Animated Solution for Mathematics - Sequence and Series: If $0<\theta ,\varphi <\frac{\pi }{2}$, $x={\sum }_{n=0}^{\infty }{\cos }^{2n}\theta$, $y={\sum }_{n=0}^{\infty }{\sin }^{2n}\varphi$ and $z={\sum }_{n=0}^{\infty }{\cos }^{2n}\theta \cdot {\sin }^{2n}\varphi$ then

Welcome, future engineer. Today we are going to peel back the layers of a problem that looks like a trigonometric nightmare but is actually a beautiful dance of algebra.

When you see infinite series like $x={\sum }_{n=0}^{\infty }{\cos }^{2n}\theta$, your first instinct might be panic. But take a breath and look at the pattern. This is not chaos; it is the classic infinite geometric progression.

The Geometric Insight

Let us expand the series for $x$:

This is a geometric series where the first term $a=1$ and the common ratio $r={\cos }^2\theta$. Since our angle $\theta$ is in the first quadrant, $0<{\cos }^2\theta <1$, which allows us to use the sum formula $S_{\infty }=\frac{a}{1-r}$.

Thus, we have:

This implies ${\sin }^2\theta =\frac{1}{x}$. Consequently, we find:

Mirroring the Logic

Now, apply the same logic to $y={\sum }_{n=0}^{\infty }{\sin }^{2n}\varphi$. This expands to $1+{\sin }^2\varphi +{\sin }^4\varphi +\dots$.

Again, $a=1$ and $r={\sin }^2\varphi$. The sum is:

This gives us ${\cos }^2\varphi =\frac{1}{y}$ and:

We now have our trigonometric building blocks expressed purely in terms of $x$ and $y$.

The Climax

Finally, we tackle $z={\sum }_{n=0}^{\infty }({\cos }^2\theta {\sin }^2\varphi {)}^n$. This is an infinite geometric series with $a=1$ and $r={\cos }^2\theta {\sin }^2\varphi$.

So, the sum is:

Substitute our expressions for ${\cos }^2\theta$ and ${\sin }^2\varphi$:

To simplify, multiply the numerator and denominator by $xy$:

Expanding the denominator gives $xy-(xy-x-y+1)=x+y-1$. Thus, we arrive at the final relationship:

Cross-multiplying yields $z(x+y-1)=xy$, which rearranges to $zx+zy-z=xy$. The final elegant relationship is: $xy=z(x+y-1)$