We need to be able to transform this equation to get rid of the trig functions.
To better explain this, this is how the problem before this one was done. (I checked the answer, i got this one right.)
$$\sin(\arccos(x))$$
We substitute $\arccos(x)$ with the symbol $\theta$.
So
$$\sin(\theta), \text{ where }\theta = \arccos(x),\text{ so }\cos(\theta)=x.$$
So if we draw a right triangle with angles A, B, and 90*. and the sides being $h$ (hypotenuse), $o$ (opposite), and $a$ (adjacent).
Angle A = $\theta$, so side $a$ is equal to $x$ and the side $h$ is equal to $1$.
With Pythagorean theorem $x^2 + o^2 = 1^2$, that means that the opposite side is equal to $\sqrt{1 – x^2}$.
With this triangle drawn, and our $\theta$ still being the same, this means that our $\sin(\theta)$ is equal to the $o/h$ of our drawn triangle. So this equations morphs into
$\sqrt{1 – x^2}/1$
How would i do this with $\sin(2\arccos(x))$ or $\tan(\arccos(x) + \arcsin(x))$?
Edit: From the answer given, i will try to solve these two equations.
$$\sin(2\arccos(x)) = 2\sin(\arccos(x))\cos(\arccos(x))$$
$$\sin(\arccos(x))$$
$$\sin(\theta), \theta = \arccos(x)$$
$$\cos(\theta) = x$$
A = $\theta$, hypotenuse = 1, adjacent = x, opposite = $\sqrt{1 – x^2}$
$$\sin(\theta) = \sqrt{1 – x^2}$$
$$\cos(\arccos(x)) = x$$
So this should be $2(\sqrt{1 – x^2})x$
Best Answer
You will need to use the idea you used, plus some trigonometric identities. For the first, you will need $\sin 2\theta=2\sin\theta\cos\theta$.
For the second, you will sort of (but not really) need $\tan(\alpha+\beta)=\dfrac{\tan \alpha+\tan \beta}{1-\tan\alpha\tan\beta}$.
This last identity is perhaps not familiar. You can obtain it by writing down the usual addition formula for $\sin(\alpha+\beta)$ and $\cos(\alpha+\beta)$.
Actually, in your particular situation, there is no need for the general identity for $\tan(\alpha+\beta)$, since there is a close relationship between $\arcsin x$ and $\arccos x$. Hint: For example, what is $\arccos(1/2)+\arcsin(1/2)$?