calculus 2 tutor worksheet 1 inverse trigonometric functions · worksheet for calculus 2 tutor,...

Calculus 2 Tutor

Worksheet 1

Inverse Trigonometric Functions

Worksheet for Calculus 2 Tutor, Section 1:

Inverse Trigonometric Functions

1. For sinx, find the:

(a) Domain: the set of values of x so that sinx is defined;

(b) Range: the set of possible outputs of y for y = sinx.

2. For sin−1 x, find the:

(a) Domain: the set of values of x so that sin−1 x is defined;

(b) Range: the set of possible outputs of y for y = sin−1 x.

3. Find all the values of θ so that:

(a) sin θ = 12

c©2018 MathTutorDVD.com 1

(b) sec θ =√2

(c) tan θ = −1

4. Evaluate the given inverse trigonometric function, or specify if there is no answer:

(a) sin−1 0

(b) cos−1 0


(c) tan−1 0

(d) cot−1 0

(e) sec−1 0

(f) csc−1 0

.

5. Evaluate the given inverse trigonometric function, or specify if there is no answer:

(a) sin−1(

1√2

)


(b) cos−1(−√32

)

(c) tan−1−√3

(d) cot−1−1

(e) sec−1√2


(f) csc−1(−2√3

3

)

6. Evaluate the composite trigonometric functions:

(a) sin sin−1 0.5

(b) cot−1 cot π4.

(c) sin cos−1√32

.


7. Evaluate the composite trigonometric functions:

(a) sin−1 sin(37π6

)

(b) csc−1 csc(−11π

4

)

8. Challenge. Express the following composite trigonometric functions in simplest form,

and specify their domain and range:

(a) sin (csc−1 x).

(b) tan−1 (cot θ) for 0 ≤ θ ≤ π2.


(c) sin−1 (cos θ)

(d) cos(sin−1 x

)

(e) sin−1(cos

(sin−1 (cos θ)

))


Answer key.

1. Domain and range of sinx:

1(a). Answer: −∞ < x <∞ . Since sinx is defined for all x, its domain is −∞ < x <∞ .

1(b). Answer: −1 ≤ y ≤ 1 . Since sinx always produces an output between −1 and 1

inclusive, its range is −1 ≤ y ≤ 1 .


2. Domain and range of sin−1 x:

2(a). Answer: −1 ≤ x ≤ 1 . The domain of sin−1 x is the set of possible values of x so

that sin θ = x for some θ. Since the range of x = sin θ is −1 ≤ x ≤ 1 as θ varies

over any angle, the domain of sin−1 x is −1 ≤ x ≤ 1 .

2(b). Answer: −π2≤ θ ≤ π

2. In fact, sin−1 x is defined to have an output between

−π2≤ θ ≤ π

2. Because sin θ is a cyclic function, we specify that the range of

sin−1 x (angle whose sine is x) has to be between −π2

and π2

to ensure that there

is a unique answer.


3. Reverse evaluation of trigonometric functions.

3(a). Answer:π

6+ 2π · n, 5π

6+ 2π · n for all integer values of n. Since

sin 0 = 0

and sin θ increases for small values of θ, the guess θ = π6

indeed yields

sinπ

6=

1

2

Then, sin θ increases until it hits 1 at θ = π2. It decreases again to zero at θ = π,

so it again is equal to 12

at

θ =5π

6

Then, sin θ is negative for π < θ < 2π. After 2π, sin θ is cyclic. Therefore, it again

satisfies sin θ = 12

atπ

6+ 2π,

5π

6+ 2π

Actually, since it is cyclic, we can add or subtract any multiple of 2π and sin θ will

have the same values. So the answer isπ

6+ 2π · n, 5π

6+ 2π · n for all integer

values of n.


