What is the sum from n=1 to infinity of (sin(200))^n ?

The sum from n=1 to infinity of (sin(200))^n is 1/(1-sin(200))-1

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sum from n=1 to infinity of (sin(200))^n

n = 1 \sum \infty (sin (200))^{n}

\frac{1}{1 - sin ( 200 )} - 1

Solution steps

n = 1 \sum \infty (sin (200))^{n}

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 (200))^{n} - (sin (200))^{0}

n = 0 \sum \infty (sin (200))^{n} = \frac{1}{1 - sin ( 200 )}

= \frac{1}{1 - sin ( 200 )} - (sin (200))^{0}

Simplify

= \frac{1}{1 - sin ( 200 )} - 1

n = 0 \sum \infty \frac{3 n}{6 + n ^{2}}

n = 1 \sum \infty \frac{4}{3 ^{n} + 8}

k = 1 \sum \infty (cos (17))^{k}

n = 1 \sum \infty \frac{8}{1 0 ^{n}}

n = 1 \sum \infty cos (\frac{15}{n})

What is the sum from n=1 to infinity of (sin(200))^n ?
The sum from n=1 to infinity of (sin(200))^n is 1/(1-sin(200))-1