cot x = 1 / tan x \cot x = 1/\tan x

Calculus Level 5

V = 2 sin ( 2 ) + 4 sin ( 4 ) + 6 sin ( 6 ) + + 180 sin ( 18 0 ) V = 2 \sin(2^\circ) + 4 \sin(4^\circ) + 6 \sin(6^\circ) + \cdots + 180 \sin (180^\circ) .

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


The answer is 5156.

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1 solution

Ossama Ismail
Feb 27, 2017

V = 2 sin 2 + 4 sin 4 + + 178 sin 178 V = 2 \sin 2 + 4 \sin 4 + \cdots + 178 \sin 178

sin ( 1 ) V = 2 sin ( 1 ) sin ( 2 ) + 4 sin ( 1 ) sin ( 4 ) + + 178 sin ( 1 ) sin ( 178 ) \sin(1) V = 2 \sin(1) \sin(2) + 4 \sin(1) \sin(4) + \cdots + 178 \sin(1) \sin(178)

We know that 2 sin ( a ) sin ( b ) = cos ( a b ) cos ( a + b ) 2 \sin (a) \sin (b) = \cos(a - b) - \cos(a + b) , then

sin ( 1 ) V = 2 sin ( 1 ) sin ( 2 ) + 2 ( 2 sin ( 1 ) sin ( 4 ) + + 89 ( 2 sin ( 1 ) sin ( 178 ) \sin(1) V = 2 \sin(1) \sin(2) + 2 (2 \sin(1) \sin(4) + \cdots + 89( 2 \sin(1) \sin(178)

sin ( 1 ) V = cos ( 1 ) cos ( 3 ) + 2 ( cos ( 3 ) cos ( 5 ) ) + + 89 ( cos ( 177 ) cos ( 179 ) ) \sin(1) V = \cos(1) - \cos(3) + 2 ( \cos(3) - \cos(5) ) + \cdots + 89( \cos(177) - \cos(179))

sin ( 1 ) V = cos ( 1 ) + cos ( 3 ) + cos ( 5 ) ) + + cos ( 175 ) + cos ( 177 ) 89 cos ( 179 ) \sin(1) V = \cos(1) + \cos(3) + \cos(5) ) + \cdots + \cos (175) + \cos (177) - 89 \cos(179)

sin ( 1 ) V = cos ( 1 ) + ( cos ( 3 ) + cos ( 177 ) ) + + ( cos ( 89 ) + cos ( 91 ) ) 89 cos ( 179 ) \sin(1) V = \cos(1) + (\cos(3) + \cos(177) ) + \cdots + ( \cos (89) + \cos (91)) - 89 \cos(179)

sin ( 1 ) V = cos ( 1 ) + 89 cos ( 1 ) = 90 cos ( 1 ) \sin(1) V = \cos (1) + 89 \cos ( 1 ) = 90 \cos( 1)

V = 90 cos ( 1 ) / sin ( 1 ) = 90 cot ( 1 ) = 90 × 57.289961 = 5156.0965 V = 90 \cos(1)/\sin(1) = 90 \cot(1) = 90 \times 57.289961 = 5156.0965

V = 5156 \lfloor V \rfloor = 5156

Why have you treated that sum as only over even x?

Brian Moehring - 4 years, 3 months ago

Yes ....i forgot to mention that

Ossama Ismail - 4 years, 3 months ago

Put a backslash " \ " in front of functions to make them proper (non-italic) sin \sin , cos \cos , and cot \cot .

Chew-Seong Cheong - 4 years, 3 months ago

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thanks >>> done

Ossama Ismail - 4 years, 3 months ago

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