If x = _n=0^(-1)^n^2n and y=_n=0^^2n, for 0

Question

If $x = \displaystyle\sum_{n=0}^{\infty}(-1)^n	an^{2n}	heta$ and $y=\displaystyle\sum_{n=0}^{\infty}\cos^{2n}	heta$, for $0<	heta<\dfrac{\pi}{4}$, then:

Accepted Answer

$$x = \sum\limits_{n = 0}^\infty  {{{\left( { - 1} \right)}^n}{{\tan }^{2n}}\theta } $$

= 1 -- tan2$$\theta $$ + tan2 4$$\theta $$ + ...

= $${1 \over {1 + {{\tan }^2}\theta }}$$ = cos2 $$\theta $$ ....(1)

$$y = \sum\limits_{n = 0}^\infty  {{{\cos }^{2n}}\theta } $$

=  1 + cos2
$$\theta $$ + cos4
$$\theta $$ + cos6
$$\theta $$ + ....

= $${1 \over {1 - {{\cos }^2}\theta }}$$ = $${1 \over {{{\sin }^2}\theta }}$$

$$ \Rightarrow $$ sin2 $$\theta $$ = $${1 \over y}$$ ...(2)

Adding (1) and (2), we get,

x + $${1 \over y}$$ = sin2 $$\theta $$ + cos2 $$\theta $$

$$ \Rightarrow $$ x + $${1 \over y}$$ = 1

$$ \Rightarrow $$ y(1 -- x) = 1

Solutions · 1