## Tuesday, March 22, 2016

### Factorize $\cos^2 x+\cos^2 2x+\cos^2 3x+\cos 2x+\cos 4x+\cos 6x$.

Factorize $\cos^2 x+\cos^2 2x+\cos^2 3x+\cos 2x+\cos 4x+\cos 6x$.

My solution:

In a problem such as this one, two inclinations may arise:

1.

Some would try to turn $\cos^2 x,\,\cos^2 2x$ and $\cos^2 3x$ so they become $\cos 2x,\,\cos 4x,\,\cos 6x$, by using the double angle formula for cosine function ($\cos 2A=2\cos^2 A-1\stackrel{\text{or}}{=}1-2\sin^2 A\stackrel{\text{or}}{=}\cos^2 A-\sin^2 A$)

2.

Others would try to turn $\cos 2x,\,\cos 4x$ and $\cos 6x$ into the expression contains $\cos^2 x,\,\cos^2 2x,\,\cos^2 3x$, either using the double/triple angle formulas or sum-to-product formulas.

But one must also realize not both approaches would lead to the answer. Even if they do, one must be the easier route than the other, therefore we need to think and choose wisely which way to begin with.

One trick that always works for problem like this is we want to deal with linear power for cosine function rather than working with the square power of the cosine functions.

That means, we want to turn $\cos^2 x,\,\cos^2 2x$ and $\cos^2 3x$ by using the double angle formula for cosine function so we have the intended expression be written all in linear function:

Note that from the double angle formula for cosine we have:

$\cos 2A=2\cos^2 A-1$

Now we rearrange it to make $\cos^2 A$ the subject, we get:

$\cos 2A+1=2\cos^2 A$

$\cos^2 A=\dfrac{\cos 2A+1}{2}$

Therefore the intended expression becomes

$\cos^2 x+\cos^2 2x+\cos^2 3x+\cos 2x+\cos 4x+\cos 6x$

$=\dfrac{\cos 2x+1}{2}+\dfrac{\cos 4A+1}{2}+\dfrac{\cos 6A+1}{2}+\cos 2x+\cos 4x+\cos 6x$

$=\dfrac{\cos 2x}{2}+\dfrac{\cos 4x}{2}+\dfrac{\cos 6x}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+\cos 2x+\cos 4x+\cos 6x$

$=\dfrac{3}{2}\left(\cos 2x+\cos 4x+\cos 6x\right)+\dfrac{3}{2}$

$=\dfrac{3}{2}\left(\cos 2x+\cos 4x+\cos 6x+1\right)$

$=\dfrac{3}{2}\left(\underbrace{\color{orange}\cos 2x+\cos 6x}_{\cos A+\cos B=2\cos\frac{A-B}{2}\cos\frac{A+B}{2}}+\underbrace{\color{purple}\cos 4x}_{\cos 2A=2\cos^2 A-1}+1\right)$

$=\dfrac{3}{2}\left(\overbrace{\color{orange}2\cos\frac{6x-2x}{2}\cos\frac{6x+2x}{2}}+\overbrace{\color{purple}2\cos^2 2x-1}+1\right)$

$=\dfrac{3}{2}\left(\color{orange}2\cos 2x\cos 4x+\color{purple}2\cos^2 2x-1\color{black}+1\right)$

$=\dfrac{3}{2}\left(\color{orange}2\cos 2x\cos 4x+\color{purple}2\cos^2 2x\right)$

$=\dfrac{3}{2}\left(2\cos 2x\right)\left(\color{orange}\cos 4x+\color{purple}\cos 2x\right)$

$=3\cos 2x\left(2\cos \dfrac{4x-2x}{2}\cos \dfrac{4x+2x}{2}\right)$

$=6\cos 2x\left(\cos x\cos 3x\right)$

$=6\cos x \cos 2x\cos 3x\right)$

and we're hence done!