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
$\displaystyle \cos(4x)= 2(2\cos^2(x)-1)^2-1$
Now expand the RHS, you should have your answer.
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.
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
$\displaystyle sin^2x=\frac{1}{2}(1-cos2x)$