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

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

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sum from n=0 to infinity of (sin(1))^n

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

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

Solution steps

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

Simplify

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

Apply Series Geometric Test:

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

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

\int \frac{1}{x ( 1 + x ^{2} ) ^{2}} d x

\int \frac{x ^{3}}{x ^{2} - 25} d x

x \to 0 lim (\frac{x ^{2} - 6}{x})

f (x) = \frac{x}{1 + x ^{2}}, (2, \frac{2}{5})

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