What is the sum from n=0 to infinity of (sin(7))^n ?

The sum from n=0 to infinity of (sin(7))^n is 1/(1-sin(7))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sum from n=0 to infinity of (sin(7))^n

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

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

Solution steps

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

Apply Series Geometric Test:

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

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

n = 1 \sum \infty 6 cos (\frac{1}{3 n})

n = 0 \sum \infty \frac{n}{4 + n ^{2}}

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

n = 1 \sum \infty ln (\frac{1}{n !})

n = 1 \sum \infty \frac{2}{n ^{2} + n ^{3}}

What is the sum from n=0 to infinity of (sin(7))^n ?
The sum from n=0 to infinity of (sin(7))^n is 1/(1-sin(7))