Find the Sum

Geometry Level 2

For f ( x ) = sin ( 2 x ) f(x) = \sin (2x) , find the value of

f ( h ) + f ( h + π 2 ) + f ( h + π ) + f ( h + 3 π 2 ) + + f ( h + 99 π ) + f ( h + 199 π 2 ) \small f(h) + f\left(h + \frac \pi 2\right) + f(h + \pi) + f\left(h + \frac {3\pi} 2\right) + \cdots + f(h+99\pi) + f\left(h + \frac {199\pi} 2\right)

where h h is a real number.

Clarification: Angles are measured in radian.


The answer is 0.

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2 solutions

Chew-Seong Cheong
Nov 28, 2020

S = f ( h ) + f ( h + π 2 ) + f ( h + π ) + f ( h + 3 π 2 ) + + f ( h + 99 π ) + f ( h + 199 π 2 ) = sin ( 2 h ) + sin ( π + 2 h ) + sin ( 2 π + 2 h ) + sin ( 3 π + 2 h ) + + sin ( 198 π + 2 h ) + sin ( 199 π + 2 h ) = sin ( 2 h ) + sin ( π + 2 h ) + sin ( 2 h ) + sin ( π + 2 h ) + sin ( 2 h ) + sin ( π + 2 h ) + + sin ( 2 h ) + sin ( π + 2 h ) = sin ( 2 h ) sin ( 2 h ) + sin ( 2 h ) sin ( 2 h ) + sin ( 2 h ) sin ( 2 h ) + sin ( 2 h ) sin ( 2 h ) + + sin ( 2 h ) sin ( 2 h ) = 0 \small \begin{aligned} S & = f(h) + f \left( h + \frac \pi 2 \right) + f(h + \pi) + f \left( h + \frac {3\pi} 2 \right) + \cdots + f(h + 99\pi) + f \left(h + \frac {199\pi} 2 \right) \\ & = \blue{\sin (2h)} + \red{\sin (\pi + 2h)} + \blue{\sin (2\pi + 2h)} + \red{\sin (3\pi + 2h)} + \cdots + \blue{\sin (198\pi + 2h)} + \red{\sin (199\pi + 2h)} \\ & = \blue{\sin (2h)} + \red{\sin (\pi + 2h)} + \blue{\sin (2h)} + \red{\sin (\pi + 2h)} + \blue{\sin (2h)} + \red{\sin (\pi + 2h)} + \cdots + \blue{\sin (2h)} + \red{\sin (\pi + 2h)} \\ & = \blue{\sin (2h)} \red{- \sin (2h)} \blue{+ \sin (2h)} \red{- \sin (2h)}\blue{+ \sin (2h)} \red{- \sin (2h)}\blue{+ \sin (2h)} \red{- \sin (2h)}+\cdots \blue{+ \sin (2h)} \red{- \sin (2h)} \\ & = \boxed 0 \end{aligned}

f ( h ) + f ( h + π 2 ) + f ( h + π ) + f ( h + 3 π 2 ) + f ( h + 2 π ) + + f ( h + 99 π ) + f ( h + 199 π 2 ) = k = 0 199 f ( h + k π 2 ) = k = 0 199 sin ( 2 h + k π ) = n = 0 99 sin ( 2 h + 2 n π ) + n = 0 99 sin [ 2 h + ( 2 n + 1 ) π ] = n = 0 99 sin ( 2 h ) + n = 0 99 sin [ 2 h + π ] = n = 0 99 sin ( 2 h ) + n = 0 99 [ sin ( 2 h ) ] = n = 0 99 sin ( 2 h ) n = 0 99 sin ( 2 h ) = 0 \begin{aligned} & f\left( h \right)+f\left( h+\frac{\pi }{2} \right)+f\left( h+\pi \right)+f\left( h+\frac{3\pi }{2} \right)+f\left( h+2\pi \right)+\ldots +f\left( h+99\pi \right)+f\left( h+\frac{199\pi }{2} \right) \\ & =\sum\limits_{k=0}^{199}{f\left( h+\frac{k\pi }{2} \right)} \\ & =\sum\limits_{k=0}^{199}{\sin \left( 2h+k\pi \right)} \\ & =\sum\limits_{n=0}^{99}{\sin \left( 2h+2n\pi \right)}+\sum\limits_{n=0}^{99}{\sin \left[ 2h+\left( 2n+1 \right)\pi \right]} \\ & =\sum\limits_{n=0}^{99}{\sin \left( 2h \right)}+\sum\limits_{n=0}^{99}{\sin \left[ 2h+\pi \right]} \\ & =\sum\limits_{n=0}^{99}{\sin \left( 2h \right)}+\sum\limits_{n=0}^{99}{\left[ -\sin \left( 2h \right) \right]} \\ & =\sum\limits_{n=0}^{99}{\sin \left( 2h \right)}-\sum\limits_{n=0}^{99}{\sin \left( 2h \right)} \\ & =\boxed{0} \\ \end{aligned}

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