Evaluate (sin x+ sin 3x +sin 5x+...+ sin17x)/(cos x+ cos 3x+cos 5x+...+cos 17x) when x=pi/24.

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You should convert the sums into product using the following formulas such that:

`sina + sin b= 2 sin((a+b)/2)cos((a-b)/2)`

`cos a + cos b = 2 cos((a+b)/2)cos((a-b)/2)`

Notice that you may form the following groups such that:

`(sin x + sin 17x) + (sin 3x + sin 15 x) + (sin 5x + sin 13 x) + (sin 7x + sin 11x) + sin 9x`

`(cos x +cos 17x) + (cos 3x +cos 15 x) + (cos 5x +cos 13 x) + (cos 7x +cos 11x) +cos 9x`

Converting the sums of sines into products yields:

`2 sin((x+17x)/2)cos((x-17x)/2) + 2 sin((3x+15x)/2)cos((3x-15x)/2) + 2 sin((5x+13x)/2)cos((5x-13x)/2) + 2 sin((7x+11x)/2)cos((7x-11x)/2) +sin 9x`

`2 sin(9x)cos(8x) + 2 sin(9x)cos(6x) + 2 sin(9x)cos(4x) + 2 sin(9x)cos(2x) + sin 9x`

Factoring out `2sin 9x`  yields:

`2sin 9x*(cos(8x) + cos(6x) + cos(4x) + cos(2x) + 1)`

Converting the sums of cosines into products yields:

`2 cos((x+17x)/2)cos((x-17x)/2) + 2 cos((3x+15x)/2)cos((3x-15x)/2) + 2 cos((5x+13x)/2)cos((5x-13x)/2) + 2 cos((7x+11x)/2)cos((7x-11x)/2) +cos 9x`

`2 cos(9x)cos(8x) + 2 cos(9x)cos(6x) + 2 cos(9x)cos(4x) + 2 cos(9x)cos(2x) +cos 9x`

Factoring out `2cos 9x`  yields:

`2cos 9x(cos(8x) + cos(6x) + cos(4x) + cos(2x) + 1)`

Substituting the products for sums in the given expression yields:

`(2sin 9x*(cos(8x) + cos(6x) + cos(4x) + cos(2x) + 1))/(2cos 9x(cos(8x) + cos(6x) + cos(4x) + cos(2x) + 1))`

Reducing like factors yields:

`(2sin 9x*(cos(8x) + cos(6x) + cos(4x) + cos(2x) + 1))/(2cos 9x(cos(8x) + cos(6x) + cos(4x) + cos(2x) + 1)) = (sin 9x)/(cos 9x) = tan 9x`

You may substitute `pi/24`  for x such that:

`tan 9x = tan 9pi/24 => tan 9pi/24 = tan 3pi/8 = tan((3pi/4)/2)`

`tan 9pi/24 = sqrt((1 - cos(3pi/4))/(1+cos(3pi/4)))`

`cos(3pi/4) = cos(pi - pi/4) = -cos(pi/4) = -sqrt2/2`

`tan 9pi/24 = sqrt ((1+sqrt2/2)/(1-sqrt2/2))`

Hence, evaluating the given expression at `x = pi/24 ` yields `(sin x + sin 3x + .... + sin 17x)/(cos x +cos 3x + ... + cos 17 x) = sqrt((2+sqrt2)/(2-sqrt2)).`

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