prove that $\displaystyle cos4x = 8cos^{4}x - 8cos^{2}x - 1 $

I have tried for quite a while, but am completely stuck now.

Can anyone have ago?

Thanks.

Printable View

- Jan 12th 2010, 12:30 PMTweetyprov trig indentity
prove that $\displaystyle cos4x = 8cos^{4}x - 8cos^{2}x - 1 $

I have tried for quite a while, but am completely stuck now.

Can anyone have ago?

Thanks. - Jan 12th 2010, 12:50 PMpickslides
You should use the identity

$\displaystyle \cos(2u)= 2\cos^2(u)-1$

Now make $\displaystyle u = 2x$

$\displaystyle \cos(2\times 2x)= 2\cos^2(2x)-1$

$\displaystyle \cos(4x)= 2\cos^2(2x)-1$

Now use it again

$\displaystyle \cos(4x)= 2(2\cos^2(x)-1)^2-1$

Now expand the RHS, you should have your answer. - Jan 12th 2010, 12:59 PMArchie Meade
Hi Tweety,

first factorise..

$\displaystyle 8Cos^{2}x(Cos^{2}x-1)-1=8Cos^{2}x(-Sin^{2}x)-1$

$\displaystyle =-8Cos^{2}xSin^{2}x-1=-8\frac{1}{2}(1+cos2x)\frac{1}{2}(1-cos2x)-1$

$\displaystyle =-2(1-Cos^{2}2x)-1=-2[1-\frac{1}{2}(1+Cos4x)]-1$

$\displaystyle =-2+1+Cos4x-1$

I guess you have a typo,

i reckon your original expression should have +1 instead of -1. - Jan 12th 2010, 01:11 PMTweety
- Jan 12th 2010, 01:17 PMArchie Meade
$\displaystyle Cos2x=2Cos^{2}x-1$

$\displaystyle Cos^{2}2x=(Cos2x)^2$ - Jan 12th 2010, 01:25 PMpickslides
- Jan 12th 2010, 01:57 PMTweety
- Jan 12th 2010, 02:21 PMArchie Meade
The method that pickslides used was very neat, Tweety.

He used the same identity at each step.

You may see the identity written as

$\displaystyle cos^{2}A=\frac{1}{2}(1+cos2A).$

Rearranging this, we get

$\displaystyle 2cos^{2}A=1+cos2A$

so

$\displaystyle cos2A=2cos^{2}A-1$

He simply used this, by letting A=2x to get cos4x

and the powers of x follow.

I started with the powers and worked back, eventually using that identity,

but i also used

$\displaystyle cos^{2}x+sin^{2}x=1$

in writing

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

The other identity i used is

$\displaystyle sin^2x=\frac{1}{2}(1-cos2x)$