Please close this thread, solved all problems...
Please use parenthesis!!
Is there a typo? -1 - 1 = -2, not -1. And if you can't factor a 2 out of everything, then you can't cancel the 2, so the last line here is wrong. The same goes for sin(x).
$\displaystyle \frac{sin(2x)}{1 - cos(2x)} = cot(x)$
$\displaystyle \frac{2sin(x)cos(x)}{1 - cos(2x)} = \frac{cos(x)}{sin(x)}$
$\displaystyle \frac{2sin(x)cos(x)}{1 - (1 - 2sin^2(x)} = \frac{cos(x)}{sin(x)}$
$\displaystyle \frac{2sin(x)cos(x)}{1 - 1 + 2sin^2(x)} = \frac{cos(x)}{sin(x)}$
$\displaystyle \frac{2sin(x)cos(x)}{2sin^2(x)} = \frac{cos(x)}{sin(x)}$
$\displaystyle \frac{cos(x)}{sin(x)} = \frac{cos(x)}{sin(x)}$
You had the right idea, but you didn't have the details right.
-Dan
Hello, Macleef!
We'll use the identity: .$\displaystyle \cos^2\!\theta \:=\:\frac{1 + \cos2\theta}{2}$ . . . twice.4) Prove: .$\displaystyle 8\cos^4\!x \:= \:\cos4x + 4\cos2x + 3$
We have: .$\displaystyle 8\left(\cos^2\!x\right)^2 \;=\;8\left(\frac{1 + \cos2x}{2}\right)^2 \;=\;\;8\left(\frac{1 + 2\cos2x + \cos^2\!2x}{4}\right) $
. . $\displaystyle = \;\;2\left(1 + 2\cos2x + \cos^2\!2x\right) \;\;=\;\;2\left(1 + 2\cos2x + \frac{1 + \cos4x}{2}\right)$
. . $\displaystyle = \;\;2 + 4\cos2x + 1 + \cos4x \;\;=\;\;{\color{blue}\cos4x + 4\cos2x + 3}$
There are two ways to get at this identity. I'll use the one you will have been likely to see:
$\displaystyle cos(2 \theta) = 2cos^2(\theta) - 1$
$\displaystyle 2cos^2(\theta) = cos(2\theta) + 1$
$\displaystyle cos^2(\theta) = \frac{cos(2\theta) + 1}{2}$
(The quick way to do this is using a half-angle identity.)
-Dan