Is it a series expansion of cot ⁡ x \cot x ?

Calculus Level 4

V = ∑ x = 1 180 x sin ⁡ ( x ∘ ) \large V = \sum_{x=1}^{180} x \sin(x^\circ)

For V V as defined above, find ⌊ V ⌋ \lfloor V \rfloor .

Notation: ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .


The answer is 10312.

Ossama Ismail by Ossama Ismail , 63, Egypt

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1 solution

V = ∑ x = 1 180 x sin ⁡ x ∘ = ∑ x = 1 179 x sin ⁡ x ∘ + 180 sin ⁡ 18 0 ∘ 0 = ∑ x = 1 89 x sin ⁡ x ∘ + 90 sin ⁡ 9 0 ∘ + ∑ x = 91 179 x sin ⁡ x ∘ = ∑ x = 1 89 x sin ⁡ x ∘ + 90 + ∑ x = 1 89 ( 180 − x ) sin ⁡ ( 180 − x ) ∘ Note that sin ⁡ ( 180 − x ) ∘ = sin ⁡ x ∘ = ∑ x = 1 89 x sin ⁡ x ∘ + 90 + ∑ x = 1 89 ( 180 − x ) sin ⁡ x ∘ = 180 ∑ x = 1 89 sin ⁡ x ∘ + 90 = 180 sin ⁡ 1 ∘ ∑ x = 1 89 sin ⁡ 1 ∘ sin ⁡ x ∘ + 90 By sin ⁡ A sin ⁡ B = 1 2 ( cos ⁡ ( A − B ) − cos ⁡ ( A + B ) ) = 90 sin ⁡ 1 ∘ ∑ x = 1 89 ( cos ⁡ ( x − 1 ) ∘ − cos ⁡ ( x + 1 ) ∘ ) + 90 = 90 sin ⁡ 1 ∘ ( cos ⁡ 0 ∘ + cos ⁡ 1 ∘ − cos ⁡ 8 9 ∘ − cos ⁡ 9 0 ∘ ) + 90 ≈ 10312.98 \begin{aligned} V & = \sum_{x=1}^{\color{#3D99F6}180} x \sin x^\circ \\ & = \sum_{x=1}^{\color{#3D99F6}179} x \sin x^\circ + 180 \cancel{\sin 180^\circ}^0 \\ & = \sum_{x=1}^{89} x \sin x^\circ + 90 \sin 90^\circ + \sum_{x=91}^{179} x \sin x^\circ \\ & = \sum_{x=1}^{89} x \sin x^\circ + 90 + \sum_{x=1}^{89} (180-x) \color{#3D99F6} \sin (180-x)^\circ & \small \color{#3D99F6} \text{Note that }\sin (180-x)^\circ = \sin x ^\circ \\ & = \sum_{x=1}^{89} x \sin x^\circ + 90 + \sum_{x=1}^{89} (180-x) \color{#3D99F6} \sin x^\circ \\ & = 180 \sum_{x=1}^{89} \sin x^\circ + 90 \\ & = \frac {180}{\sin 1^\circ} \sum_{x=1}^{89} {\color{#3D99F6} \sin 1^\circ \sin x^\circ} + 90 & \small \color{#3D99F6} \text{By }\sin A \sin B = \frac 12 \left(\cos (A-B) - \cos (A+B)\right) \\ & = \frac {\color{#3D99F6}90}{\sin 1^\circ} \sum_{x=1}^{89} {\color{#3D99F6} \left(\cos (x-1)^\circ - \cos (x+1)^\circ \right)} + 90 \\ & = \frac {90}{\sin 1^\circ} \left(\cos 0^\circ + \cos 1^\circ - \cos 89^\circ - \cos 90^\circ \right) + 90 \\ & \approx 10312.98 \end{aligned}

⟹ ⌊ V ⌋ = 10312 \implies \lfloor V \rfloor = \boxed{10312}

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Nice solution. Because you solved it, now I believe the answer is correct. I have entered into Mathwalfram alpha and it gave me a different answer. I must have done something wrong somewhere.

Hana Wehbi - 4 years, 1 month ago

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I got the right answer (see below).

Chew-Seong Cheong - 4 years, 1 month ago

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