I'm trying to prove this trigonometry identity, and I can solve it to the up until the last step, where I can't figure out which identity is being used to solve it.

This is the identity.

$$\tan\theta \cdot \cos^2\theta = \frac 12 \sin2\theta $$

Now I've solved the equation up to getting $$\sin\theta \cdot \cos\theta$$

Can someone help me out with how to get from that to

$$\frac 12 \sin2\theta $$

Thank you!

  • 1
    $\begingroup$ Have you studied the double angle identities? Like $\sin(2\theta),\cos(2\theta),\tan(2\theta)$? $\endgroup$ – John Wayland Bales Jul 9 '16 at 4:03
  • $\begingroup$ Actually, I've been trying those, but I can't seem to get to the last step with them, I don't know what I've been missing. $\endgroup$ – Chet Spalsky Jul 9 '16 at 4:06
  • $\begingroup$ As @JohnWaylandBales has pointed out, what you need to proceed is the double angle identity for $\sin(2\theta)$. In fact, the identity says: $\sin(2\theta)=2\sin(\theta)\cos(\theta)$. $\endgroup$ – Karthik Jul 9 '16 at 4:08
  • $\begingroup$ $\sin(x+y)=\sin x\cos y+\cos x\sin y$ if you replace $x=\theta$ and $y=\theta$, then $\sin(\theta + \theta)=\sin\theta\cos\theta+\cos\theta\sin\theta <=>\sin2\theta=2\sin\theta\cos\theta$ $\endgroup$ – julio godoy Jul 9 '16 at 5:00

For all $t$ not an odd multiple of $\pi/2$, one has $$\tan t \cos^2 t = \sin t \cos t = \frac12(2\sin t \cos t)=\frac12(\sin 2t)$$

This assumes you know that $\sin 2t =2\sin t \cos t$.

  • $\begingroup$ if you assume that he knows that "$\sin 2t =2\sin t \cos t$" then you could remove the last sentence of your post. $\endgroup$ – miracle173 Jul 9 '16 at 5:03
  • $\begingroup$ @miracle173: This is a trivial identity. I'm pointing out to OP the main step. $\endgroup$ – MPW Jul 9 '16 at 5:07

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.