Question

Find value cos[2 sin^-1(-2/3)]

Answer 1

0.1 reacurring

Answer 2

Can you explain that step by step?

Answer 3

First use the double angle identity:
\[\cos 2x=\cos^2x-\sin^2x\]
In this case, \(x=\sin^{-1}\left(-\dfrac{2}{3}\right)\).
This means you have
\[\begin{align*}\cos\left(2\sin^{-1}\left(-\frac{2}{3}\right)\right)&=\cos^2\left(\sin^{-1}\left(-\frac{2}{3}\right)\right)-\sin^2\left(\sin^{-1}\left(-\frac{2}{3}\right)\right)\\\\
&=\left[\cos\left(\sin^{-1}\left(-\frac{2}{3}\right)\right)\right]^2-\left[\sin\left(\sin^{-1}\left(-\frac{2}{3}\right)\right)\right]^2
\end{align*}\]

Answer 4

Now the second term is pretty easy to simplify, since \(\sin(\sin^{-1}x)=x\), so you have
\[\left[\cos\left(\sin^{-1}\left(-\frac{2}{3}\right)\right)\right]^2-\left[-\frac{2}{3}\right]^2\]
To simplify the other term, we use some trigonometry. Suppose we draw a triangle that satisfies the given sine. That is, we first write
\[\cos\left(\sin^{-1}\left(-\frac{2}{3}\right)\right)=\cos\theta\]
Basically, we make a substitution that let's us say
\[\sin\theta=-\frac{2}{3}\]
Now let's draw the triangle that satisfies this:

There are actually two triangles that satisfy this trig ratio, but we only consider the one in the fourth quadrant. (The reason for this has to do with the domain of the inverse sine function.)