If $\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$ prove that $\cos\alpha=\frac{2-m^2}{m}$

Question

If $$\frac{\cos(\alpha -3\theta)}{\cos^3 \theta}=\frac{\sin(\alpha -3\theta)}{\sin^3 \theta}=m$$
prove that
$$\cos\alpha=\frac{2-m^2}{m}$$

My approach:

$$\cos^2(\alpha-3\theta)+\sin^2(\alpha-3\theta)=m^2(\sin^6\theta+\cos^6\theta)$$ $$\Rightarrow \frac{1}{m^2}=\sin^6\theta+cos^6\theta=1-\frac{3}{4}\sin^22\theta$$ $$\Rightarrow \sin^22\theta=\frac{4}{3}᛫\frac{m^2-1}{m}$$

I can not proceed further, please help.

Answer 1

$$\sin\alpha\sin3\theta+\cos\alpha\cos3\theta-m\cos^3\theta=0$$

$$\sin\alpha\cos3\theta-\cos\alpha\sin3\theta-m\sin^3\theta=0$$

$$-\dfrac{\sin\alpha}{m(\cos^3\theta\sin3\theta+\cos3\theta\sin^3\theta)}=-\dfrac{\cos\alpha}{m(\cos3\theta\cos^3\theta-\sin3\theta\sin^3\theta)}=-1$$

Using $\sin3\theta,\cos3\theta$ formula

$\dfrac{4\sin\alpha}m=(\cos3\theta+3\cos\theta)\sin3\theta+\cos3\theta(3\sin\theta-\sin3\theta)=3\sin(3\theta+\theta)$

Similarly, $\dfrac{4\cos\alpha}m=(\cos3\theta+3\cos\theta)\cos3\theta-\sin3\theta(3\sin\theta-\sin3\theta)=1-3\cos(3\theta+\theta)$

Use $\cos^24\theta+\sin^24\theta=1$

Answer 2

Hint for your last equation: $$\sin(2\theta)=2sin(\theta)\cos(\theta)$$ so $$\sin(2\theta)^2=4\sin^2(\theta)\cos^2(\theta)$$ and you can eliminate $\theta$ So you get $$4\left(1-\cos(\theta)^2\right)\cos(\theta)^2=\frac{4}{3}\frac{m^2-1}{m}$$ Solve this for $\cos(\theta)$ I know this, when you get $\theta$ then you can compute $\alpha$ with the equations above! Expanding the term $$\cos(\alpha-3\theta)\sin(\theta)^3-\sin(\alpha-3\theta)\cos(\theta)^3$$ we get $$4\, \left( \sin \left( \theta \right) \right) ^{3}\cos \left( \alpha \right) \left( \cos \left( \theta \right) \right) ^{3}-3\, \left( \sin \left( \theta \right) \right) ^{3}\cos \left( \alpha \right) \cos \left( \theta \right) +4\, \left( \sin \left( \theta \right) \right) ^{4}\sin \left( \alpha \right) \left( \cos \left( \theta \right) \right) ^{2}- \left( \sin \left( \theta \right) \right) ^{4 }\sin \left( \alpha \right) -4\, \left( \cos \left( \theta \right) \right) ^{6}\sin \left( \alpha \right) +3\, \left( \cos \left( \theta \right) \right) ^{4}\sin \left( \alpha \right) +4\, \left( \cos \left( \theta \right) \right) ^{5}\cos \left( \alpha \right) \sin \left( \theta \right) - \left( \cos \left( \theta \right) \right) ^{ 3}\cos \left( \alpha \right) \sin \left( \theta \right)=0 $$ This term can we solve for $$\alpha$$ $$\alpha=-\arctan \left( 3\,{\frac {\cos \left( \theta \right) \sin \left( \theta \right) \left( 2\, \left( \sin \left( \theta \right) \right) ^{2}-1 \right) }{6\, \left( \sin \left( \theta \right) \right) ^{4}-6 \, \left( \sin \left( \theta \right) \right) ^{2}+1}} \right) $$ and now you can use your $$\theta$$

Answer 3

An approach using factoring expressions is the following: define $ \quad x := e^{i\alpha}, \quad y := e^{i\theta}, \quad $ and $$ a := \frac{\cos(\alpha -3\theta)}{\cos^3 \theta}, \, b := \frac{\sin(\alpha -3\theta)}{\sin^3 \theta}, \, c := \cos \alpha - \frac{2-a^2}a, \, d := \cos \alpha - \frac{2-b^2}b, \, e := a-b, $$ $$ F_ 1 \!:=\! 3 x^2 - y^4 + x^2 y^4 - 3 y^8, \, F_ 2 \!:=\! 3 x^2 - y^2 + x^2 y^2 - 3 y^4, \, F_ 3 \!:=\! 3 x^2 + y^2 - x^2 y^2 - 3 y^4. $$ Now $\, e = 8 F_ 1 y^2/(x (1-y^4)^3) \,$ and thus if $\, F_ 1=0 \,$ then $\,a=b=m.\,$ Factorization gives $$ c = F_1F_2/(2 x (1+y^2)^3(x^2+y^6) \quad \text{and} \quad d = F_1F_3/(2 x (1-y^2)^3(x^2-y^6))$$ so we have $\, c=d=0\,$ and we are done.