3(b). Answer: ±π4+ 2π · n for all integer values of n. This problem is the same as

solving for

cos θ =1√2

This is true for

θ =π

4

It is also true for

θ =−π4

There are no other times where the graph of cos θ crosses the line 1√2

between

θ = −π2

and θ = π2. Also, between −π and −π

2and π

2and π, the function cos θ

is negative. Therefore, ±π4

are the only solutions between −π and π. Since sec θ

like cos θ is cyclic with period 2π, the answer is ±π4+ 2π · n for all integer values

of n.


3(c). Answer:3π

4+ π · n for all integer values of n. To solve tan θ = −1 is the same

as solving for

sin θ = − cos θ

Since sin θ = cos θ at π4, then sin θ = − cos θ in the middle of other quadrants: that

is,3π

4

and

−π4

Since tan θ is cyclic with period π, the answer is3π

4+ π · n for all integer values

of n. The solution −π4

is the boxed answer evaluated at n = −1. Since the period

of tan θ is only π, not 2π, both of the answers between−π and π can be expressed

in this one form.


4. Inverse trigonometric functions evaluated at x = 0.

4(a). Answer: 0 . To solve θ = sin−1 0, we take sin θ = 0. The angle θ = 0 has

sin θ = 0, so θ = 0 is a solution to sin θ = 0. It is within the required boundaries

−π2≤ θ ≤ π

2, so the answer is sin−1 0 = 0 .

4(b). Answer:π

2. To solve θ = cos−1 0, we take cos θ = 0. The angle θ = π

2has

cos θ = 0, so θ = π2

is a solution to cos θ = π2. It is within the boundaries 0 ≤ θ ≤ π,

so the answer is cos−1 0 =π

2.

4(c). Answer: 0 . To solve θ = tan−1 0, we take tan θ = 0. The angle θ = 0 has

tan θ = sin θcos θ

= 0, so θ = 0 is a solution to tan θ = 0. It is within the boundaries

−π2< θ < π

2, so the answer is tan−1 0 = 0 .

4(d). Answer:π

2. To solve θ = cot−1 0, we take cot θ = 0. The angle θ = π

2has

cot θ = cos θsin θ

= 01= 0, so θ = π

2is a solution to cot θ = 0. It is within the boundaries


0 < θ < π, so the answer is cot−1 0 =π

2.

4(e). Answer: no solution . To solve θ = sec−1 0, we want to see if there is a θ so that

sec θ = 0. That is, we want to find θ so that 1cos θ

= 0. However, this is impossible

- there is no value of cos θ so that 1cos θ

= 0. There is no solution because 0 is

outside of the domain of sec−1 x.

4(f). Answer: no solution . To solve θ = csc−1 0, we want to see if there is a θ so that

csc θ = 0. That is, we want to find θ so that 1sin θ

= 0. However, this is impossible -

there is no value of sin θ so that 1sin θ

= 0. There is no solution because 0 is outside

of the domain of csc−1 x.


5. Inverse trigonometric functions evaluated for other x.

5(a). Answer: θ =π

4. We want to find θ so that sin θ = 1√

2and −π

2≤ θ ≤ π

2. The

answer θ =π

4meets those criteria as sin π

4= 1√

2.

5(b). Answer: θ =5π

6. We want to find θ so that cos θ = −

√32

and 0 ≤ θ ≤ π. The

answer θ =5π

6meets those criteria.

5(c). Answer: θ =π

3. We want to find θ so that tan θ =

√3 and −π

2< θ < π

2. Since

tan θ =sin θ

cos θ

when sin θ =√32

and cos θ = 12

we will have tan θ =√3. Both of these hold true

when θ =π

3, which is also within the required range of θ.


5(d). Answer: θ =3π

4. We want to find θ so that cot θ = −1 and 0 < θ < π. Since

cot θ =cos θ

sin θ

we are looking to find θ so that

cos θ = − sin θ.

This is true when θ = −π4, θ = 3π

4, etc, but only one of these is within the desired

range of values of θ, so the answer is θ =3π

4.

