
∴L.H.S. = tan 4x = tan 2(2x)

Q24 :Prove that cos 4x = 1 – 8sin 2 x cos 2 x
L.H.S. = cos 4x
= cos 2(2x)
= 1 – 2 sin 2 2x [cos 2A = 1 – 2 sin 2 A]
= 1 – 2(2 sin x cos x) 2 [sin2A = 2sin A cosA]
= 1 – 8 sin 2 x cos 2 x
= R.H.S.

Q25 :Prove that: cos 6x = 32 cos 6 x – 48 cos 4 x + 18 cos 2 x – 1
L.H.S. = cos 6 x
= cos 3(2x)
= 4 cos 3 2x – 3 cos 2x [cos 3A = 4 cos 3 A – 3 cos A]
= 4 [(2 cos 2 x – 1) 3 – 3 (2 cos 2 x – 1) [cos 2x = 2 cos 2 x – 1]
= 4 [(2 cos 2 x) 3 – (1) 3 – 3 (2 cos 2 x) 2 + 3 (2 cos 2 x)] – 6cos 2 x + 3
= 4 [8cos 6 x – 1 – 12 cos 4 x + 6 cos 2 x] – 6 cos 2 x + 3
= 32 cos 6 x – 4 – 48 cos 4 x + 24 cos 2 x – 6 cos 2 x + 3
= 32 cos 6 x – 48 cos 4 x + 18 cos 2 x – 1
= R.H.S

Exercise 3.4 : Solutions of Questions on Page Number : 78
Q1 :Find the principal and general solutions of the equation

Therefore, the principal solutions are x = and

Therefore, the general solution is

Q2 :Find the principal and general solutions of the equation

Therefore, the principal solutions are x = and
Therefore, the general solution is, where n ∈ Z

Q3 :Find the principal and general solutions of the equation

Therefore, the principal solutions are x = and
Therefore, the general solution is

Q4 :Find the general solution of cosec x = -2
cosec x= -2

Therefore, the principal solutions are x =
Therefore, the general solution is

Q5 :Find the general solution of the equation cos 4x = cos 2x

Q6 :Find the general solution of the equation cos 3x + cos x – cos 2x = 0
cos 3x + cos x – cos 2x = 0

Q7 :Find the general solution of the equation sin 2x + cos x = 0
sin 2x + cos x = 0

Therefore, the general solution is

Q8 :Find the general solution of the equation sec 2 2x = 1 – tan 2x
sec2 2x = 1 – tan 2x

Therefore, the general solution is

Q9 :Find the general solution of the equation sin x + sin 3x + sin 5x =0
sin x + sin 3x + sin 5x =0

Therefore, the general solution is

Exercise Miscellaneous : Solutions of Questions on Page Number : 81
Q1 :Prove that:

Q2 :Prove that: (sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0
L.H.S.= (sin 3x + sin x) sin x + (cos 3x – cos x) cos x
= RH.S.

Q3 :Prove that:
L.H.S. =

Q4 :Prove that:
L.H.S. =

Q5 :Prove that:
It is known that
∴L.H.S. =

Q6 :Prove that:
It is known that
L.H.S. =
= tan 6x
= R.H.S.

Q7 :Prove that:
L.H.S. =

Q8 : , x in quadrant II
Here, x is in quadrant II.
i.e.,

Therefore, are all positive.

As x is in quadrant II, cosx is negative.

Thus, the respective values of are

Q9 :Find for , x in quadrant III
Here, x is in quadrant III.

Therefore, and are negative, where as is positive.

Thus, the respective values of are

Q10 :Find for , x in quadrant II
Here, x is in quadrant II.

Trigonometric identities

Pythagorean identity The following identity is commonly used:

Inverse trigonometric identities The following identities are commonly used:

Addition formulas The following identities are commonly used:

 Name Formula Cosine addition $cos(a+b)=cos(a)cos(b)-sin(a)sin(b)$ Sine addition $sin(a+b)=sin(a)cos(b)+sin(b)cos(a)$ Tangent addition $displaystyle an(a+b)=frac< an(a)+ an(b)><1- an(a) an(b)>$

Product-to-sum and sum-to-product identities The following identities are commonly used:

