If $a\geq 0$, expand $\cos^a x$ in terms of $\cos kx$, $\sin mx$
$$\cos^a x=\left(\frac{e^{ix}+e^{-ix}}2\right)^a$$
Since $a$ is a non negative real number, so by General Binomial theorem
$$\cos^a x=\frac1{2^a}\sum_{k=0}^\infty\binom ake^{ikx}e^{-i(a-k)x}$$
where $\binom ak=a(a-1)(a-2)…(a-k+1)$
$$\cos^a x=\frac1{2^a}\sum_{k=0}^\infty\binom ake^{i(2k-a)x}$$
Any help would be appreciated.
Best Answer
The question seemingly asks for the Fourier expansion of $\cos^a x$ for $|x|\leqslant\pi/2$ (that is, of $|\cos x|^a$ for $x\in\mathbb{R}$): $$|\cos x|^a=\sum_{n=0}^\infty c_n\cos 2nx,$$ the coefficients of which can be computed using the formula $$\int_{-\pi/2}^{\pi/2}\cos^a x\cos bx\,dx=\frac\pi{2^a}\frac{\Gamma(a+1)}{\Gamma\left(\frac{a+b}2+1\right)\Gamma\left(\frac{a-b}2+1\right)}$$ seen here and in the linked questions. The result is $$c_0=\frac1{\sqrt\pi}\frac{\Gamma\left(\frac{a+1}2\right)}{\Gamma\left(\frac{a}2+1\right)},\quad c_n=2c_0\frac{\prod_{k=0}^{n-1}(a-2k)}{\prod_{k=1}^n(a+2k)}.\quad(n>0)$$
Obviously, if $a$ is an even integer, the series is finite.