1. ## prov 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.

2. Originally Posted by Tweety
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.
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

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

3. 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,

4. Originally Posted by pickslides
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

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

Thank you,

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

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

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

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

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

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

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

### prov of tregonometry

Click on a term to search for related topics.