Trigonometric Identities: Given that $\frac{2\cos(a)+3\sin(a)}{3\cos(a)-2\sin(a)}=-2$ find $\sin(2a)$

Question

Given: $\dfrac{2\cos(a)+3\sin(a)}{3\cos(a)-2\sin(a)}=-2$

$\sin(2a)$ = ?

So what I did first was expand $\sin(2a)$ to $2\sin(a)\cos(a)$

Then I went back to the given and simplified that, and got that

$\sin(a)=8\cos(a)$

Then I plugged that into the first expanded expression I got, and got

$16\cos^2(a)$

Somehow this is wrong? What am I doing wrong.

Answer 1

HINT

We have that

$$\frac{2\cos(a)+3\sin(a)}{3\cos(a)-2\sin(a)}=-2 \iff \frac{2+3\tan a}{3-2\tan a}=-2 \implies \tan a=?$$

then use that by Tangent half-angle substitution

$$\sin 2a = \frac{2\tan a}{1+\tan^2 a}$$

Answer 2

If I'm not wrong, then you should just do this after finding out, that $\sin(a) = 8\cos(a)$

$\tan(a) = 8$

$1 + \tan^2(a) = 1/\cos^2(a)$, so $\sin(2a) = 2\sin(a)\cos(a)=2\tan(a)\cos^2(a) = \dfrac{2\tan(a)}{1 + \tan^2(a)}$

Answer 3

you have $$\sin(a)=8\cos(a)$$ and we know that $$\sin ^ 2 ( \alpha) + \cos ^2 (\alpha ) =1$$

That gives you $$ \cos^2 (\alpha )=1/{65} $$ and $$ \sin^2 (\alpha )=64/{65} $$

Thus$$\sin(\alpha) \cos (\alpha) = \frac {8}{65}$$

That makes $$\sin ( 2 \alpha) = \frac {16}{65}$$