Best solutions to - "Find the value of 1sin⁡2∘+2sin⁡4∘+3sin⁡6∘+…+90sin⁡180∘"

minikim38

Answered question

2022-01-24

Find the value of

1 {\sin 2}^{\circ} + 2 {\sin 4}^{\circ} + 3 {\sin 6}^{\circ} + \dots + 90 {\sin 180}^{\circ}

Ian Adams

Skilled2022-01-26Added 163 answers

An approach. One may write

\sum_{k = 1}^{n} k \sin (k a) = \sum_{k = 1}^{n} \frac{d}{d a} (- \cos (k a))

= - \frac{d}{d a} \sum_{k = 0}^{n} \cos (k a)

= - \frac{d}{d a} (\frac{1}{2} + \frac{\sin [(n + \frac{1}{2}) a]}{2 \sin \frac{a}{2}})

= \frac{(n + 1) \sin (n a) - n \sin ((n + 1) a)}{4 \sin^{2} \frac{a}{2}}

then one may take

a := 2^{\circ}, n := 90

star233

Skilled2022-01-26Added 403 answers

Hints: Your sum is related with:

S = \sum_{n = 1}^{90} n \sin (\frac{π n}{90}) = \sum_{n = 0}^{89} (90 - n) \sin (\frac{π n}{90})

that fulfills

2 S = 90 \sum_{n = 1}^{89} \sin (\frac{π n}{90}) = 90 \cdot \cot (\frac{π}{180})

RizerMix

Expert2022-01-27Added 656 answers

Use

\sin (θ) = \sin (180^{\circ} - θ)

. Let the summation as S. Then

2 S = \sum_{k = 1}^{90} (k \sin (2 k^{\circ}) + (90 - k) \sin ((180 - 2 k)^{\circ}))

= 90 \sum_{k = 1}^{90} \sin (2 k^{\circ})

Now use

- 2 \sin (2 k^{\circ}) \sin (1^{\circ}) = \cos ((2 k + 1)^{\circ}) - \cos ((2 k - 1)^{\circ})

, then

S = 45 \sum_{k = 1}^{90} \sin (2 k^{\circ}) = - \frac{45}{2 \sin (1^{\circ})} \sum_{k = 1}^{90} (\cos ((2 k + 1)^{\circ}) - \cos ((2 k - 1)^{\circ})

= \frac{45 \cos (1^{\circ})}{\sin (1^{\circ})} = 45 \cot (1^{\circ})

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