What is the sum from n=0 to infinity of (sin(100))^n ?

The sum from n=0 to infinity of (sin(100))^n is 1/(1-sin(100))

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sum from n=0 to infinity of (sin(100))^n

n = 0 \sum \infty (sin (100))^{n}

\frac{1}{1 - sin ( 100 )}

Solution steps

n = 0 \sum \infty (sin (100))^{n}

Apply Series Geometric Test:

\frac{1}{1 - sin ( 100 )}

= \frac{1}{1 - sin ( 100 )}

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What is the sum from n=0 to infinity of (sin(100))^n ?
The sum from n=0 to infinity of (sin(100))^n is 1/(1-sin(100))