Show that $\displaystyle \cos(5a) = 16\cos^5a - 20\cos^3a + 5\cos(a)$.
Thanks in advance.
Hi =)
I'd suggest you do this way :
$\displaystyle \cos 5a=\cos (4a+a)=\cos 4a \sin a-\sin 4a \cos a=\cos (2*2a) \sin a-\sin(2*2a) \cos a$
Then, use the appropriate formulae :
$\displaystyle \cos 2x=2 \cos^2x$
$\displaystyle \sin 2x=2 \cos x \sin x$
$\displaystyle \sin^2x=1-\cos^2x$
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Or you can start from $\displaystyle 16\cos^5a - 20\cos^3a + 5\cos(a)$
And expand while substituting : $\displaystyle \cos a=\frac{e^{ia}+e^{-ia}}{2}$
Some smooth approaches have already been given. Here's yet another approach (but only if you're familiar with complex numbers):
$\displaystyle \cos (5a) = Re \left[ e^{i(5a)}\right] = Re \left[ \left( e^{ia} \right)^5 \right] = Re \left[ (\cos (a) + i \sin (a) )^5 \right] .....$
For FP3, use De Moivre's Theorem. $\displaystyle z^n = r^n[cos\theta+isin\theta]^n = r^n[cos(n\theta)+isin(n\theta)]$
Now binomial theorem gives $\displaystyle [cos\theta+isin\theta]^5 =$
$\displaystyle [cos^5\theta+5cos^4\theta(i)sin\theta+10cos^3\theta (i^2)sin^2\theta+$
$\displaystyle 10cos^2\theta(i^3)sin^3\theta+5cos\theta(i^4)sin^4 \theta+(i^5)sin^5\theta]$
But you know that $\displaystyle i^2=-1, i^3=-i, i^4=1, i^5=i$ and so on
Considering only the real part of the expansion
$\displaystyle Re[cos\theta+isin\theta]^5 = [cos^5\theta - 10 cos^3\theta sin^2\theta + 5 cos\theta sin^4\theta]$
Consider
$\displaystyle sin^2\theta = 1 - cos^2\theta$
$\displaystyle sin^4\theta = (1 - cos^2\theta)^2$
$\displaystyle [cos^5\theta - 10 cos^3\theta (1 - cos^2\theta) + 5 cos\theta (1 - cos^2\theta)^2] = [16cos^5\theta - 20 cos^3\theta + 5cos\theta]$
(sorry I ran out of time there, working should be easy)