# Finding $\alpha$ for $\sin(4 \alpha + \frac{\pi}{6}) = \sin (2\alpha + \frac{\pi}{5})$

I try to find $\alpha$ for $\sin(4 \alpha + \frac{\pi}{6}) = \sin (2\alpha + \frac{\pi}{5})$.

Left side:

$$\sin(4 \alpha + \frac{\pi}{6}) =$$ $$= \sin4\alpha \times \cos \frac{\pi}{6} + \cos 4\alpha \times \sin\frac{\pi}{6} =$$

$$= \sin4\alpha \times \frac{\sqrt{3}}{2} + \cos 4\alpha \times \frac{1}{2} =$$

$$= 2\sin2\alpha \times \cos2\alpha \times \frac{\sqrt{3}}{2} + (\cos^2 2\alpha - \sin^2 2\alpha) \times \frac{1}{2} =$$

$$= 2\sin\alpha \times \cos \alpha \times \cos2\alpha \times \sqrt{3} + \frac{\cos^2 2\alpha - \sin^2 2\alpha}{2} =$$

$$= 2\sin\alpha \times \cos \alpha \times (\cos^2\alpha - \sin^2\alpha) \times \sqrt{3} + \frac{(\cos^2 \alpha - \sin^2\alpha)(\cos^2 \alpha - \sin^2\alpha) - \sin^2 2\alpha}{2} =$$

$$= 2\sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{(\cos^2 \alpha - \sin^2\alpha)(\cos^2 \alpha - \sin^2\alpha) - (2\sin\alpha \times \cos\alpha)(2\sin\alpha \times \cos\alpha)}{2} =$$

$$= 2\sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{(\cos^2 \alpha - \sin^2\alpha)(\cos^2 \alpha - \sin^2\alpha)}{2} - 2\sin^2\alpha \times \cos^2\alpha =$$

$$= 2 \times \sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{\cos^4 \alpha - (\sin^2\alpha \times \cos^2\alpha) - (\sin^2\alpha \times \cos^2\alpha) + \sin^4 \alpha}{2} - 2\sin^2\alpha \times \cos^2\alpha =$$

$$= 2 \sqrt{3}\sin\alpha \cos \alpha (\cos^2\alpha - \sin^2\alpha) + \frac{(\cos^2\alpha - \sin^2\alpha)^2}{2} - 2\sin^2\alpha \times \cos^2\alpha =$$

$$= 2\sqrt{3}(\sin\alpha \times \cos^3 \alpha - \sin^3\alpha \times \cos\alpha) + \frac{(1 - 2\sin^2\alpha)^2}{2} - 2\sin^2\alpha \times \cos^2\alpha =$$

$$= 2\sqrt{3}(\sin\alpha \times \cos^3 \alpha - \sin^3\alpha \times \cos\alpha) + \frac{1}{2}-2\sin^2\alpha+2\sin^4\alpha - 2\sin^2\alpha \times \cos^2\alpha =$$

$$= 2\sin\alpha(\sqrt{3}\times \cos^3 \alpha - \sqrt{3}\sin^2\alpha \times \cos\alpha + \frac{1}{4\sin\alpha}-\sin\alpha + \sin^3\alpha - \sin\alpha \times \cos^2\alpha) =$$

Right side:

$$\sin(2\alpha + \frac{\pi}{5}) = \sin2x \times \cos\frac{\pi}{5} + \cos2\alpha \times \sin\frac{\pi}{5} =$$

$$= 2\sin\alpha \cos\alpha \times \cos\frac{\pi}{5} + (\cos^2\alpha - \sin^2\alpha)\sin\frac{\pi}{5} =$$

$$= 2\sin\alpha \cos\alpha \times \cos\frac{\pi}{5} + (1 - 2\sin^2\alpha)\sin\frac{\pi}{5} =$$

$$= 2\sin\alpha \cos\alpha \times \cos\frac{\pi}{5} + \sin\frac{\pi}{5} - 2\sin^2\alpha \times \sin\frac{\pi}{5} =$$

$$= 2\sin\alpha(\cos\alpha \times \cos\frac{\pi}{5} + \frac{\sin\frac{\pi}{5}}{2\sin\alpha} - \sin\alpha \times \sin\frac{\pi}{5})$$

I can't get any further on either. Bringing them together I get:

$$2\sin\alpha(\sqrt{3}\times \cos^3 \alpha - \sqrt{3}\sin^2\alpha \times \cos\alpha + \frac{1}{4\sin\alpha}-\sin\alpha + \sin^3\alpha - \sin\alpha \times \cos^2\alpha) = 2\sin\alpha(\cos\alpha \times \cos\frac{\pi}{5} + \frac{\sin\frac{\pi}{5}}{2\sin\alpha} - \sin\alpha \times \sin\frac{\pi}{5})$$

$$\iff \sqrt{3}\times \cos^3 \alpha - \sqrt{3}\sin^2\alpha \times \cos\alpha + \frac{1}{4\sin\alpha}-\sin\alpha + \sin^3\alpha - \sin\alpha \times \cos^2\alpha = \cos\alpha \times \cos\frac{\pi}{5} + \frac{\sin\frac{\pi}{5}}{2\sin\alpha} - \sin\alpha \times \sin\frac{\pi}{5}$$

Where do I go from here? Om am I already dead wrong?

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You really made your life hard for yourself by pursuing this method so far. Considering the graph symmetries of the sine graph shows that $\sin \theta = \sin \phi$ if and only if either $\theta = \phi + 2n\pi$ or $\theta = -\phi + (2n+1)\pi$ for some $n \in \mathbb{Z}$ $-$ do this and your life will be made much easier! Impressive effort, though, I'll give you that. – Clive Newstead Sep 13 '12 at 15:37

$\sin(4\alpha+\frac{\pi}{6})=\sin(2\alpha+\frac{\pi}{5})$

$\sin(4\alpha+\frac{\pi}{6})-\sin(2\alpha+\frac{\pi}{5}) = 0$

Consider the formula: $\sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$,

and $A\cdot B=0\Leftrightarrow A=0\vee B=0$

$2\cos\frac{6\alpha+\frac{11\pi}{30}}{2}\sin\frac{2\alpha-\frac{\pi}{30}}{2}=0$

$\cos\frac{6\alpha+\frac{11\pi}{30}}{2}\sin\frac{2\alpha-\frac{\pi}{30}}{2}=0$

$\cos\frac{6\alpha+\frac{11\pi}{30}}{2}=0\vee\sin\frac{2\alpha-\frac{\pi}{30}}{2}=0$

$\frac{6\alpha+\frac{11\pi}{30}}{2}=m\Rightarrow\cos m=0\Rightarrow m=\frac{\pi}{2}+k\pi$

Hence we have:

$\frac{6\alpha+\frac{11\pi}{30}}{2}=\frac{\pi}{2}+k\pi\Rightarrow\alpha=\frac{k\pi}{3}+\frac{19\pi}{180}$, $k\in Z$

$\frac{2\alpha-\frac{\pi}{30}}{2}=n\Rightarrow \sin n=0\Rightarrow n=k\pi$

Hence we have:

$\frac{2\alpha-\frac{\pi}{30}}{2}=k\pi\Rightarrow\alpha=k\pi+3$, $k\in Z$

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I don't quite get your final line. When calculating $\frac{2\alpha + \frac{\pi}{30}}{2}$ I get $\alpha = n\pi + \frac{\pi}{60}$. Am I missing a step or something? – Quispiam Sep 13 '12 at 23:38

Hint:

if $\sin A=\sin B$,then we have $A=k\pi +(-1)^kB$,it is easy to prove.

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