# What identity was used in this Trigonometry problem?

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!

• Have you studied the double angle identities? Like $\sin(2\theta),\cos(2\theta),\tan(2\theta)$? – John Wayland Bales Jul 9 '16 at 4:03
• 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. – Chet Spalsky Jul 9 '16 at 4:06
• As @JohnWaylandBales has pointed out, what you need to proceed is the double angle identity for $\sin(2\theta)$. In fact, the identity says: $\sin(2\theta)=2\sin(\theta)\cos(\theta)$. – Karthik Jul 9 '16 at 4:08
• $\sin(x+y)=\sin x\cos y+\cos x\sin y$ if you replace $x=\theta$ and $y=\theta$, then $\sin(\theta + \theta)=\sin\theta\cos\theta+\cos\theta\sin\theta <=>\sin2\theta=2\sin\theta\cos\theta$ – 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 $\sin 2t =2\sin t \cos t$.
• if you assume that he knows that "$\sin 2t =2\sin t \cos t$" then you could remove the last sentence of your post. – miracle173 Jul 9 '16 at 5:03