# Trig iden: Quadruple angle formulas

• Jul 23rd 2007, 06:51 PM
andrewsx
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!
• Jul 23rd 2007, 07:18 PM
tukeywilliams
Is it:

1. $\displaystyle \frac{2 \cos^{2} 4x-1}{\sin 4x \cos 4x}$ and

2. $\displaystyle 1 - 4 \sin^{2}x + 4 \sin^{4} x$?
• Jul 23rd 2007, 08:00 PM
Jhevon
We use the addition formulas for sine and cosine here.

Recall:

$\displaystyle \cos 2x = 2 \cos^2 x - 1 = 1 - 2 \sin^2 x$

$\displaystyle \Rightarrow \cos 8x = 2 \cos^2 4x - 1$

$\displaystyle \sin 2x = 2 \sin x \cos x$

$\displaystyle \Rightarrow \sin 8x = 2 \sin 4x \cos 4x$

Quote:

Originally Posted by andrewsx
How do you tackle these?
1. 2cos^2 4x-1
sin4xcos4x

$\displaystyle \frac {2 \cos^2 4x - 1}{\sin 4x \cos 4x}$

$\displaystyle = \frac {\cos 8x}{ \frac {1}{2} \sin 8x}$

$\displaystyle = 2 \frac {\cos 8x }{\sin 8x}$

$\displaystyle = 2 \cot 8x$

Quote:

2. 1-4sin^2x+4sin^4x
$\displaystyle 1 - 4 \sin^2 x + 4 \sin^4 x$ ............Note that this is quadratic in $\displaystyle \sin^2 x$, so let's foil

$\displaystyle = \left( 1 - 2 \sin^2 x \right) \left(1 - 2 \sin^2 x \right)$

$\displaystyle = \left( 1 - 2 \sin^2 x \right)^2$

$\displaystyle = \cos^2 2x$