1. ## prov trig indentity

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

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

Can anyone have ago?

Thanks.

2. Originally Posted by Tweety
prove that $cos4x = 8cos^{4}x - 8cos^{2}x - 1$

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

Can anyone have ago?

Thanks.
You should use the identity

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

Now make $u = 2x$

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

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

Now use it again

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

3. Hi Tweety,

first factorise..

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

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

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

$=-2+1+Cos4x-1$

I guess you have a typo,

4. Originally Posted by pickslides
You should use the identity

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

Now make $u = 2x$

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

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

Now use it again

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

Thank you,

Could you please explain the last step, what happened to 2x? How did it become just 'x'?

5. $Cos2x=2Cos^{2}x-1$

$Cos^{2}2x=(Cos2x)^2$

6. Originally Posted by Tweety
Thank you,

Could you please explain the last step, what happened to 2x? How did it become just 'x'?
Archie Meade has explained the substitution I have used, is it clear?

7. Originally Posted by pickslides
Archie Meade has explained the substitution I have used, is it clear?
Yes thank you!

8. The method that pickslides used was very neat, Tweety.

He used the same identity at each step.

You may see the identity written as

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

Rearranging this, we get

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

so

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

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

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

in writing

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

The other identity i used is

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