Trigonometry: Evaluating Inverse Trigonometric Functions
 This is a free lesson from our course in Trigonometry In this lesson you'll learn with the help of several examples with solution how to evaluate expressions which involve inverse trigonometric functions and values of sin-1 1, cos-1 1/2, tan-1 3, cos-1 3/2, cos-1 1 and sin-1 2/2. When you evaluate an inverse trigonometric function you are asking what angle measure, say , did you plug into the trig function (remember regular, and not inverse!) to get x. Theorems : � y = sin-1 x is equivalent to sin y = x with -1 < x < 1 and - / 2 <= y <= / 2 � y = cos-1 x is equivalent to cos y = x with -1 < x < 1 and 0 <= y <= � y = tan-1 x is equivalent to tan y = x with - / 2 < y < / 2 (More text below video...)
Other useful lessons:
 Meaning - Inverse Trigonometric Functions Solving Trigonometric Equations
(Continued from above) E.g.: sin-1(- 3 / 2)
Let, y = sin-1(- 3 / 2).
According to theorem 1 above, this is equivalent to sin y = - 3 / 2 , with - / 2 <= y <= / 2
sin ( /3) = 3 / 2.
You also know that sin(-x) = - sin x. So sin (- / 3) = - 3 / 2
Comparing the last expression with the equation sin y = - 3 / 2, we conclude that y = - / 3

The values of inverse functions can be written in both radians and degrees. All this will be explained to you by the instructor with the help of video and solving problems related to inverse trigonometric functions.

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