What is the sum from n=1 to infinity of sin^n(8) ?

The sum from n=1 to infinity of sin^n(8) is 1/(1-sin(8))-1

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sum from n=1 to infinity of sin^n(8)

n = 1 \sum \infty sin^{n} (8)

\frac{1}{1 - sin ( 8 )} - 1

Solution steps

n = 1 \sum \infty sin^{n} (8)

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} (8) - sin^{0} (8)

n = 0 \sum \infty sin^{n} (8) = \frac{1}{1 - sin ( 8 )}

= \frac{1}{1 - sin ( 8 )} - sin^{0} (8)

Simplify

= \frac{1}{1 - sin ( 8 )} - 1

n = 1 \sum \infty \frac{1}{5 ^{n} + 3}

n = 1 \sum \infty \frac{n !}{10 4 ^{n}}

n = 1 \sum \infty \frac{2}{n ( n + 4 )}

n = 1 \sum \infty (- 0.6)^{n}

n = 1 \sum \infty \frac{2}{n ^{3} + 5}

What is the sum from n=1 to infinity of sin^n(8) ?
The sum from n=1 to infinity of sin^n(8) is 1/(1-sin(8))-1