prove-n-0-sinx-2n-sec-2-x-

Question Number 156795 by tabata last updated on 15/Oct/21

p r o v e \underset{n = 0}{\sum^{\infty}} {(s i n x)}^{2 n} = {s e c}^{2} x ? ? ?

Answered by FongXD last updated on 15/Oct/21

we have : \underset{n = 0}{\sum^{\infty}} {(sinx)}^{2 n} = \underset{n = 0}{\sum^{\infty}} {(\sin^{2} x)}^{n}

since - 1 ⩽ sinx ⩽ 1, \Rightarrow 0 ⩽ \sin^{2} x < 1, \forall x \in R - {\frac{π}{2} (4 k \pm 1), k \in Z}

therefore, \underset{n = 0}{\sum^{\infty}} {(\sin^{2} x)}^{n} is the infinite sum of the geometric sequence

whose first term is {(\sin^{2} x)}^{0} = 1 and common ratio r = \sin^{2} x

Formula : S_{\infty} = \frac{u_{1}}{1 - q}

we get : \underset{n = 0}{\sum^{\infty}} {(\sin^{2} x)}^{n} = \frac{1}{1 - \sin^{2} x} = \frac{1}{\cos^{2} x} = \sec^{2} x ✓

therefore . \begin{matrix} \underset{n = 0}{\sum^{\infty}} {(sinx)}^{2 n} = \sec^{2} x is indeed true \end{matrix}

Commented by Ruuudiy last updated on 17/Oct/21

Thanks Ser . \underset{n = 0}{\sum^{\infty}} {(\sin x)}^{2 n} = \sec^{2} x is indeed true .

Answered by som(math1967) last updated on 15/Oct/21

1 + {s i n}^{2} x + {s i n}^{4} x + \dots \infty

= \frac{1}{1 - {s i n}^{2} x} = \frac{1}{{c o s}^{2} x} = {s e c}^{2} x