Inverse trig domain

• March 21st 2010, 06:48 PM
Stroodle
Inverse trig domain
Hi there,

Just wanted to check if my answer to this question is correct. I need to find the implied domain of $sin^{-1}\left [ 2cos(x+\frac{\pi}{2})\right ]$

I'm getting $x\in \left [ -\frac{5\pi}{6} , -\frac{\pi}{6}\right ]$

Thanks.
• March 21st 2010, 09:42 PM
Quote:

Originally Posted by Stroodle
Hi there,

Just wanted to check if my answer to this question is correct. I need to find the implied domain of $sin^{-1}\left [ 2cos(x+\frac{\pi}{2})\right ]$

I'm getting $x\in \left [ -\frac{5\pi}{6} , -\frac{\pi}{6}\right ]$

Thanks.

We must have

$-1\leq 2cos\left(x+\frac{\pi}{2}\right)\leq 1$

$-\frac{1}{2}\leq cos\left(x+\frac{\pi}{2}\right)\leq\frac{1}{2}$

Taking the inverse throughout gives:

$-\frac{\pi}{3}\leq x+\frac{\pi}{2}\leq \frac{\pi}{3}$

$\frac{-5\pi}{6}\leq x \leq -\frac{\pi}{6}$

This gives me the same domain you obtained. So you are correct (Clapping)
• March 21st 2010, 09:57 PM
Stroodle
Awesome. Thanks for the confirmation.
• March 22nd 2010, 02:39 AM
Hello everyone

Hold on a moment!

Think about the graph of $y = \cos \theta$. When $\cos\theta = \tfrac12,\; \theta = \tfrac{\pi}{3}, ...$
and when $\cos\theta = -\tfrac12, \;\theta = \tfrac{2\pi}{3}, ...$

So:
$-\frac12\le\cos\left(x+\frac{\pi}{2}\right)\le\frac 12$

$\Rightarrow \frac{\pi}{3}\le x+\frac{\pi}{2}\le\frac{2\pi}{3}$
I think you'd better have another look at it!