How to handle the "$2$" in an expression such as "$\sin\left(2\arcsin\frac{x}{4}\right)$"

Question

When an inverse trig function has a constant in front of it being multiplied how do you look at it? Do you distribute the 2 to the whole Pythagorean theorem? I can't really wrap my head around how to view these types of problems, as I've only known how to solve them without a constant between.

In particular, I'm wondering how this answer is achieved:

$$\sin\left(2\arcsin\left(\frac{x}{4}\right)\right) = \frac{x\sqrt{1-\frac{x^2}{16}}}{2}$$

Answer 1

He used the double angle theorem: $sin(2x) = 2 sin(x) cos(x)$ he also used the fact that $cos(arcsin(x)) = \sqrt{1-x^2}$.

Answer 2

$\sin{2x}=2\sin{x}\cos{x}$

Now, your input is just a function....

Answer 3

Let $$\alpha = \arcsin \dfrac {x}{4}$$ so that $$\sin \alpha = \dfrac {x}{4}$$ and $$\cos \alpha = \sqrt {1-\dfrac {x^2}{16}}$$

However, $$\sin 2 \alpha = 2 \sin \alpha \cos \alpha$$ so we have $$2 *\dfrac {x}{4}*\sqrt {1-\dfrac {x^2}{16}}$$ and reducing we get $$\dfrac{x \sqrt{1-\dfrac {x^2}{16}}}{2}$$

Answer 4

You shouldn't look at the factor $2$ as being "in front" of the arc sine, but inside the sine.

Then

$$\sin\left(2\arcsin\frac x4\right)=2\sin\left(\arcsin\frac x4\right)\cos\left(\arcsin\frac x4\right)=2\frac x4\sqrt{1-\left(\frac x4\right)^2}.$$