# $\sin(2\arccos(x))$, please help me understand how to do these kind of problems.

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 $2(\sqrt{1 - x^2})x$

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Arturo beat me to the edit! I was about halfway done when you finished. –  Joe Mar 29 '12 at 3:12

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 $\arccos(1/2)+\arcsin(1/2)$?

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Thank you so much! Would you mind solving it for me (or at least show me the correct answer) so i know i am doing it right? –  Steven Rogers Mar 29 '12 at 3:02
@Steven: Why don't you try solving it yourself using this information and update the question with your work? –  anon Mar 29 '12 at 3:06
Okay, i will do that right now, thank you! –  Steven Rogers Mar 29 '12 at 3:08
For the first, we get $2\sin\theta\cos\theta$. And $\sin\theta=\sqrt{1-x^2}$, so you should get $2x\sqrt{1-x^2}$. For the tangent question, we can compute using the formula. But note that if $x$ is between $0$ and $1$, the angles $\arcsin x$ and $\arccos x$ add up to $\pi/2$ (radians), and the tangent function is not defined at $\pi/2$. If $x$ is negative, again from the definitions of $\arcsin x$ and $\arccos x$, the sum is $\pi/2$, so the tangent function is not defined. –  André Nicolas Mar 29 '12 at 3:15
Thank you so much everyone! I totally get it now! –  Steven Rogers Mar 29 '12 at 3:24