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I have the following trigonometric equation:

$$2\sin(\alpha - 45)\sin(2\alpha) = \sin(\alpha + 45)\sin(\alpha)$$

Is it possible to find $\alpha$?

Please also include each step in your solution.

EDIT: Sorry if I haven't mentioned- yes, it is a solution I reached to as a part of an assignment I was given (school). All I wish to know if I can pull $\alpha$ from what I found.

Thanks in advance.

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Is this a homework question? What have you tried so far? – Clive Newstead Sep 16 '12 at 14:51
We have just learned some basic identities, which I have already tried using them, without any success. – Novak Sep 16 '12 at 15:00
Hint: you will require the following identities.$$ \sin(\alpha + \beta) = \sin\alpha \cos \beta + \sin \beta \cos \alpha \cdots(1)$$ $$ \sin(2\alpha) = 2\sin\alpha\cos a\cdots(2) $$ $$\sin(\alpha - \beta) = \sin\alpha \cos\beta - \sin\beta\cos\alpha\cdots(3)$$Note that there are a lot of solutions for this equation, so these identities will just help you to simplify, since the solutions cannot be found without technology. – Parth Kohli Sep 16 '12 at 15:21
$2\sin(\alpha-45^{\circ})2\sin \alpha \cos \alpha=\sin(\alpha+45^{\circ})\sin \alpha $ $\implies \sin \alpha(4\sin(\alpha-45^{\circ})\cos \alpha-\sin(\alpha+45^{\circ}))=0$ If $ \sin \alpha=0, \alpha=n\pi$ where $n\pi$ any integer. – lab bhattacharjee Sep 16 '12 at 15:34
@ParthKohli: That's what I typed as an answer. – Gigili Sep 16 '12 at 15:35
up vote 2 down vote accepted

The formulas you need to use:

  • $\sin(x-y)=\sin x\cos y-\cos x\sin y$
  • $\sin 2x=2\sin x\cos x$
  • $\sin(x+y)=\sin x\cos y+\cos x\sin y$

$$2\sin(\alpha - 45)\sin(2\alpha) = \sin(\alpha + 45)\sin(\alpha)$$ $$(2(\sin\alpha\cos45-\cos\alpha\sin45))(2\sin\alpha\cos\alpha)=(\sin\alpha\cos 45+\cos\alpha\sin 45)\sin\alpha$$ Here you can cancel out $\sin\alpha$ from both sides of the equation but you'll need to point out we assumed $\sin\alpha \neq 0$. $$4\sin\alpha\cos\alpha(\sin\alpha-\cos45))=(\sin\alpha\cos 45+\cos\alpha\sin 45)\sin\alpha$$ $$4(\sin\alpha-\cos\alpha)(\cos\alpha)=(\sin\alpha+\cos\alpha)$$

$$\sin2\alpha-\cos2\alpha-1=\frac 12(\sin\alpha+\cos\alpha)$$ $$2\cos 2\alpha-2\sin2\alpha+\cos\alpha+\cos\alpha+2=0$$ And here is the solutions. It doesn't seem possible to further simplify it.

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Well, I wanted to make the complete solution invisible (>!) but it doesn't work. – Gigili Sep 16 '12 at 15:34
Thanks for the answer, I will now try to solve it using your advices. – Novak Sep 16 '12 at 20:14
$$(2(\sin\alpha\cos45-\cos\alpha\sin45))(2\sin\alpha\cos\alpha)=(\sin\alpha\cos 45+\cos\alpha\sin 45)\sin\alpha$$ $$\implies 2(\sin\alpha \frac{1}{\sqrt 2} -\cos\alpha \frac{1}{\sqrt 2})(2\sin\alpha\cos\alpha)=(\sin\alpha \frac{1}{\sqrt 2}+\cos\alpha\frac{1}{\sqrt 2})\sin\alpha$$ $$\implies 4(\sin\alpha -\cos\alpha )(\sin\alpha\cos\alpha)=(\sin\alpha +\cos\alpha)\sin\alpha$$ – lab bhattacharjee Sep 17 '12 at 4:26
I think you divided both sides by $\sin\alpha$, which quantity could be zero. – Gerry Myerson Sep 17 '12 at 6:35
@GerryMyerson, yes, $\sin\alpha=0$ is a solution, which I've commented in the problem itself. I think, we need to prove/disprove the existence of the other solutions. – lab bhattacharjee Sep 17 '12 at 14:10

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