What is the sum from n=1 to infinity of (sin(9))^n ?

The sum from n=1 to infinity of (sin(9))^n is 1/(1-sin(9))-1

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sum from n=1 to infinity of (sin(9))^n

n = 1 \sum \infty (sin (9))^{n}

\frac{1}{1 - sin ( 9 )} - 1

Solution steps

n = 1 \sum \infty (sin (9))^{n}

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= n = 0 \sum \infty (sin (9))^{n} - (sin (9))^{0}

n = 0 \sum \infty (sin (9))^{n} = \frac{1}{1 - sin ( 9 )}

= \frac{1}{1 - sin ( 9 )} - (sin (9))^{0}

Simplify

= \frac{1}{1 - sin ( 9 )} - 1

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What is the sum from n=1 to infinity of (sin(9))^n ?
The sum from n=1 to infinity of (sin(9))^n is 1/(1-sin(9))-1