 Name Formula Product-to-sum $displaystylecos(a)cos(b)=frac<1><2>(cos(a-b)+cos(a+b))$ $displaystylesin(a)sin(b)=frac<1><2>(cos(a-b)-cos(a+b))$ $displaystylesin(a)cos(b)=frac<1><2>(sin(a+b)+sin(a-b))$ $displaystylecos(a)sin(b)=frac<1><2>(sin(a+b)-sin(a-b))$ $displaystyle an(a) an(b)=frac$ Sum-to-product $displaystylecos(a)+cos(b)=2cosleft(frac<2> ight)cosleft(frac<2> ight)$ $displaystylecos(a)-cos(b)=-2sinleft(frac<2> ight)sinleft(frac<2> ight)$ $displaystylesin(a)+sin(b)=2sinleft(frac<2> ight)cosleft(frac<2> ight)$ $displaystylesin(a)-sin(b)=2sinleft(frac<2> ight)cosleft(frac<2> ight)$

Symmetry identities The following identities are commonly used:

 By $alpha=0$ By $alpha=frac<4>$ By $alpha=frac<2>$ $displaystylecosleft(- heta ight)=cos( heta)$ $displaystylecosleft(frac<2>- heta ight)=sin( heta)$ $displaystylecosleft(pi- heta ight)=-cos( heta)$ $displaystylesinleft(- heta ight)=-sin( heta)$ $displaystylesinleft(frac<2>- heta ight)=cos( heta)$ $displaystylesinleft(pi- heta ight)=sin( heta)$ $displaystyle anleft(- heta ight)=- an( heta)$ $displaystyle anleft(frac<2>- heta ight)=frac<1>< an( heta)>$ $displaystyle anleft(pi- heta ight)=- an( heta)$

Shift identities The following identities are commonly used:

 By $frac<2>$ By $pi$ $displaystylecosleft( heta+frac<2> ight)=-sin( heta)$ $displaystylecosleft( heta+pi ight)=-cos( heta)$ $displaystylesinleft( heta+frac<2> ight)=cos( heta)$ $displaystylesinleft( heta+pi ight)=-sin( heta)$ $displaystyle anleft( heta+frac<2> ight)=-frac<1>< an( heta)>$ $displaystyle anleft( heta+pi ight)= an( heta)$

Introduction to the Six Trigonometric Functions (Ratios)

In right triangle Trigonometry there are six possible ratios (functions). A ratio is a comparison of two numbers (or sides of a triangle) by division. The Greek letter, θ, will be used to represent the reference angle in the right triangle.

Opposite refers to the side of the triangle that is opposite of the reference angle

Adjacent refers to the side of the triangle that is adjacent to the reference angle.(the adjacent side will always form one side of the reference angle.

The hypotenuse is the side of the triangle that is always opposite the right angle.

These six ratios represent all the ways to compare two sides of a right triangle. Notice that cosecant is the reciprocal of sine, secant is the reciprocal of cosine and cotangent is the reciprocal of tangent. The hypotenuse will never vary in its location however, the opposite and adjacent side will be determined by the reference angle.

To link to this Introduction to the Six Trigonometric Functions (Ratios) page, copy the following code to your site:

Directions: Using the integers -9 to 9, at most one time each, fill in the &hellip

I am not sure if here is an appropriate place for correct answers, but here are some other combinations my students found, starting from the blank on the left and working to the blank on the right:

9, 4, 6, 3, 2 ,7
5, 4, 6, 7, 2, 3
5, 1, 2, 9, 8, 4
5, 3, 4, 9, 7, 2
5, 3, 4, 8, 6, 2
6, 1, 2, 4, 3, 5
6, 1, 2, 8, 7, 5
6, 1, 2, 9, 8, 5
4, 3, 2, 9, 8, 1

3, 1, 2, 8, 9, 4
9, 4, 2, 8, 7, 5
5, 4, 6, 2, 1, 3
9, 1, 2, 4, 3, 8
4, 2, 6, 8, 7, 3
3, 1, 6, 8, 5, 2
9, 4, 2, 7, 6, 5
7, 1, 2, 9, 8, 6
7, 3, 2, 9, 8, 4
9, 6, 2, 5, 4, 3
5, 4, 6, 7, 2, 3

3, 4, 6, 9, 8, 1
4, 3, 6, 9, 8, 3
9, 8, 6, 7, 2, 5
8, 2, 4, 5, 3, 6
8, 2, 6, 5, 4, 7
3, 2, 6, 7, 4, 1
4, 2, 6, 3, 8, 5