non-calc inverse sine problem

Question

Show that $$2\sin^{-1}\left(\frac23\right)= \sin^{-1}{\left(\frac{4\sqrt5}9\right)}$$

I don't know how to get started on this question. this is non-calc

Answer 1

Let's translate this into normal sine to make things easier.

If $\theta$ is an angle such that $\sin\theta=\frac23$, show that $\sin2\theta=\frac{4\sqrt5}9$.

Can you take it from there?

Answer 2

We can use that $\sin(2\theta)=2\sin \theta \cos \theta$ then

$$2\arcsin \frac23=a \implies \sin\left(2\arcsin \frac23\right)=2\cdot \frac23\cos\left(\arcsin \frac23\right)=\sin a$$

and since $\arcsin$ has range between $-\pi/2$ and $\pi/2$

$$\cos\left(\arcsin \frac23\right)=\sqrt{1-\sin^2\left(\arcsin \frac23\right)}=\frac{\sqrt 5}3$$

then

$$\sin a=\frac{4\sqrt 5}9 \implies a=\arcsin \frac{4\sqrt 5}9$$