Prove that $$\frac{1}{\sqrt{1-\sin^2{x}}}=\sum_{n=0}^{\infty} \frac{(2n)!(\sqrt{\sin{x}})^{4n}}{4^n (n!)^2}$$
Asked
Active
Viewed 57 times
-2
-
1Do you mean: $$\frac{1}{\sqrt{1-\sin^2{x}}}=\sum_{n=0}^{\infty} \frac{(2n)!(\sqrt{\sin{x}})^{4n}}{4^n (n!)^2}$$ – projectilemotion Apr 01 '17 at 16:33
-
Yes☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝☝ – creedoos Apr 01 '17 at 16:48
-
Ok, the ambiguity of the problem is now settled. What are your thoughts and what have you tried to solve the problem? That way, we don't repeat information you already know. – projectilemotion Apr 01 '17 at 16:53
1 Answers
0
Hint. One may use the generalized binomial theorem (see here) to get, $$\begin{eqnarray*} \frac{1}{\sqrt{1-u}}=(1-u)^{-1/2}=\sum_{n=0}^\infty \frac{{2n\choose n}}{2^{2n}}u^{n} &, \qquad |u|<1, \end{eqnarray*}$$ then one may put $u:=\sin^2 x$.
Olivier Oloa
- 120,989