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.


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 $$\sum_{n=0}^\infty (\cos^2(\theta))^n=\sum_{n=0}^\infty\cos^{2n}(\theta)=\frac{1}{1-\cos^2(\theta)}.$$

  • $\begingroup$ 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

$$\begin{align}x &= \sum^{\infty}_{n =0} \cos^{2n}\theta \\ y &= \sum^{\infty}_{n =0} \sin^{2n}\theta\\ z &= \sum^{\infty}_{n =0} \cos^{2n}\theta \sin^{2n}\theta \end{align}$$

$$x=1+{\cos^2\theta}+{\cos^4\theta}+\cdots\infty \;terms$$ 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) $$xy+z=\csc^2\theta\cdot\sec^2\theta+\dfrac {1}{1-\cos^2\theta\cdot\sin^2\theta}$$

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

  • $\begingroup$ $x\neq1+\dfrac {1}{\cos^2\theta}+\dfrac {1}{\cos^4\theta}+\cdots+\infty$, 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 $\endgroup$ – iostream007 Jul 4 '13 at 4:45
  • 1
    $\begingroup$ 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 $\endgroup$ – iostream007 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.