Series representation of $|\cos\theta|^{1/2}+|\sin\theta|^{1/2}$

Question

Which real values of $\theta$ allows us to represent $|\cos\theta|^{1/2}+|\sin\theta|^{1/2}$ as a series of the form $$1+a_1\sin 2\theta+a_2\sin^22\theta+\cdots+a_k\sin^k2\theta+a_{k+1}\sin^{k+1}2\theta+\cdots ?$$

In that case how can we find constants $a_1, a_2, a_3, \cdots ?$

Answer 1

At $\theta = 0$, the $|\sin(\theta)|^{1/2}$ gives you a singularity like $|\theta|^{1/2}$. But if your series converges on some interval, it is $f(\sin(2\theta))$ where $f$ is an analytic function in a neighbourhood of $0$, and this can't have that square root singularity at $0$. So the answer is: nowhere.

EDIT: If you let $\theta = x^2$, then for $0 < \theta < \pi$ you have $$\eqalign{&\sin(\theta)^{1/2} = \sin(x^2)^{1/2}\cr &= x-{\frac{1}{12}}{x}^{5}+{\frac{1}{1440}}{x}^{9}-{\frac{1}{24192}}{x}^ {13}-{\frac{67}{29030400}}{x}^{17}-{\frac{1}{5677056}}{x}^{21}-{\frac{ 64397}{4649508864000}}{x}^{25}-{\frac{113249}{100429391462400}}{x}^{29 }+\ldots} $$ while for $0 < \theta < \pi/2$ $$ \eqalign{&\cos(\theta)^{1/2} = \cos(x^2)^{1/2} \cr &= 1-{\frac{1}{4}}{x}^{4}-{\frac{1}{96}}{x}^{8}-{\frac{19}{5760}}{x}^{12 }-{\frac{559}{645120}}{x}^{16}-{\frac{29161}{116121600}}{x}^{20}-{ \frac{2368081}{30656102400}}{x}^{24}-{\frac{276580459}{11158821273600} }{x}^{28}+\ldots}$$

Answer 2

$\sin^2 \theta = \frac {1}{2} (1 - \cos 2\theta)\\ |\sin \theta|^\frac 12 = (\frac {1}{2})^\frac 14 (1-\cos 2\theta)^\frac 14\\ |\cos \theta|^\frac 12 = (\frac {1}{2})^\frac 14 (1+\cos 2\theta)^\frac 14$

We can do a binomial expansion:

$|\sin \theta|^\frac 12 = (\frac {1}{2})^\frac 14(1^\frac 14 - \frac 14 (1^{-\frac {3}{4}}) \cos 2\theta - \frac {3}{2!4^2}(1^{-\frac74}) \cos^2 2\theta -\frac{3\cdot7}{3!4^3}\cos^3 2\theta - \frac{3\cdot7\cdot\ 11}{4!4^4}\cos^4 2\theta\cdots)$

$|\cos \theta|^\frac 12$ will look very similar, with the exception of some sign switching. When you add them together.

$|\sin \theta|^\frac 12 + |\cos \theta|^\frac12= 2(\frac {1}{2})^\frac 14(1 - \frac {3}{2\cdot4^2} \cos^2 2\theta - \frac{3\cdot7\cdot\ 11}{4!4^4}\cos^4 2\theta- \frac{3\cdot7\cdot\ 11\cdot 15\cdot 19}{6!4^6}\cos^6 2\theta\cdots)$

$\cos^2 2\theta = 1 - \sin^2 2\theta$ And I am feeling a little too exhausted to go through all of those substitutions. But it will work.