If $x=\displaystyle\sum_{n=0}^{\infty}\cos^{2n} heta$, $y=\displaystyle\sum_{n=0}^{\infty}\sin^{2n} heta$ and $z=\displaystyle\sum_{n=0}^{\infty}\cos^{2n} heta\sin^{2n} heta$, where $0

Question

If $x=\displaystyle\sum_{n=0}^{\infty}\cos^{2n}	heta$, $y=\displaystyle\sum_{n=0}^{\infty}\sin^{2n}	heta$ and $z=\displaystyle\sum_{n=0}^{\infty}\cos^{2n}	heta\sin^{2n}	heta$, where $0<	heta<\dfrac{\pi}{2}$, then show that $xyz=x+y+z$.

[Hint: Use the formula $1+x+x^2+x^3+\cdots=\dfrac{1}{1-x}$, where $|x|<1$.]

Accepted Answer

> [!TLDR] > > Each of $x,y,z$ is an infinite geometric series with common ratio less than $1$ in absolute value (since $0< heta<\pi/2$), summing via $1+r+r^2+\cdots=\frac{1}{1-r}$; substituting these closed forms into both $xyz$ and $x+y+z$ shows the two sides reduce to the identical expression. **Step 1. Sum $x$.** $x=\displaystyle\sum_{n=0}^\infty(\cos^2 heta)^n=\dfrac{1}{1-\cos^2 heta}=\dfrac{1}{\sin^2 heta}=\csc^2 heta$ (hint's formula with $r=\cos^2 heta$, $0<\cos^2 heta<1$ since $0< heta<\pi/2$). **Step 2. Sum $y$.** Similarly $y=\displaystyle\sum_{n=0}^\infty(\sin^2 heta)^n=\dfrac{1}{1-\sin^2 heta}=\dfrac{1}{\cos^2 heta}=\sec^2 heta$. **Step 3. Sum $z$.** $z=\displaystyle\sum_{n=0}^\infty(\cos^2 heta\sin^2 heta)^n=\dfrac{1}{1-\sin^2 heta\cos^2 heta}$ (again geometric, ratio $\sin^2 heta\cos^2 heta\in(0,1)$). **Step 4. Compute $xyz$.** $$xyz=\csc^2 heta\cdot\sec^2 heta\cdot\frac{1}{1-\sin^2 heta\cos^2 heta}=\frac{1}{\sin^2 heta\cos^2 heta\left(1-\sin^2 heta\cos^2 heta ight)}.$$ …

$\theta$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
$\sin\theta$	$0$	$\frac12$	$\frac{1}{\sqrt2}$	$\frac{\sqrt3}{2}$	$1$
$\cos\theta$	$1$	$\frac{\sqrt3}{2}$	$\frac{1}{\sqrt2}$	$\frac12$	$0$
$\tan\theta$	$0$	$\frac{1}{\sqrt3}$	$1$	$\sqrt3$	undefined

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