5(e). Answer: θ =π

4. We want to find θ so that sec θ =

√2 and 0 < θ < π

2or

π ≤ θ < 3π2. Translating into a problem about the more familiar cosine, we are

looking to find θ so that

cos θ =1√2

This is true at θ = π4

and θ = −π4

, but only the first is within the desired range of

values of θ, so the answer is θ =π

4.

5(f). Answer: θ =4π

3. We want to find θ so that csc θ = −2

√3

3and 0 < θ < π

2or

π < θ ≤ 3π2. Translating into a problem about the more familiar sine, we are

looking to find θ so that

sin θ = − 3

2√3


Fortunately, this fraction simplifies to −√32. The values of θ so that

sin θ = −√3

2

are θ = −π3, 4π

3, 5π

3, etc, but only one of these is within the desired range, so the

answer is θ =4π

3.


6. Composite trigonometric functions.

6(a). Answer: 0.5 . Since 0.5 is within the domain of sin−1, there is some value of θ so

that sin θ = 0.5. The function sin−1 finds that value of θ. Whatever that value of θ

is, we are about to take the sine of it, so it doesn’t really matter what it is since we

already know that sin θ = 0.5. Thus, the answer is 0.5 . We also could calculate

sin−1 0.5 = π6

and then sin π6= 0.5.

6(b). Answer:π

4. We know that cot π

4will be in the domain of cot−1, since there is

an angle where cotangent can be computed to yield cot π4

- namely,π

4! We also

could compute cot π4= 1 and then solve cot−1 1 = π

4, which is within the range

0 < θ < π as required for cotangent.

6(c). Answer:1

2. We calculate cos−1

√32

= π6, which is within the range 0 ≤ θ ≤ π.

Then we calculate sin π6=

1

2.


7. Composite trigonometric functions with the domain shifted.

7(a). Answer:π

6. The answer is not just 37π

6because that would be outside of the

defined range of sin−1 x, which is

−π2≤ θ ≤ π

2

However, since sin is periodic with period 2π, we can write

sin37π

6= sin

(6π +

π

6

)= sin

π

6

Then, sin−1 sin π6=π

6.

7(b). Answer:5π

4.The answer is not just −11π

4because that would be outside of the

defined range of csc−1 x, which is 0 < θ < π2

or π < θ ≤ 3π2. However, since csc is

periodic with period 2π, we can write csc −11π4

as

csc

(16π

4− 11π

4

)

Then, the answer is csc−1 csc 5π4=

5π

4.


8. Composite trigonometric functions with variables.

8(a). Answer:1

x. The domain of this function is −∞ < x ≤ −1 and 1 ≤ x < ∞. For

any x we can calculate csc−1 x except for −1 < x < 1. If

csc−1 x = θ

then

x = csc θ

which means

sin θ =1

x

Since x ≤ −1 or x ≥ 1, then1

xis indeed between −1 and 1 as is required for

the range of sin θ. We can also solve this problem by drawing a diagram:

Plot for 8a

If the angle θ is csc−1 x, then csc θ = hypotenuseopposite = x. Then, the hypotenuse is x

and the opposite side is 1. The angle’s sine is then the opposite divided by the

hypotenuse, which is 1x. The range of this over the given domain is

−1 ≤ 1

x≤ 1

which is the range of sin.


8(b). Answer:π

2− θ . The domain of cot θ is θ 6= πn for integer n because there

sin = 0. In this domain, cot θ could evaluate to be any real number x, and then

there is a value of tan−1 x for any value of x. So the domain of the function is

θ 6= πn. We are working with the sub-domain from 0 to π2. Over this domain, we

draw a triangle to visualize.

Plot for 8b

If the angle is θ, the hypotenuse is 1 and the sides are sin θ and cos θ. The problem

asks for an angle φ so that tanφ = cot θ. The opposite angle π2− θ has

tan(π2− θ

)=

cos θ

sin θ= cot θ

Therefore, the solution isπ

2− θ . The range of this is between 0 and π

2.


