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.


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

  • 2
    $\begingroup$ 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 '15 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 '15 at 2:04
  • $\begingroup$ @frank000 I'm familiar with that identity, but don't exactly know how to prove it... $\endgroup$ – McB Oct 12 '15 at 2:04
  • $\begingroup$ $x=3\theta$, $y=\theta$ $\endgroup$ – user175968 Oct 12 '15 at 2:05
  • $\begingroup$ You could also use complex numbers for a very short and easy prove that both sides are identical. $\endgroup$ – Stefan Gruenwald Oct 12 '15 at 4:07

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


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 $\cos 2\theta$.








  • $\begingroup$ Fourth and subsequen lines should be augmented by "or $\sin \theta = 0$" since division by zero is undefined. $\endgroup$ – Eric Towers Oct 12 '15 at 2:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.