Inverse Functions And Logarithms 3

Inverse Trigonometric Functions

We know that the sine function y=sin x is not one-to-one (use the Horizontal Line Test). But the function f(x)=sin x,  is one-to-one so we can find the inverse of it

$${\sin }^{-1}y=x\iff y=\sin x,\frac{-\pi }{2}⩽x⩽\frac{\pi }{2}$$

$$Thus,-1⩽y⩽1$$

\[Note:{\sin ^{ - 1}}x \ne \frac{1}{{\sin x}}\]

Remember:

So now we know that the range of sine is between -1 and 1 

and the range of inverse sine is between 

for that, we should know the facts that

1. $${\sin }^{-1}(\sin x)=x,for\frac{-\pi }{2}⩽x⩽\frac{\pi }{2}$$
2. $$\sin ({\sin }^{-1}x)=x,for-1⩽x⩽1$$

the same thing we can do with cos and tan 

$${\cos }^{-1}(\cos x)=x,for0⩽x⩽\pi$$

$$\cos ({\cos }^{-1}x)=x,for-1⩽x⩽1$$

$${\tan }^{-1}x=y\iff \tan y=x,and\frac{-\pi }{2}⩽y⩽\frac{\pi }{2}$$