Show the sum to infinity of a geometric progression with cos theta and sin theta

Key concept: This is a neat geometric progression question: once you identify the first term and ratio, the infinite-sum formula does most of the work. The only extra step is rewriting the fraction in a friendlier trig form.

Step by step

Let the progression be $\cos\theta,\ \cos\theta\sin\theta,\ \cos\theta\sin^2\theta,\ \dots$

So:

• first term $a=\cos\theta$
• common ratio $r=\sin\theta$

Because the progression converges, we must have $|\sin\theta|<1.$

For a convergent geometric series, $S_\infty=\frac{a}{1-r}.$ Hence $S_\infty=\frac{\cos\theta}{1-\sin\theta}.$

Now multiply top and bottom by $1+\sin\theta$: $S_\infty=\frac{\cos\theta(1+\sin\theta)}{1-\sin^2\theta}.$ Using $1-\sin^2\theta=\cos^2\theta$,

\frac{1+\sin\theta}{\cos\theta}.$$ Split the fraction: $$S_\infty=\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}=\sec\theta+\tan\theta.$$ So the required result is $$\boxed{S_\infty=\sec\theta+\tan\theta}.$$ ### Pitfall alert The usual slip is forgetting to check convergence. The formula $\frac{a}{1-r}$ only works when $|r|<1$. Another trap is stopping at $\frac{\cos\theta}{1-\sin\theta}$ and not simplifying it to the trig identity the question wants. ### Try different conditions If the ratio were $-\sin\theta$ instead, the sum would become $$\frac{\cos\theta}{1+\sin\theta}=\sec\theta-\tan\theta,$$ after rationalising. So the sign of the ratio completely changes the final trig form. ### Further reading geometric progression, infinite series, trigonometric identity