# Thread: So confused..... domain and range of composite trig functions

1. ## So confused..... domain and range of composite trig functions

How can I identify the domain and range of problems such as sin(cos^-1(2x)) without using a calculator???

Other problems would be sin(sin^-1(x-1/2)) or cos^-1(2sin(x)).

Please help asap because I have a math test tomorrow and this is the only concept I do not know.

Thanks.

2. Hello HubridNoxx

Welcome to Math Help Forum!
Originally Posted by HubridNoxx
How can I identify the domain and range of problems such as sin(cos^-1(2x)) without using a calculator???

Other problems would be sin(sin^-1(x-1/2)) or cos^-1(2sin(x)).

Please help asap because I have a math test tomorrow and this is the only concept I do not know.

Thanks.
I assume that by $\sin^{-1}(x)$, you mean $\arcsin(x)$, etc, in which case my advice is to get rid of the inverse functions as soon as you can. They're nasty things!

Then, in the first one, let:
$y = \arccos(2x)$
So the domain is $-1\le 2x \le 1$, or $-\tfrac12 \le x \le \tfrac12$. And, assuming we're just taking the principle values, the range of values of $y$ is $0\le y \le\pi$.

Now if we re-write the above equation, we get:
$2x = \cos y$

$\Rightarrow \sin(\arccos(2x)) = \sin y=\sqrt{1-\cos^2y} = \sqrt{1-4x^2}$
noting that we only need the positive square root, because for $0\le y \le \pi,\; \sin y \ge 0$.

Therefore range of values of $\sin(\arccos(2x))\;( =\sqrt{1-4x^2})$ for $-\tfrac12le \le \tfrac12$ is $[0, 1]$.

See if you can work out the others in the same way.

3. Thanks, this should help me on my test.

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