<s>[INST] Show that $\sum_{n=1}^\infty 2^{2n}\sin^4\frac a{2^n}=a^2-\sin^2a$ Show that $$\sum_{n=1}^\infty 2^{2n}\sin^4\frac a{2^n}=a^2-\sin^2a$$
I am studying for an exam and I bumped into this question. It's really bothering me because I really don't have any clue what to do. Does it have anything to do with the Cauchy condensation? Can somebody help me I would really appreciate it
 [/INST] Solution
Notice that
\begin{align*}
2^{2n}\sin^4\frac a{2^n}&=2^{2n}\cdot\sin^2\frac a{2^n}\cdot\sin^2\frac a{2^n}\\&=2^{2n}\cdot\sin^2\frac a{2^n}\cdot\left(1-\cos^2\frac a{2^n}\right)\\&=2^{2n}\cdot\sin^2\frac a{2^n}-2^{2n}\cdot\sin^2\frac a{2^n}\cos^2\frac a{2^n}\\&=2^{2n}\cdot\sin^2\frac a{2^n}-2^{2n-2}\sin^2\frac a{2^{n-1}}.
\end{align*}
Hence, the partial sum 
\begin{align*}
&\sum_{n=1}^{m}\left(2^{2n}\sin^4\frac a{2^n}\right)\\=&2^2\sin^2\frac a{2}-\sin^2a+2^4\sin^2\frac a{2^2}-2^2\sin^2\frac a{2}+ \cdots+ 2^{2m}\sin^2\frac a{2^m}-2^{2m-2}\sin^2\frac a{2^{m-1}}\\=&2^{2m}\sin^2\frac a{2^m}-\sin^2a.
\end{align*}
Let $m \to \infty$. We obtain $$\sum_{n=1}^{\infty}\left(2^{2n}\sin^4\frac a{2^n}\right)=\lim_{m \to \infty}\left(\frac{\sin\dfrac a{2^m}}{\dfrac{a}{2^m}}\cdot a\right)^2-\sin^2 a=(1\cdot a)^2-\sin^2 a=a^2-\sin^2 a.$$
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