# Show that $\cos(6x)= 32\cos^6x -48\cos^4x +18\cos^2x -1$

After writing down $\cos6x$= $Re (\cos x + i\sin x)^6$, I used the binomial theorem to expand the expression. Very soon it got really tedious and after trying $5$ times, fruitlessly, to arrive at the given expression, I gave up. Is there a shorter way around this?

This can be done with a few trig identities. We start with the double angle formula for cosine:

$$\cos(6x) = \cos(2\cdot 3x) = 2\cos^2(3x) - 1$$

Now we need to find an expression for $\cos(3x)$ before we proceed: $$\cos(3x) = \cos(2x + x) = \cos(2x)\cos(x) - \sin(2x)\sin(x)$$

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

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

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

$$=4\cos^3(x) - 3\cos(x).$$

Thus,

$$\cos(6x) = 2(4\cos^3(x) - 3\cos(x))^2 - 1$$

$$=32 \cos^6(x) - 48 \cos^4(x) + 18 \cos^2(x) - 1.$$

This was accomplished using the following:

Cosine Sum Identity: $\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$

Pythagorean Identity: $\sin^2(x) + \cos^2(x) = 1$

Cosine Double Angle Identity (or Cosine Sum Identity): $\cos(2x) = 2\cos^2(x) - 1$

Sine Double Angle Identity: $\sin(2x) = 2\sin(x)\cos(x)$

It might be tedious, but really there's no reason you shouldn't get to the answer.

Taking (for reduced finger strain) $c=\cos x$ and $s=\sin x$, and noting that $s^2 = 1-c^2$

\begin{align} \cos 6x &= \Re((c+is)^6) \\ &= \Re(c^6+6ic^5s-15c^4s^2-20ic^3s^3 +15c^2s^4+6ics^5-s^6)\\ &= c^6-15c^4s^2+15c^2s^4-s^6\\ &= c^6-15c^4(1-c^2)+15c^2(1-2c^2+c^4)-(1-3c^2+3c^4-c^6)\\ &= c^6(1+15+15+1)+c^4(-15-30-3) + c^2(15+3) -1\\ &= 32c^6-48c^4+18c^2-1\\ &= 32\cos^6 x-48\cos^4 x+18\cos^2 x-1\\ \end{align}

as required.

the easier way to do this is to derive the chebyshev recurrence relation for $$a_n = \cos nt, x = \cos t$$ by the sum to product formula. that is, $$a_{n+2} +a_n = \cos (nt+2t) + \cos(nt) = 2\cos t \cos(nt) = 2xa_n, a_0 = 1, a_1 = x.$$

now we can compute $$a_2 = 2x^2 - 1, a_3 = 2x(2x^2-1) - x=4x^3 - 3x,\\a_4=8x^4-6x^2-(2x^2-1) = 8x^4-8x^2 + 1\\a_5 = 16x^5-16x^3+2x-(4x^3-3x) = 16x^5-20x^3+5x\\a_6 = 32x^6-40x^4+10x^2-(8x^4-8x^2+1)=32x^6-48x^4+18x^2-1.$$

Perhaps a little bit less tedious then a binomial expansion to the power of $6$, you could use the triple and double angle identities $\cos 3x=4\cos^3x-3\cos x$ and $\cos 2x=2\cos^2x-1$, so that we have \begin{align}\cos (6x)&=\cos(3\cdot 2x)\\&=4\cos^3(2x)-3\cos(2x)\\&=4(2\cos^2x-1)^3-3(2\cos^2x-1)\\&=4(8\cos^6x-12\cos^4x+6\cos^2x-1)-6\cos^2x+3\\&=32\cos^6x-48\cos^4x+18\cos^2x-1\end{align}

• are you saying finding $a_2, a_3, a_4, a_5$ and $a_6$ tedious?
– abel
Commented May 22, 2015 at 20:18
• I have edited my answer - I meant less tedious that the OP's binomial expansion to the power of $6$. It was not aimed at your answer, as I did not have time to read it! Commented May 22, 2015 at 20:20
• i was just kidding. i can see that if you only want just one, your method is shorter(+1).
– abel
Commented May 22, 2015 at 20:43

Without actually calculating anything you knew that your approach would eventually express $\cos(6x)$ as a degree $6$ polynomial in $\cos(x)$. With a bit more consideration you can realize that it is actually a degree 3 polynomial in $\cos^2(x)$ by even-ness. Now to figure out what polynomial you just need to plug in $4$ values of $\cos^2(x)$ and solve for the coefficients (or use Lagrange interpolation).

