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

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

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

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

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

Solution steps

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

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^{n} (9) - sin^{0} (9)

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

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

Simplify

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

n = 1 \sum \infty \frac{5}{n ^{n}}

n = 1 \sum \infty \frac{ln ( 4 n )}{n}

n = 1 \sum \infty \frac{3 ^{n}}{3 + n !}

n = 1 \sum \infty (\frac{4}{5})^{n - 1}

n = 0 \sum \infty arctan (3 n)

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