How to express a trigonometic equation in $\sin 2\theta $ and $\cos 2\theta $?

Question

How do I express the given equation in $\sin 2\theta $ and $\cos 2\theta $ in terms of x?

$x + 3 = 7\sin \theta $ with $\frac{\pi }{2}{\text{ < }}\theta {\text{ < }}\pi $

for $\sin 2\theta $ i got $\frac{{( - 2x + 6)\sqrt { - {x^2} - 6x + 40} }}{7}$

for $\cos 2\theta $ i got $\frac{{ - 12x - 49}}{{49}}$

i think im missing something, can someone double check

Answer 1

For $\sin 2\theta $ you forgot to multiply the denominators when you multiply fractions.

$\cos 2\theta $ is completely off. Did you use the identity $\cos 2\theta = 1 - 2{\sin ^2}\theta $ ?

Answer 2

HINT: $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$

$\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$

$\cos^2(\theta) + \sin^2(\theta) = 1$

$\frac{\pi}{2} < \theta < \pi$

$\cos(\theta) < 0$