# 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 $m^2(1-3c^2s^2)=1$.