8(c). Answer:π

2− θ − 2πn or −3π

2+ θ − 2πn The domain of cos θ is all real num-

bers. Since the range of cosine is between −1 and 1, there will always be a value

of sin−1 cos θ, so the domain of the composite function is also all real numbers. To

evaluate the function, we can again draw a triangle to visualize.

Plot for 8c

If the triangle has angles θ and φ, and sides sin θ and cos θ, the angle φ = π2− θ

will have

sinφ =cos θ

1= cos θ

Then,

sin−1 (cos θ) =π

2− θ

if it is between −π2

and π2, that is, if θ is between 0 and π. For θ between π and 2π,

we have

cos θ = cos(θ − 2π) = cos(2π − θ)

where the argument is once again between 0 and π. Then,

sin−1 (cos 2π − θ) = π

2− (2π − θ) = −3π

2+ θ

So the answer will be of the typeπ

2− θ − 2πn or −3π

2+ θ − 2πn for some in-

teger n. It is the first if θ can be reduced by 2πn to be between 0 and π, and the

second if θ can be reduced by 2πn to be between π and 2π.


8(d). cos(sin−1 x

). The domain of sin−1 x is −1 ≤ x ≤ 1. Then, whatever the value of

sin−1 x = θ, cos θ will be defined. So the domain of the function is −1 ≤ x ≤ 1. We

can draw a triangle with angle sin−1 x = θ, opposite side x, and hypotenuse 1.

Plot for 8d

The sine of that angle is sin sin−1 x = x, which is indeed the calculation of op-

posite over hypotenuse. Then, by the Pythagorean theorem, the adjacent side

is√1− x2. Then, cos θ =

√1−x21

=√1− x2 . The range of the function is also

between 0 and 1 inclusive.


8(e). Answer: θ + πn or πn− θ , whichever is between 0 and π2

for some value of

n. For any value of θ, cos θ will be defined and will have value between −1 and

1 inclusive. Then, we will be able to take the arcsine of this value to produce an

angle between −π2

and π2. We will be able to take the cosine of this angle to get a

value between 0 and 1. We will be able to take the arcsine of this angle to get a

value between 0 and π2. So this is defined over all the real numbers. To solve this

problem in general, first we solve it for θ between 0 and 2π. We break it into four

cases:

1. First, we solve it for 0 < θ < π2. For 0 ≤ θ ≤ π

2, we can solve this prob-

lem in two ways. Using the third challenge question, we can write it as

sin−1(cos

(π2− θ

)). Then, applying the second part of this problem again, we

can write it as π2−

(π2− θ

)= θ Alternatively, using the previous part of this

problem, we can write it as sin−1(√

1− cos2 θ)= sin−1 sin θ = θ.

2. Next, we can solve the problem for π2≤ θ ≤ π. Using the third challenge ques-

tion, we can write it as sin−1(cos

(π2− θ

)). This is the same as sin−1

(cos

(θ − π

2

)),

which is π − θ by the third challenge question.

3. Next, we can solve the problem for π ≤ θ ≤ 3π2. Using the third challenge

question, this can be written as sin−1 cos(−3π

2+ θ

). Again by the third chal-

lenge question, this is θ − π.

4. Finally, we can solve the problem for 3π2≤ θ ≤ 2π. Using the third challenge

question, this can be written as sin−1 cos(−3π

2+ θ

). Again by the third chal-

lenge question, this is 2π − θ.

After checking for continuity at the points 0, π2, π, 3π

2, we can verify that the answer

is a piecewise function that has value θ′, π−θ′, θ′−π, 2π−θ′ where θ′ is θ reduced

to an angle between 0 and 2π, and where each function holds only on its respec-

tive quadrant. Graphing this function, we note that it can also be expressed as

a piecewise function that has value θ′, π − θ′, where θ′ is θ reduced to an angle


between 0 and π.


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