Prove the general formula for powers of cosine…
Prove the general formula for powers of cosine: $$ \cos^nx=2^{1-n}\left[\cos nx+n\cos(n-2)x+\frac{n(n-1)}{1\cdot2}\cos(n-4)x+\cdots+A\right] $$ where $A=\dfrac12\dbinom{n}{n/2}$ if $n$ is even, and $A=\dbinom{n}{(n-1)/2}$ if $n$ is odd.
Explanation of Method
Write $\cos x=\dfrac{e^{ix}+e^{-ix}}2$, raise to the $n$-th power, and expand with the binomial theorem. Symmetric terms $e^{i(n-2k)x}+e^{-i(n-2k)x}$ pair into $2\cos((n-2k)x)$; for even $n$ there is one unpaired constant term in the middle, for odd $n$ every term pairs.
Step 1 · Set up the power
$$ \cos^nx=\left(\frac{e^{ix}+e^{-ix}}2\right)^n=\frac{(e^{ix}+e^{-ix})^n}{2^n}, \qquad (e^{ix}+e^{-ix})^n=\sum_{k=0}^n\binom nk e^{i(n-2k)x} $$
Case 1 · $n=2m$ even
All terms pair except the middle one ($k=m$, giving exponent $0$):
$$ (e^{ix}+e^{-ix})^{2m}=2\sum_{k=0}^{m-1}\binom{2m}k\cos\big((2m-2k)x\big)+\binom{2m}m $$
Dividing by $2^{2m}$:
$$ \cos^{2m}x=2^{1-2m}\left[\sum_{k=0}^{m-1}\binom{2m}k\cos\big((2m-2k)x\big)+\frac12\binom{2m}m\right] $$
i.e. $\ \cos^{2m}x=2^{1-2m}\left[\cos2mx+2m\cos(2m-2)x+\dfrac{2m(2m-1)}2\cos(2m-4)x+\cdots+\dfrac12\dbinom{2m}m\right]$, matching the general formula with $A=\frac12\binom{2m}m$.
Case 2 · $n=2m+1$ odd
Every term pairs (no fixed point):
$$ (e^{ix}+e^{-ix})^{2m+1}=2\sum_{k=0}^{m}\binom{2m+1}k\cos\big((2m+1-2k)x\big) $$
Dividing by $2^{2m+1}$:
$$ \cos^{2m+1}x=2^{-2m}\sum_{k=0}^{m}\binom{2m+1}k\cos\big((2m+1-2k)x\big) $$
i.e. $\ \cos^{2m+1}x=2^{-2m}\left[\cos(2m+1)x+(2m+1)\cos(2m-1)x+\cdots+\dbinom{2m+1}m\cos x\right]$, matching the general formula with $A=\binom{2m+1}m$.
Step 2 · Unified statement
$$ \boxed{\cos^nx=2^{1-n}\left[\cos nx+n\cos(n-2)x+\frac{n(n-1)}2\cos(n-4)x+\cdots+A\right]}, \qquad A=\begin{cases}\frac12\dbinom{n}{n/2}, & n \text{ even}\\[4pt]\dbinom{n}{(n-1)/2}, & n\text{ odd}\end{cases} $$
Verification
| $n$ | Formula gives | Known identity |
|---|---|---|
| $2$ | $\frac12[\cos2x+1]$ | $\cos^2x=\frac{1+\cos2x}2$ ✓ |
| $3$ | $\frac14[\cos3x+3\cos x]$ | standard triple-angle reduction ✓ |
| $4$ | $\frac18[\cos4x+4\cos2x+3]$ | $\cos^4x=\frac{3+4\cos2x+\cos4x}8$ ✓ |
The general formula was checked symbolically for $n=1,\ldots,5$.
Result
Any positive-integer power of cosine reduces to a finite sum of cosines of multiple angles.