• Mar 6th 2012, 05:11 AM
alyosha2
The 'a' is supposed to be theta. I'm not sure why it's comming out as 'a'.

$\frac{2sin^2(\frac{1}{2}\theta)}{\theta} = (\frac{2sin^2(\frac{1}{2}\theta)}{\frac{1}{2}\thet a})^2 * (\frac{1}{2}\theta)$

I can't see how these equate.
• Mar 6th 2012, 03:44 PM
skeeter
Quote:

Originally Posted by alyosha2
The 'a' is supposed to be theta. I'm not sure why it's comming out as 'a'.

$\frac{2sin^2(\frac{1}{2}\theta)}{\theta} = (\frac{2sin^2(\frac{1}{2}\theta)}{\frac{1}{2}\thet a})^2 * (\frac{1}{2}\theta)$

I can't see how these equate.

it's not an identity, but does have solutions ...

let $u = \frac{\theta}{2}$

$\frac{\sin^2{u}}{u} = \frac{4\sin^4{u}}{u}$

... note that $u \ne 0 \implies \theta \ne 0$

$4\sin^4{u} - \sin^2{u} = 0$

$\sin^2{u}(4\sin^2{u} - 1) = 0$

$\sin^2{u} = 0$ ... $u = k\pi ; k \ne 0 \implies \theta = 2k \pi ; k \ne 0$

$\sin{u} = \frac{1}{2}$ ... $u = \frac{\pi}{6} \pm 2k\pi ; \frac{5\pi}{6} \pm 2k\pi \implies \theta = \frac{\pi}{3} \pm 2k\pi ; \frac{5\pi}{3} \pm 2k\pi$