How do you tackle these?
1. 2cos^2 4x-1
sin4xcos4x
2. 1-4sin^2x+4sin^4x
The answers to the first is 2cot8x and the second is cos^22x. I would like to see the step by step process.
Help would be appreciated thanks!
We use the addition formulas for sine and cosine here.
Recall:
$\displaystyle \cos 2x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x$
$\displaystyle \Rightarrow \cos 8x = 2 \cos^2 4x - 1$
$\displaystyle \sin 2x = 2 \sin x \cos x$
$\displaystyle \Rightarrow \sin 8x = 2 \sin 4x \cos 4x$
$\displaystyle \frac {2 \cos^2 4x - 1}{\sin 4x \cos 4x} $
$\displaystyle = \frac {\cos 8x}{ \frac {1}{2} \sin 8x}$
$\displaystyle = 2 \frac {\cos 8x }{\sin 8x}$
$\displaystyle = 2 \cot 8x$
$\displaystyle 1 - 4 \sin^2 x + 4 \sin^4 x$ ............Note that this is quadratic in $\displaystyle \sin^2 x$, so let's foil2. 1-4sin^2x+4sin^4x
$\displaystyle = \left( 1 - 2 \sin^2 x \right) \left(1 - 2 \sin^2 x \right)$
$\displaystyle = \left( 1 - 2 \sin^2 x \right)^2$
$\displaystyle = \cos^2 2x$