Page 19 - Trigonometry
P. 19
Example 6:
Derivative of an inverse trigonometric function. Let,
−1
= sin ( )
Then
−1
sin( ) = sin[sin ( )] =
Differentiating,
sin( ) =
cos( ) = 1
Dividing by cos(θ),
1 1
= =
cos( ) √1 − sin ( )
2
This becomes,
1
=
√1 − 2
Leaving,
−1 1
sin ( ) = 1
√1− 2
As you can see x < 1, to verify this result consider three
test values for = 0.49, 0.265, 0.857. Set calculator
15
