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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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    @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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    @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