1
$\begingroup$

$$\frac{\sin(t)}{\cos(\frac t2)} =2\sin(\frac t2)$$

I'm not really sure how to tackle this. I've tried expressing $\cos(\frac t2)$ as $\sin(\frac \pi2 - \frac t2)$ and $\sin(t)$ as $\sin(2\pi + t)$ to get: $$\frac{\sin(2\pi + t)}{\sin(\frac \pi2 - \frac t2)}$$

and then applying the $\sin(A\pm B)$ identity to top and bottom and cancelling down, but that just leads me back to the original expression.

I also noted that $2\sin(\frac t2)$ is sort of similar to the identities $2\sin(A)\cos(B)$ and $2\sin(A)\cos(A)$, so I multiplied top and bottom by $2\cos(t)$ to get: $$\frac{2\sin(t)\cos(t)}{2\cos(t)cos(\frac t2)}$$ and then from there, $$\frac{\sin2A}{\cos(\frac{3t}{2})+\cos(\frac t2)}$$ however, I don't know where to go from there or if that's even how to do it. I'm sure there must be a simple way to solve this identity! I'd appreciate any help.

Jack

$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

Hint: $\sin (2x)=2\sin(x)\cos(x)$

$\endgroup$
2
  • 1
    $\begingroup$ Ohh, I got it! I assumed it had something to do with that identity but your hint made me sure. So, sin(t) = 2 sin(t/2) cos (t/2), from which 2 sin(t/2) obviously equals sin(t) / cos(t/2). Thanks! $\endgroup$
    – Jack
    Aug 20, 2015 at 23:37
  • $\begingroup$ You're welcome! If you want, you can accept the answer so others see that this question has been resolved. $\endgroup$
    – Reveillark
    Aug 20, 2015 at 23:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .