# Cos(x+pi/3)+cos(x-pi/3)=cosx Have the show the statement and the rules? Thanks again

Apr 7, 2018

Recall that $\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B$ and $\cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$.

$\cos x \cos \left(\frac{\pi}{3}\right) - \sin x \sin \left(\frac{\pi}{3}\right) + \cos x \cos \left(\frac{\pi}{3}\right) + \sin x \sin \left(\frac{\pi}{3}\right) = \cos x$

$2 \cos x \cos \left(\frac{\pi}{3}\right) = \cos x$

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

$\cos x = \cos x$

$L H S = R H S$

As required.

Hopefully this helps!

Apr 7, 2018

#### Explanation:

.

$\cos \left(x + \frac{\pi}{3}\right) + \cos \left(x - \frac{\pi}{3}\right) = \cos x$

We have a Sum to Product formula that gives us:

$\cos \alpha + \cos \beta = 2 \cos \left(\frac{\alpha + \beta}{2}\right) \cos \left(\frac{\alpha - \beta}{2}\right)$

Using this formula, we get:

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

Apr 7, 2018

See below.

#### Explanation:

Identities:

$\textcolor{red}{\boldsymbol{\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B}}$

$\textcolor{red}{\boldsymbol{\cos \left(A - B\right) = \cos A \cos B + \sin A \sin B}}$

I'll do these in two parts, because of space:

$\cos \left(x + \frac{\pi}{3}\right) = \cos \left(x\right) \cos \left(\frac{\pi}{3}\right) - \sin \left(x\right) \sin \left(\frac{\pi}{3}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \cos \left(x\right) \left(\frac{1}{2}\right) - \sin \left(x\right) \left(\frac{\sqrt{3}}{2}\right)$

$\cos \left(x - \frac{\pi}{3}\right) = \cos \left(x\right) \cos \left(\frac{\pi}{3}\right) + \sin \left(x\right) \sin \left(\frac{\pi}{3}\right)$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \cos \left(x\right) \left(\frac{1}{2}\right) + \sin \left(x\right) \left(\frac{\sqrt{3}}{2}\right)$

$\cos \left(x + \frac{\pi}{3}\right) + \cos \left(x - \frac{\pi}{3}\right)$

$= \cos \left(x\right) \left(\frac{1}{2}\right) - \sin \left(x\right) \left(\frac{\sqrt{3}}{2}\right) + \cos \left(x\right) \left(\frac{1}{2}\right) + \sin \left(x\right) \left(\frac{\sqrt{3}}{2}\right)$

$\cos \left(x\right) \left(\frac{1}{2}\right) - \cancel{\sin \left(x\right) \left(\frac{\sqrt{3}}{2}\right)} + \cos \left(x\right) \left(\frac{1}{2}\right) + \cancel{\sin \left(x\right) \left(\frac{\sqrt{3}}{2}\right)}$

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

$\cancel{2} \cos \left(x\right) \cancel{\left(\frac{1}{2}\right)}$

$\cos x$

As required:

$L H S \equiv R H S$

Apr 7, 2018

Well, I don't know the rules on your Rule Menu, so I'm using mine.

LHS $\implies \cos \left(x + \frac{\pi}{3}\right) + \cos \left(x - \frac{\pi}{3}\right)$

Applying, color(red)(cos(A+B) = cosAcosB - sinAsinB and color(magenta)(cos(A-B) = cosAcosB + sinAsinB

color(white)(dd

=> color(red)(cosxcos(pi/3) -sinxsin(pi/3)) + color(magenta)(cosxcos (pi/3) +sinxsin(pi/3)

$\implies \cancel{2} \cos x \left(\frac{1}{\cancel{2}}\right)$ color(white)(wwwwwwwww $\left[\text{as } \cos \left(\frac{\pi}{3}\right) = \frac{1}{2}\right]$

$\implies \cos x$