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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 Goh Aug 26 '17 at 1:12

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


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