What is the sum from n=0 to infinity of (sin(3))^n ?

The sum from n=0 to infinity of (sin(3))^n is 1/(1-sin(3))

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sum from n=0 to infinity of (sin(3))^n

n = 0 \sum \infty (sin (3))^{n}

\frac{1}{1 - sin ( 3 )}

Solution steps

n = 0 \sum \infty (sin (3))^{n}

Apply Series Geometric Test:

\frac{1}{1 - sin ( 3 )}

= \frac{1}{1 - sin ( 3 )}

n = 1 \sum \infty - \frac{1}{4 ^{n}}

n = 0 \sum \infty \frac{15}{6 ^{n}}

n = 0 \sum \infty \frac{3 n !}{n ^{2}}

n = 0 \sum \infty (- 1)^{n} \cdot 1^{n}

n = 3 \sum \infty \frac{1}{( n ^{2} - 4 )}

What is the sum from n=0 to infinity of (sin(3))^n ?
The sum from n=0 to infinity of (sin(3))^n is 1/(1-sin(3))