# Question fdeca

Oct 26, 2014

$\cos \left(x\right)$

#### Explanation:

Recall our trig identities:

$\sin \left(2 x\right) = 2 \sin \left(x\right) \cos \left(x\right)$

cos(2x) = cos^2(x) – sin^2(x) 

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

Looking at our problem:

$\cos \left(2 x\right) \cos \left(x\right) + \sin \left(2 x\right) \sin \left(x\right)$

Let's substitute for $\sin \left(2 x\right)$. This will give us:

$\cos \left(2 x\right) \cos \left(x\right) + \left(2 \sin \left(x\right) \cos \left(x\right)\right) \cdot \sin \left(x\right)$

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

Now, let's substitute ${\sin}^{2} \left(x\right)$ using the identity ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

This gives us:

$\cos \left(2 x\right) \cos \left(x\right) + 2 {\sin}^{2} \left(x\right) \cos \left(x\right)$

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

Now, let's substitute $\cos \left(2 x\right)$ using the identity cos(2x) = cos^2(x) – sin^2(x) #

This gives us:

$\cos \left(2 x\right) \cos \left(x\right) + 2 \left(1 - {\cos}^{2} \left(x\right)\right) \cdot \cos \left(x\right)$

$= \left({\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)\right) \cdot \cos \left(x\right) + 2 \left(1 - {\cos}^{2} \left(x\right)\right) \cdot \cos \left(x\right)$

Now, let's substitute the sin^2(x) inside the parenthesis using the identity, ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

This gives us:

$\left({\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)\right) \cdot \cos \left(x\right) + 2 \left(1 - {\cos}^{2} \left(x\right)\right) \cdot \cos \left(x\right)$

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

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

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

Now, multiplying through with our $\cos \left(x\right)$'s:

$\left(2 {\cos}^{2} \left(x\right) - 1\right) \cdot \cos \left(x\right) + 2 \left(1 - {\cos}^{2} \left(x\right)\right) \cdot \cos \left(x\right)$

$= 2 {\cos}^{3} \left(x\right) - \cos \left(x\right) + 2 \left(\cos \left(x\right) - {\cos}^{3} \left(x\right)\right)$

$= 2 {\cos}^{3} \left(x\right) - \cos \left(x\right) + 2 \cos \left(x\right) - 2 {\cos}^{3} \left(x\right)$

The $2 {\cos}^{3} \left(x\right)$ and the $- 2 {\cos}^{3} \left(x\right)$ cancel out,

and we are left with:

$- \cos \left(x\right) + 2 \cos \left(x\right)$

which equals:

$\cos \left(x\right)$