What is the sum from n=1 to infinity of 12sin^n(70) ?

The sum from n=1 to infinity of 12sin^n(70) is 12(1/(1-sin(70))-1)

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

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

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

Solution steps

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

Apply the constant multiplication rule:

\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}))

Simplify

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

n = 0 \sum \infty \frac{n}{5 n ^{2} + 1}

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

n = 0 \sum \infty (\frac{4 π}{e})^{n}

k = 0 \sum \infty \frac{k}{3 k + 1}

n = 0 \sum \infty (\frac{8}{7})^{n}

What is the sum from n=1 to infinity of 12sin^n(70) ?
The sum from n=1 to infinity of 12sin^n(70) is 12(1/(1-sin(70))-1)