What is the sum from n=0 to infinity of sin^n(2) ?

The sum from n=0 to infinity of sin^n(2) is 1/(1-sin(2))

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sum from n=0 to infinity of sin^n(2)

n = 0 \sum \infty sin^{n} (2)

\frac{1}{1 - sin ( 2 )}

Solution steps

n = 0 \sum \infty sin^{n} (2)

Apply Series Geometric Test:

\frac{1}{1 - sin ( 2 )}

= \frac{1}{1 - sin ( 2 )}

\int \frac{2}{e ^{- x} + 1} d x

\int e^{2 x + 6} d x

\int ln (x^{8}) d x

\frac{d}{d x} (6^{6 - x^{2}})

y = - 2 x + 7

What is the sum from n=0 to infinity of sin^n(2) ?
The sum from n=0 to infinity of sin^n(2) is 1/(1-sin(2))