Prove the identity $\displaystyle 1 - cos\;2\;\theta = \frac{2}{1 + cot^2\;\theta}$
I'm a little lost where to begin on this one. Does anyone have any tips for proving these identities?
R.H.S.$\displaystyle = \frac{2}{1 + \cot^2 \theta} = \frac{2}{1 + \frac{\cos^2 \theta}{\sin^2 \theta}} = \frac{2 \sin^2 \theta}{\sin^2 \theta + \cos^2 \theta} = 2 \sin^2 \theta$
$\displaystyle = -(1 - 2 \sin^2 \theta) + 1 = -\cos(2 \theta) + 1 =$L.H.S.
I knew to do the last couple of bits because L.H.S.$\displaystyle =1 - \cos (2 \theta) = 1 - (\cos^2 \theta - \sin^2 \theta) = 1 - \cos^2 \theta + \sin^2 \theta = 2 \sin^2 \theta$.
I did $\displaystyle 1 - cos\;2\theta = \frac{2}{csc^2\;\theta}$
$\displaystyle 1 - cos\;2\theta = 2sin^2\;\theta$
$\displaystyle 1 - cos\;2\theta = 2(1 - cos^2\;\theta)$
$\displaystyle 1 - cos\;2\theta = 2 - 2cos^2\;\theta$
$\displaystyle 1 - (1 - 2sin^2\;\theta) = 2 - 2(1 - sin^2\;\theta)$
$\displaystyle 2sin^2\;\theta = 2 - 2 + 2sin^2\;\theta$
$\displaystyle 2sin^2\;\theta = 2sin^2\;\theta$
Hows it look compared to yours?
Well, that's a much better way.
I'd make just a couple of small changes so that the solution agrees with the standard setting out of an identity proof:
R.H.S.$\displaystyle = \frac{2}{csc^2\;\theta}$
$\displaystyle = 2 \sin^2\;\theta$
$\displaystyle = 2(1 - \cos^2\;\theta)$
$\displaystyle = 2 - 2 \cos^2\;\theta$ And here's where we reach the fork in the road because you need to keep working with the R.H.S.
$\displaystyle = 1 + (1 - 2 \cos^2\;\theta)$
$\displaystyle = 1 - (2 \cos^2\;\theta - 1)$
$\displaystyle = 1 - \cos (2 \theta )$
= L.H.S.
Do you see the style required?
Well, I guess you can ..... but it's not good form in my opinion. Usually if you want to do that, you can instead use it to see how the LHS (or RHS) needs to continue.
In my opinion, trig identities should follow the format:
This side (LHS or RHS as required) = whatever it is
= blah blah
= yada yada
= ho hum ho hum
.
.
.
= the other side (RHS or LHS as required).