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

First, though, you'll find this page useful.

Then, in the first one, let: $\displaystyle y = \arccos(2x)$

So the domain is $\displaystyle -1\le 2x \le 1$, or $\displaystyle -\tfrac12 \le x \le \tfrac12$. And, assuming we're just taking the principle values, the range of values of $\displaystyle y$ is $\displaystyle 0\le y \le\pi$.

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

$\displaystyle \Rightarrow \sin(\arccos(2x)) = \sin y=\sqrt{1-\cos^2y} = \sqrt{1-4x^2}$

noting that we only need the positive square root, because for $\displaystyle 0\le y \le \pi,\; \sin y \ge 0$.

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

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

Grandad