sum from n=1 to infinity of sin^n(8)

Lösung

n = 1 \sum \infty sin^{n} (8)

\frac{1}{1 - sin ( 8 )} - 1

Dezimale

92.96947 \dots

Schritte zur Lösung

n = 1 \sum \infty sin^{n} (8)

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} (8) - sin^{0} (8)

n = 0 \sum \infty sin^{n} (8) = \frac{1}{1 - sin ( 8 )}

= \frac{1}{1 - sin ( 8 )} - sin^{0} (8)

Vereinfache

= \frac{1}{1 - sin ( 8 )} - 1