Question

Evaluate sin[2 arccos(-8/17 )].

Answer 1

\[\cos^{-1}(\frac{-8}{7})\approx2.06\]

Can you solve it now?

Answer 2

Remember the mnemonic SOH-CAH-TOA
\[\sin\theta=\frac{opposite}{hypotenuse}\]
\[\cos\theta=\frac{adjacent}{hypotenuse}\]

Answer 3

Let x = arccos(-8/17)
sin(2x)=2sinxcosx

Answer 4

\[2sinxcosx=2(\frac{15}{17})(\frac{-8}{17})=\frac{-240}{289}=\sin2(\arccos\frac{-8}{17})\]