What is the sum from n=1 to infinity of sin^n(7) ?

The sum from n=1 to infinity of sin^n(7) is 1/(1-sin(7))-1

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sum from n=1 to infinity of sin^n(7)

n = 1 \sum \infty sin^{n} (7)

\frac{1}{1 - sin ( 7 )} - 1

Solution steps

n = 1 \sum \infty sin^{n} (7)

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} (7) - sin^{0} (7)

n = 0 \sum \infty sin^{n} (7) = \frac{1}{1 - sin ( 7 )}

= \frac{1}{1 - sin ( 7 )} - sin^{0} (7)

Simplify

= \frac{1}{1 - sin ( 7 )} - 1

n = 0 \sum \infty 6 (\frac{1}{7})^{n - 1}

n = 0 \sum \infty 2^{- n - 1}

n = 2 \sum \infty n^{3}

n = 0 \sum \infty \frac{n}{10 n ^{2} + 1}

n = 0 \sum \infty \frac{1}{5 ^{n} - n}

What is the sum from n=1 to infinity of sin^n(7) ?
The sum from n=1 to infinity of sin^n(7) is 1/(1-sin(7))-1