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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!

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    $\begingroup$ Have you studied the double angle identities? Like $\sin(2\theta),\cos(2\theta),\tan(2\theta)$? $\endgroup$ Commented Jul 9, 2016 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. $\endgroup$ Commented Jul 9, 2016 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: $\sin(2\theta)=2\sin(\theta)\cos(\theta)$. $\endgroup$ Commented Jul 9, 2016 at 4:08
  • $\begingroup$ $\sin(x+y)=\sin x\cos y+\cos x\sin y$ if you replace $x=\theta$ and $y=\theta$, then $\sin(\theta + \theta)=\sin\theta\cos\theta+\cos\theta\sin\theta <=>\sin2\theta=2\sin\theta\cos\theta$ $\endgroup$ Commented Jul 9, 2016 at 5:00

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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 $\sin 2t =2\sin t \cos t$.

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  • $\begingroup$ if you assume that he knows that "$\sin 2t =2\sin t \cos t$" then you could remove the last sentence of your post. $\endgroup$ Commented Jul 9, 2016 at 5:03
  • $\begingroup$ @miracle173: This is a trivial identity. I'm pointing out to OP the main step. $\endgroup$ Commented Jul 9, 2016 at 5:07

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