sum from n=1 to infinity of 12sin^n(70)

解

n = 1 \sum \infty 12 sin^{n} (7 0^{\circ})

12 (\frac{1}{1 - sin ( 7 0 ^{\circ} )} - 1)

十進法表記

186.98062 \dots

解答ステップ

n = 1 \sum \infty 12 sin^{n} (7 0^{\circ})

定数乗算の規則を適用する:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 12 \cdot n = 1 \sum \infty sin^{n} (7 0^{\circ})

n = k \sum \infty (a_{n}) = n = 0 \sum \infty (a_{n}) - a_{0} - a_{1} - \dots a_{k - 1}

= 12 (n = 0 \sum \infty sin^{n} (7 0^{\circ}) - sin^{0} (7 0^{\circ}))

n = 0 \sum \infty sin^{n} (7 0^{\circ}) = \frac{1}{1 - sin ( 7 0 ^{\circ} )}

= 12 (\frac{1}{1 - sin ( 7 0 ^{\circ} )} - sin^{0} (7 0^{\circ}))

簡素化

= 12 (\frac{1}{1 - sin ( 7 0 ^{\circ} )} - 1)