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

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

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sum from n=1 to infinity of sin^n(3)

n = 1 \sum \infty sin^{n} (3)

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

Decimal

0.16430 \dots

Solution steps

n = 1 \sum \infty sin^{n} (3)

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} (3) - sin^{0} (3)

n = 0 \sum \infty sin^{n} (3) = \frac{1}{1 - sin ( 3 )}

= \frac{1}{1 - sin ( 3 )} - sin^{0} (3)

Simplify

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

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