Moreover though this polynomial is unique, so if you are given a polynomial (like you gave me) in order to check it you just need to plug in 4 values for $\cos^2(x)$. I recommend $x = 0, \pi/2, \pi/3,$ and $\pi/4$ make the calculations easy.

Let $e^{ix} =\cos(x)+i\sin(x) =c+is$.

$\begin{array}\\ e^{2inx} &=(c+is)^{2n}\\ &=\sum_{k=0}^{2n} \binom{2n}{k} c^k s^{2n-k}i^{2n-k} \end{array}$

$\begin{array}\\ e^{-2inx} &=(c-is)^{2n}\\ &=\sum_{k=0}^{2n} \binom{2n}{k} c^k s^{2n-k}(-i)^{2n-k} \end{array}$

$\begin{array}\\ 2\cos(2nx) &=e^{2inx}+e^{-2inx}\\ &=\sum_{k=0}^{2n} \binom{2n}{k} c^k s^{2n-k}(i^{2n-k}+(-i)^{2n-k}) \end{array}$

If $k$ is odd, $k = 2j+1$,

$\begin{array}\\ i^{2n-k}+(-i)^{2n-k} &=i^{2n-2j-1}+(-i)^{2n-2j-1}\\ &=(-1)^{n-j}(-i)+(-1)^{n-j}(i)\\ &=0\\ \end{array}$

so

$\begin{array}\\ 2\cos(2nx) &=\sum_{k=0}^{n} \binom{2n}{2k} c^{2k} s^{2n-2k}(i^{2n-2k}+(-i)^{2n-2k})\\ &=\sum_{k=0}^{n} \binom{2n}{2k} c^{2k} s^{2n-2k}((-1)^{n-k}+(-1)^{n-k})\\ &=\sum_{k=0}^{n} \binom{2n}{2k}2 c^{2k} s^{2n-2k}(-1)^{n-k}\\ &=\sum_{k=0}^{n} \binom{2n}{2k}2 c^{2n-2k} s^{2k}(-1)^{k}\\ &=\sum_{k=0}^{n} \binom{2n}{2k}2 (-1)^kc^{2n-2k} (s^2)^{k}\\ &=2\sum_{k=0}^{n} \binom{2n}{2k} (-1)^kc^{2n-2k} (1-c^2)^{k}\\ &=2\sum_{k=0}^{n} \binom{2n}{2k} (-1)^k(c^2)^{n-k}\sum_{j=0}^k \binom{k}{j}(-1)^{k-j} (c^2)^{k-j}\\ &=2\sum_{k=0}^{n} \binom{2n}{2k} (-1)^k(c^2)^{n-k}\sum_{j=0}^k \binom{k}{j}(-1)^{j} (c^2)^{j}\\ &=2\sum_{k=0}^{n}\sum_{j=0}^k \binom{2n}{2k}\binom{k}{j} (-1)^{k+j}(c^2)^{n-k+j} \\ &=2\sum_{k=0}^{n}\sum_{j=0}^k \binom{2n}{2k}\binom{k}{k-j} (-1)^{k+(k-j)}(c^2)^{n-j} \\ &=2\sum_{j=0}^n\sum_{k=j}^{n} \binom{2n}{2k}\binom{k}{k-j} (-1)^{j}(c^2)^{n-j} \\ &=2\sum_{j=0}^n(-1)^{j}(c^2)^{n-j}\sum_{k=j}^{n} \binom{2n}{2k}\binom{k}{k-j} \\ \end{array}$

so $\cos(2nx) =\sum_{j=0}^n(-1)^{j}(c^2)^{n-j}\sum_{k=j}^{n} \binom{2n}{2k}\binom{k}{k-j}$.

This is, modulo mistakes on my part, the kind of representation you want.

A simple (not enlightening) answer: $\cos(nx)=T_n(\cos(x))$, where $T_n(x)$ is the $n$th Chebyshev polynomial.