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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?2012-09-16
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    We have just learned some basic identities, which I have already tried using them, without any success.2012-09-16
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    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](http://www.wolframalpha.com/input/?i=2sin%28%CE%B1%E2%88%9245%29sin%282%CE%B1%29%3Dsin%28%CE%B1%2B45%29sin%28%CE%B1%29).**2012-09-16
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    $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.2012-09-16
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    @ParthKohli: That's what I typed as an answer.2012-09-16
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    @Gigili luckily, I typed it a minute before you :)2012-09-16
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    @ParthKohli: Heh, yes! A stitch in time saves nine!2012-09-16

1 Answers 1

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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.2012-09-16
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    Thanks for the answer, I will now try to solve it using your advices.2012-09-16
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    $$(2(\sin\alpha\cos45-\cos\alpha\sin45))(2\sin\alpha\cos\alpha)=(\sin\alpha\cos 45+\cos\alpha\sin 45)\sin\alpha$$ $$\implies 2(\sin\alpha \frac{1}{\sqrt 2} -\cos\alpha \frac{1}{\sqrt 2})(2\sin\alpha\cos\alpha)=(\sin\alpha \frac{1}{\sqrt 2}+\cos\alpha\frac{1}{\sqrt 2})\sin\alpha$$ $$\implies 4(\sin\alpha -\cos\alpha )(\sin\alpha\cos\alpha)=(\sin\alpha +\cos\alpha)\sin\alpha$$2012-09-17
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    I think you divided both sides by $\sin\alpha$, which quantity could be zero.2012-09-17
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    @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.2012-09-17