# Question bdbc2

May 23, 2016

See below.

#### Explanation:

To prove that: ${\cos}^{4} \left(x\right) = \frac{3}{8} + \frac{1}{2} \cos \left(2 x\right) + \frac{1}{8} \cos \left(4 x\right)$

First we have that:

${\cos}^{4} \left(x\right) = {\cos}^{2} \left(x\right) {\cos}^{2} \left(x\right)$

We can now make use of trig identity:

${\cos}^{2} \left(x\right) = \frac{1}{2} \cos \left(2 x\right) + \frac{1}{2}$

So we can say that:

${\cos}^{4} \left(x\right) = \left(\frac{1}{2} \cos \left(2 x\right) + \frac{1}{2}\right) \left(\frac{1}{2} \cos \left(2 x\right) + \frac{1}{2}\right)$

Expanding these brackets gives us:

$\frac{1}{4} {\cos}^{2} \left(2 x\right) + \frac{1}{2} \cos \left(2 x\right) + \frac{1}{4}$

Now using the identity again we can say that

${\cos}^{2} \left(2 x\right) = \frac{1}{2} \cos \left(4 x\right) + \frac{1}{4}$

Substituting this into our expression gives:

cos^4(x) = 1/4(1/2cos(4x)+1/2)+1/2cos(2x)+1/4)#

$= \frac{1}{8} \cos \left(4 x\right) + \frac{1}{8} + \frac{1}{2} \cos \left(2 x\right) + \frac{1}{4}$

$= \frac{1}{8} \cos \left(4 x\right) + \frac{1}{2} \cos \left(2 x\right) + \frac{3}{8}$

As required.