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I am curious to see whether anybody can give me a proof that takes less steps.

Here is how I did it:

$$\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$$

LHS $$\eqalign{\sin(2\theta + \theta) + \sin\theta &= \sin2\theta\cos\theta + \cos2\theta\sin\theta + \sin\theta\\ &= \sin2\theta\cos\theta + (\cos^2\theta - \sin^2\theta)\sin\theta + \sin\theta\\ &= \sin2\theta\cos\theta + \sin\theta(\cos^2\theta - \sin^2\theta + 1)\\ &= \sin2\theta\cos\theta + \sin\theta(2\cos^2\theta)\\ &= \sin2\theta\cos\theta + 2\sin\theta\cos^2\theta\\ &= \sin2\theta\cos\theta + \cos\theta(\sin\theta\cos\theta + \sin\theta\cos\theta)\\ &= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\ &= \sin2\theta\cos\theta + \cos\theta(\sin2\theta)\\ &= 2\sin2\theta\cos\theta.}$$

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Use the trigonometric identity about $\sin A + \sin B= 2\sin \frac{A+B}{2} \cos\frac{A-B}{2}$.

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Use the sum of angles identities for the $\sin$ function:- $$\sin(3\theta)=\sin(2\theta+\theta)=\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta)$$ $$\sin(\theta)=\sin(2\theta-\theta)=\sin(2\theta)\cos(\theta)-\cos(2\theta)\sin(\theta)$$

Adding both equations results in $$\sin(3\theta)+\sin(\theta)=2\sin(2\theta)\cos(\theta)$$

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  • $\begingroup$ Great, this is the kind of thing I was looking for. Thank you. $\endgroup$
    – Kermit the Hermit
    Jun 17, 2014 at 16:44
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Yes, you can differentiate both sides and check whether they are equal: $$\eqalign{ \dfrac{\mathrm d}{\mathrm d\theta}\big[\sin3\theta+\sin\theta\big]&\overset?=\dfrac{\mathrm d}{\mathrm d\theta}\big[2\sin2\theta\cos\theta\big] \\ 3\cos3\theta+\cos\theta&\overset?= 2\big[\tfrac12( \cos\theta+3\cos3\theta)\big]\tag{$\overset{\rm Chain}{\underset{\sf rule}{}}\overset{\,+}{}\overset{\rm Prod.}{\underset{\sf rule}{}}$}\\ 3\cos3\theta+\cos\theta&\overset?= \cos\theta+3\cos3\theta. \quad\checkmark }$$ So you can conclude that they're the same. ;-)

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  • $\begingroup$ To be precise, from that you deduce that they differ by a constant, but that constant can be simply determined by evaluating at a point. $\endgroup$
    – Dario
    Jun 17, 2014 at 16:07
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    $\begingroup$ @Dario Yes my proof wasn't complete, I leaved it to Midni. $\endgroup$
    – Hakim
    Jun 17, 2014 at 16:09
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    $\begingroup$ I plan on studying calculus after finishing this pre-calculus material which this is part of. So I will thank you after I understand what your solution means haha. I like your blog, you seem quite interesting ;) $\endgroup$
    – Kermit the Hermit
    Jun 17, 2014 at 16:53
  • $\begingroup$ @Midni Haha many thanks for your very kind words! $\overset{\cdot\cdot}\smile$ $\color{white}{\text{even if I didn't update the blog for a long time ^^}}$ $\endgroup$
    – Hakim
    Jun 17, 2014 at 17:09
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$$\sin(3\theta)+\sin(\theta)=\Im[e^{3i\theta}]+\Im[e^{i\theta}]=\Im[e^{3i\theta}+e^{i\theta}]=$$ $$=\Im[e^{2i\theta}(e^{i\theta}+e^{-i\theta})]=2\cos(\theta)\Im[e^{2i\theta}]=2\cos(\theta)\sin(2\theta)\ .$$

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    $\begingroup$ I forgot to mention in my post that it should be done using only elementary trigonometric identities(hence me having no idea what your solution means haha). Thank you for your post nonetheless. $\endgroup$
    – Kermit the Hermit
    Jun 17, 2014 at 16:42
  • $\begingroup$ I suppose this approach is not required at all after the proofs presented above. $\endgroup$
    – aghost
    Jun 17, 2014 at 18:55

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