1. ## cos5a Expansion

Show that $\cos(5a) = 16\cos^5a - 20\cos^3a + 5\cos(a)$.

2. Hi =)

Originally Posted by Air
Show that $\cos(5a) = 16\cos^5a - 20\cos^3a + 5\cos(a)$.

I'd suggest you do this way :

$\cos 5a=\cos (4a+a)=\cos 4a \sin a-\sin 4a \cos a=\cos (2*2a) \sin a-\sin(2*2a) \cos a$

Then, use the appropriate formulae :

$\cos 2x=2 \cos^2x$

$\sin 2x=2 \cos x \sin x$

$\sin^2x=1-\cos^2x$

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Or you can start from $16\cos^5a - 20\cos^3a + 5\cos(a)$

And expand while substituting : $\cos a=\frac{e^{ia}+e^{-ia}}{2}$

3. Originally Posted by Air
Show that $\cos(5a) = 16\cos^5a - 20\cos^3a + 5\cos(a)$.

Some smooth approaches have already been given. Here's yet another approach (but only if you're familiar with complex numbers):

$\cos (5a) = Re \left[ e^{i(5a)}\right] = Re \left[ \left( e^{ia} \right)^5 \right] = Re \left[ (\cos (a) + i \sin (a) )^5 \right] .....$

4. Originally Posted by Air
Show that $\cos(5a) = 16\cos^5a - 20\cos^3a + 5\cos(a)$.

For FP3, use De Moivre's Theorem. $z^n = r^n[cos\theta+isin\theta]^n = r^n[cos(n\theta)+isin(n\theta)]$

Now binomial theorem gives $[cos\theta+isin\theta]^5 =$

$[cos^5\theta+5cos^4\theta(i)sin\theta+10cos^3\theta (i^2)sin^2\theta+$
$10cos^2\theta(i^3)sin^3\theta+5cos\theta(i^4)sin^4 \theta+(i^5)sin^5\theta]$

But you know that $i^2=-1, i^3=-i, i^4=1, i^5=i$ and so on

Considering only the real part of the expansion

$Re[cos\theta+isin\theta]^5 = [cos^5\theta - 10 cos^3\theta sin^2\theta + 5 cos\theta sin^4\theta]$

Consider
$sin^2\theta = 1 - cos^2\theta$
$sin^4\theta = (1 - cos^2\theta)^2$

$[cos^5\theta - 10 cos^3\theta (1 - cos^2\theta) + 5 cos\theta (1 - cos^2\theta)^2] = [16cos^5\theta - 20 cos^3\theta + 5cos\theta]$
(sorry I ran out of time there, working should be easy)

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# cos5a expand

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