# Inverse Trigo Functions

• March 7th 2010, 02:52 AM
acc100jt
Inverse Trigo Functions
For the inverse cosine function, we have
$\cos x=y$ implies $x=\cos^{-1}y$ and $0 \leq x \leq \pi$

Let's say if I want to find the inverse of $g(x)=\cos x$ where $-2\pi \leq x \leq -\pi$

Is the following correct?

$g(x)= \cos x = \cos(x+2\pi)$

Let $y=g(x)$
$y=cos(x+2\pi)$
$x+2\pi=\cos^{-1}y$ since $0 \leq x+2\pi \leq \pi$
$x=\cos^{-1}y-2\pi$

Hence, $g^{-1}(x)=\cos^{-1}x-2\pi$ where $-1 \leq x \leq 1$

Is there any other way of doing this?

What about finding the inverse of
$h(x)=\cos x$ where $-\pi \leq x \leq 0$?

Many thanks for those who can help :)
• March 7th 2010, 05:41 AM
skeeter
Quote:

Originally Posted by acc100jt
For the inverse cosine function, we have
$\cos x=y$ implies $x=\cos^{-1}y$ and $0 \leq x \leq \pi$

Let's say if I want to find the inverse of $g(x)=\cos x$ where $-2\pi \leq x \leq -\pi$

Is the following correct?

$g(x)= \cos x = \cos(x+2\pi)$

Let $y=g(x)$
$y=cos(x+2\pi)$
$x+2\pi=\cos^{-1}y$ since $0 \leq x+2\pi \leq \pi$
$x=\cos^{-1}y-2\pi$

Hence, $g^{-1}(x)=\cos^{-1}x-2\pi$ where $-1 \leq x \leq 1$

Is there any other way of doing this?

What about finding the inverse of
$h(x)=\cos x$ where $-\pi \leq x \leq 0$?

Many thanks for those who can help :)

what you're doing is fine. I just find the resulting angle from the defined range of $y = \arccos{x}$ and think about where it would be in another interval on the unit circle.

for $-\pi \le x \le 0$

$y = -\arccos{x}$