Question

How do you solve sin(2arccos7/13)

Answer 1

First, recall the double angle identity for sine:
\[\sin\left(2\arccos\frac{7}{13}\right)=2\sin\left(\arccos\frac{7}{13}\right)\cos\left(\arccos\frac{7}{13}\right)\]
The strategy is to regard the arccos part as an angle itself:
\[\theta=\arccos\frac{7}{13}~~\iff~~\cos\theta=\frac{7}{13}\]
This gives you the following type of triangle:

Find the missing side, then you'll have everything you need to find the sine and cosine of \(\theta\).