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Verify the following trig identity: $$\sin(3\theta)-\sin\theta = 2\cos(2\theta)\sin\theta$$

Here is my work so far.

$\sin(3\theta)-\sin\theta = 2\cos(2\theta)\sin\theta$

LHS:$$\sin(\theta+2\theta)-\sin\theta$$ $$\sin\theta \cos(2\theta)+\sin(2\theta)\cos\theta-\sin\theta$$ $$\sin\theta \cos(2\theta)+(2\sin\theta \cos\theta)\cos\theta-\sin\theta$$

Where do I go from here? I think I should leave the first term in the line above as is, and try and manipulate the second two terms to equal $\sin\theta \cos(2\theta)$, then the LHS will add together to equal $2\cos(2\theta)\sin\theta$, and the identity will be verified. How do you suggest I get there?

Any hints or advice would be appreciated.

EDIT:

I ended up verifying this identity using the identity frank000 mentioned in comments. Thanks to everyone for the input, it was all very helpful.

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    $\begingroup$ Do you know how to prove $\sin(x)-\sin(y)=2\cos(\frac{x+y}{2})\sin(\frac{x-y}{2})$ in general? $\endgroup$
    – user175968
    Oct 12, 2015 at 2:03
  • $\begingroup$ Just for this question $sin(3x)=3sin(x)-4sin^3(x)$, $cos(2x)=1-2sin^2(x)$ $\endgroup$
    – user175968
    Oct 12, 2015 at 2:04
  • $\begingroup$ @frank000 I'm familiar with that identity, but don't exactly know how to prove it... $\endgroup$
    – McB
    Oct 12, 2015 at 2:04
  • $\begingroup$ $x=3\theta$, $y=\theta$ $\endgroup$
    – user175968
    Oct 12, 2015 at 2:05
  • $\begingroup$ You could also use complex numbers for a very short and easy prove that both sides are identical. $\endgroup$ Oct 12, 2015 at 4:07

3 Answers 3

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You're almost done! In the second and third term, factor the $\sin \theta$ and you'll get $2 \cos^2 \theta -1$ which you might recognize as $\cos 2\theta$.

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hint: also use $\sin(\theta) = \sin(2\theta - \theta) = \sin(2\theta)\cos(\theta)-\sin(\theta)\cos(2\theta)$

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$\sin(3θ)−\sinθ=2\cos(2θ)\sinθ$

$\sinθ\cos(2θ)+\sin(2θ)\cosθ−sinθ=2(\cos^2θ-\sin^2θ)\sinθ$

$\sinθ(\cos^2θ-\sin^2θ)+(2\sinθ\cosθ)\cosθ−\sinθ=2(\cos^2θ-\sin^2θ)\sinθ$

$\cos^2θ-\sin^2θ−1=-2\sin^2θ$

$-\sin^2θ-\sin^2θ=-2\sin^2θ$

$-2\sin^2θ=-2\sin^2θ$

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  • $\begingroup$ Fourth and subsequen lines should be augmented by "or $\sin \theta = 0$" since division by zero is undefined. $\endgroup$ Oct 12, 2015 at 2:54

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