Question

How to find the exact value of inverse trigonometric function.

Answer 1

Hint: Inverse trigonometric functions are nothing but the reverse functions of the original trigonometric functions. Now we know the output value of trigonometric functions for each input angle. Hence for each trigonometric function we have output values corresponding to input angles. For inverse function we will use the same correspondence to find the required angle from the given input values

Complete answer:
Let us first understand functions and inverse functions.
A function is basically a relation in which each input has a unique output.
Hence a function is a Binary relation between two sets such that every element of the first set is associated with one element of another set.
If a function takes elements of set A to set B then A is called the domain of the function and B is called the co-domain or range of function.
Now let us understand what inverse functions are.
An inverse function is a function which reverses the original function.
For example if we have a function f such that f(x) = y.
Then if g is the inverse function of f then g(y) = x.
Hence if we have a function f which takes elements of set A to set B. Then the inverse function g of f is a function which takes elements of set B to set A.
Now we know the trigonometric functions which are sin, cos, tan, cot, sec, cosec.
Similarly for each trigonometric functions we have their inverse functions ${{\sin }^{-1}},{{\cos }^{-1}},....$
Now finding the values of inverse functions is quite easy if we know the trigonometric values.
For example we have $\cos \left( \dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ .
Then ${{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)=\dfrac{\pi }{4}$ .
Hence we have to find an angle such that the angle gives the corresponding values of the input.

Note: Note that the inverse function ${{\cos }^{-1}}x\ne \dfrac{1}{{{\cos }^{-1}}x}$ . Also note that we have $\cos \left( \dfrac{\pi }{4} \right)=\cos \left( \dfrac{5\pi }{4} \right)=\cos \left( 2n\pi +\dfrac{\pi }{4} \right)=\dfrac{1}{\sqrt{2}}$ Hence in general we have ${{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}} \right)=2n\pi +\dfrac{\pi }{4}$ . To avoid multiple answers we will restrict the range to be $\left[ 0,2\pi \right]$