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Problem :

For $\displaystyle 0 < \theta < \frac{\pi}{2}$ if $$\begin{align}x &= \sum^{\infty}_{n =0} \cos^{2n}\theta \\ y &= \sum^{\infty}_{n =0} \sin^{2n}\theta\\ z &= \sum^{\infty}_{n =0} \cos^{2n}\theta \sin^{2n}\theta \end{align}$$

then options are :

(a) $xyz = xz+y$

(b) $xyz = xy+z$

(c) $xy^2 =y^2+x$

I have solution of this however I have one doubt in that, it is mention that :

$\displaystyle x=\sum^{\infty}_{n =0} \cos^{2n}\theta = \frac{1}{1-\cos^2\theta}$ (How it is derived... or what about this result.). Please guide on this... thanks.

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This is just the geometric series in disguise: for any number $s$ with the property that $|s|<1$, $$\sum_{n=0}^\infty s^n=\frac{1}{1-s}.$$ By restricting $0<\theta<\frac{\pi}{2}$, we guarantee that $|\cos^2(\theta)|<1$, so that $$\sum_{n=0}^\infty (\cos^2(\theta))^n=\sum_{n=0}^\infty\cos^{2n}(\theta)=\frac{1}{1-\cos^2(\theta)}.$$

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  • $\begingroup$ We have taken sum of geometric progression where sum to infinity = $\frac{a}{1-r}$ where a is first term and r is common ratio. Are we considering first term as 1 here, however it is less than 1 but approximately equal to 1. Please suggest. $\endgroup$
    – Sachin
    Jul 4 '13 at 4:37
  • $\begingroup$ @sultan: First term for $n=0$ is $(\cos^2\theta)^0=1$ $\endgroup$
    – Aang
    Jul 4 '13 at 4:39
  • $\begingroup$ sorry I got it.. Thanks.. $\endgroup$
    – Sachin
    Jul 4 '13 at 4:41
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$$\begin{align}x &= \sum^{\infty}_{n =0} \cos^{2n}\theta \\ y &= \sum^{\infty}_{n =0} \sin^{2n}\theta\\ z &= \sum^{\infty}_{n =0} \cos^{2n}\theta \sin^{2n}\theta \end{align}$$

$$x=1+{\cos^2\theta}+{\cos^4\theta}+\cdots\infty \;terms$$ this is infinite geometric series : $S_{\infty}=\dfrac {a}{1-r}\;\;\,|r|<1$ ,here $\cos\theta<1, when\; 0<\theta<\dfrac \pi2 \;so\; \cos^2\theta<1$ $$x=\dfrac {1}{1-\cos^2\theta}\implies \dfrac {1}{\sin^2\theta} \implies x=\csc^2\theta$$ here $a=1$ and r=$\cos^2\theta$

similarly for y $a=1$ and $r=\sin^2\theta$ $$y=\dfrac {1}{1-\sin^2\theta}\implies \dfrac {1}{\cos^2\theta}\implies y=\sec^2\theta$$ and $$z=\dfrac {1}{1-\cos^2\theta\cdot\sin^2\theta}$$ so option (b) $$xy+z=\csc^2\theta\cdot\sec^2\theta+\dfrac {1}{1-\cos^2\theta\cdot\sin^2\theta}$$

$$xy+z=\dfrac {\csc^2\theta\cdot\sec^2\theta}{1-\cos^2\theta\cdot\sin^2\theta}$$ $$xy+z=xyz$$

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  • $\begingroup$ $x\neq1+\dfrac {1}{\cos^2\theta}+\dfrac {1}{\cos^4\theta}+\cdots+\infty$, but $1+\cos^2\theta+\cos^4\theta+\cdots$ $\endgroup$
    – Aang
    Jul 4 '13 at 4:41
  • $\begingroup$ @Avatar thanks for locating my mistake.I didn't see $\endgroup$ Jul 4 '13 at 4:45
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    $\begingroup$ There is $+\infty$ at the end of the series, which mustn't be there as $|\cos^{2n}\theta|<1$ $\forall n\in \Bbb N$ $\endgroup$
    – Aang
    Jul 4 '13 at 4:49
  • $\begingroup$ To write $+\infty$ there I mean series goes upto $\infty$ terms $\endgroup$ Jul 4 '13 at 5:01
  • $\begingroup$ so you can just write $\infty$ without a $+$ since $+$ suggests the addition operation of series with $\infty$ $\endgroup$
    – Aang
    Jul 4 '13 at 14:41

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