Problems on inverse trigonometric functions are solved and detailed solutions are presented. Also exercises with answers are presented at the end of this page. We first review some of the theorems and properties of the inverse functions.

**Theorems**

1.
**y = arcsin x is equivalent to sin y = x**

with **-1 ≤ x ≤ 1 and - π / 2 ≤ y ≤ π / 2**

2.
**y = arcos x is equivalent to cos y = x**

with **-1 ≤ x ≤ 1 and 0 ≤ y ≤ π**

3.
**y = arctan x is equivalent to tan y = x**

with **- π / 2 < y < π / 2**

**Question 1:** Find the exact value of

1. arcsin(- √3 / 2)

2. arctan(- 1 )

3. arccos(- 1 / 2)

**Solution to question 1:**

1. arcsin(- √3 / 2)

Let y = arcsin(- √3 / 2). According to theorem 1 above, this is equivalent to

sin y = - √3 / 2 , with - π / 2 ≤ y ≤ π / 2

From table of special angles sin (π /3) = √3 / 2.

We also know that sin(-x) = - sin x. So

sin (- π / 3) = - √3 / 2

Comparing the last expression with the equation sin y = - √3 / 2, we conclude that

y = - π / 3

2. arctan(- 1 )

Let y = arctan(- 1 ). According to 3 above

tan y = - 1 with - π / 2 < y < π / 2

From table of special angles tan (π / 4) = 1.

We also know that tan(- x) = - tan x. So

tan (-π / 4) = - 1

Compare the last statement with tan y = - 1 to obtain

y = - π/4

3. arccos(- 1 / 2)

Let y = arccos(- 1 / 2). According to theorem 2 above

cos y = - 1 / 2 with 0 ≤ y ≤ π

From table of special angles cos (π / 3) = 1 / 2

We also know that cos(π - x) = - cos x. So

cos (π - π/3) = - 1 / 2

Compare the last statement with cos y = - 1 / 2 to obtain

y = π - π / 3 = 2 π / 3

**Question 2:** Simplify cos(arcsin x )

**Solution to question 2:**

Let z = cos ( arcsin x ) and y = arcsin x so that z = cos y. According to theorem 1 above y = arcsin x may also be written as

sin y = x with - π / 2 ≤ y ≤ π / 2

Also

sin^{2}y + cos^{2}y = 1

Substitute sin y by x and solve for cos y to obtain

cos y = + or - √ (1 - x^{2})

But - π / 2 ≤ y ≤ π / 2 so that cos y is positive

z = cos y = cos(arcsin x) = √ (1 - x ^{2})

**Question 3:** Simplify csc ( arctan x )

**Solution to question 3:**

Let z = csc ( arctan x ) and y = arctan x so that z = csc y = 1 / sin y. Using theorem 3 above y = arctan x may also be written as

tan y = x with - π / 2 < y < π / 2

Also

tan^{2}y = sin^{2}y / cos^{2}y = sin^{2}y / (1 - sin^{2}y)

Solve the above for sin y

sin y = + or - √ [ tan^{2}y / (1 + tan^{2}y) ]

= + or - | tan y | / √ [ (1 + tan^{2}y) ]

For - π / 2 < y ≤ 0 sin y is negative and tan y is also negative so that | tan y | = - tan y and

sin y = - ( - tan y ) / √ [ (1 + tan^{2}y) ] = tan y / √ [ (1 + tan^{2}y) ]

For 0 ≤ y < π/2 sin y is positive and tan y is also positive so that | tan y | = tan y and

sin y = tan y / √ [ (1 + tan^{2}y) ]

Finally

z = csc ( arctan x ) = 1 / sin y = √ [ (1 + x^{2}) ] / x

**Question 4:** Evaluate the following

1. arcsin( sin (7 π / 4))

2. arccos( cos (4 π / 3 ))

**Solution to question 4:**

1.
arcsin( sin ( y ) ) = y only for - π / 2 ≤ y ≤ π / 2. So we first transform the given expression noting that sin (7 π / 4) = sin (-π / 4) as follows

arcsin( sin (7 π / 4)) = arcsin( sin (- π / 4))

- π / 4 was chosen because it satisfies the condition - π / 2 ≤ y ≤ π / 2. Hence

arcsin( sin (7 π / 4)) = - π / 4

2.
arccos( cos ( y ) ) = y only for 0 ≤ y ≤ π . We first transform the given expression noting that cos (4 π / 3) = cos (2 π / 3) as follows

arccos( cos (4 π / 3)) = arccos( cos (2 π / 3))

2 π / 3 was chosen because it satisfies the condition 0 ≤ y ≤ π . Which gives

arccos( cos (4 π / 3)) = 2 π / 3

**Exercises**

1. Evaluate arcsin( sin (13 π / 4))

2. Simplify sec ( arctan x )

3. Find the exact value of arccos(- √3 / 2)

**Answers to Above Exercises**

1. - π / 4

2. √(x^{2} + 1)

3. 5 π / 6