Dear Uncle Colin,

I recently came across a problem in which I had to integrate $\cos^3(x)$. Somewhere in my mind, I recall that the thing to do is to make it into something involving $\cos(3x)$, but I couldn’t put the details together. Could you help?

-- Not A Very Inspired Expression Rearranger


But of course! In fact, I can show you two ways to accomplish it: the cloggerish way I learnt, and a much nicer way.

The cloggerish way

The standard way is to say $\cos(3x) = \cos(2x+x) = \cos(2x)\cos(x) - \sin(2x)\sin(x)$.

From there, you replace $\cos(2x)$ with $2\cos^2(x) -1$ and $\sin(2x)$ with $2\sin(x)\cos(x)$, leaving you with:

$\cos(3x) = \left(2 \cos^2(x) -1\right)\cos(x) - 2\cos(x)\sin^2(x)$

We can replace $\sin^2(x)$ with $1 - \cos^2(x)$:

$\cos(3x) = \left(2 \cos^2(x) -1\right)\cos(x) - 2\cos(x)\left(1-\cos^2(x)\right)$

Then expand everything:

$\cos(3x) = 2\cos^3(x) - \cos(x) - 2\cos(x) + 2 \cos^3(x) = 4\cos^3(x) - 3\cos(x)$

With that identity in place, you can rearrange to get $\cos^3(x)$ on its own:

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

The nicer way

In your formula book, assuming you’re doing a qualification that has such a thing, you’ll find a group of trig identities that are rarely, if ever, examined. These include $2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) = \cos(A) + \cos(B)$, and this is the one we’re interested in: it turns a product of cosines into a sum.

For example, if you consider $2\cos(x)\cos(x)$, you find that if $A=2x$ and $B=0$, you get $\cos(2x) + 1$ – which rearranges, if you want it to, as $\cos(2x) = 2\cos^2(x) - 1$.

In general, you can turn $\cos(P)\cos(Q)$ into $\frac 12 \left( \cos(P+Q) + \cos(P-Q) \right)$, which is extremely handy here.

$\cos^3(x) = \left[\cos(x)\cos(x)\right] \cos(x)$, obviously, so it’s $\frac 12 \left( \cos(2x) + 1 \right) \cos(x)$, or $\frac{1}{2} \cos(2x)\cos(x) + \frac{1}{2} \cos(x)$.

Applying the same rule to the first term gives $\cos^3(x) = \frac{1}{4} \left(\cos(3x) + \cos(x)\right) + \frac{1}{2}\cos(x) = \frac{1}{4} \left( \cos(3x) + 3 \cos(x) \right)$.

The Ninja Way

“Who let you in?”

“Oh, of course. You don’t need letting in.”

”$\cos^3(x) = \left(1 - \sin^2(x)\right)\cos(x)$”, he said. “Or $\cos(x) - \sin^2(x) \cos(x)$.”

“First term’s trivial. Second one’s function-derivative.”

And he was gone.

Hope that’s helpful, NAVIER!

-- Uncle Colin