# Finding the relation between function x,y,z - trigo problem

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

• 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. Jul 4 '13 at 4:37
• @sultan: First term for $n=0$ is $(\cos^2\theta)^0=1$
– Aang
Jul 4 '13 at 4:39
• sorry I got it.. Thanks.. 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) $$xy+z=\csc^2\theta\cdot\sec^2\theta+\dfrac {1}{1-\cos^2\theta\cdot\sin^2\theta}$$

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

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