Trigonometry pre calculus level good question [duplicate]

Here is a trigonometry problem.

Given $$\frac{\cos(\alpha-3\theta)}{\cos^3(\theta)}=\frac{\sin(\alpha-3\theta)}{\sin^3(\theta)} = m$$ Show that $$m^2+m\cos(\alpha) = 2.$$

I tried to convert $\sin^3(x)$ into $\sin(3x)$ and similarly to cosine term, but couldn't get the answer. Please tell how to proceed further and any other way to solve it.

marked as duplicate by Claude Leibovici, Martin Sleziak, hardmath, Simply Beautiful Art, Siong Thye GohAug 26 '17 at 1:12

• Its a another question actually – Kamal Aug 25 '17 at 4:31
• @ChaseRyanTaylor It is $m^2+m\cos(\alpha)$ instead of $\theta$ – Isaac Browne Aug 25 '17 at 4:32
• Ah, I see, that tiny detail makes all the difference. – Chase Ryan Taylor Aug 25 '17 at 4:33
• – lab bhattacharjee Aug 25 '17 at 9:09

From the equation we have following:

$m \sin^3\theta = \sin\alpha \cos3\theta-\cos\alpha \sin3\theta \cdots (1)$

$m \cos^3\theta = \cos\alpha \cos3\theta+\sin\alpha \sin3\theta \cdots (2)$

$\cos3\theta \times (2)-\sin3\theta \times (1) \rightarrow m(\cos^3\theta\cos3\theta-\sin^3\theta\sin3\theta)=\cos\alpha$

$\cos3\theta \times (1) + \sin3\theta \times (2) \rightarrow m(\cos^3\theta\sin3\theta+\sin^3\theta\cos3\theta)=\sin\alpha$

Using $\cos3\theta=4\cos^3\theta-3\cos\theta$ and $\sin3\theta=3\sin\theta-4\sin^3\theta$, we have (from here I will write $\cos\theta$ and $\sin\theta$ as $c$ and $s$, respectively)

$m(4(c^6+s^6)-3(c^4+s^4))=\cos\alpha \cdots (3)$

$m(3cs(c^2-s^2))=\sin\alpha \cdots (4)$

Since $4(c^6+s^6)-3(c^4+s^4)=1-6c^2s^2$, it is enough to prove $m^2+m\times m(1-6c^2s^2)=2$, or $m^2(1-3c^2s^2)=1$.

Now $(3)^2+(4)^2$ leads to $m^2(1-6c^2s^2)^2+m^2(9c^2s^2(c^2-s^2)^2)=1$. Since $(c^2-s^2)^2=1-4c^2s^2$, we have $m^2(1-12c^2s^2+36c^4s^4+9c^2s^2(1-4c^2s^2))=1$, or $m^2(1-3c^2s^2)=1$.