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.
Theorem
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
