# Write sum sin x+sin2x+···+sinnx as 1 product?

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You need to perform the following multiplication by `sin (x/2)` , such that:

`S*sin(x/2) = sin(x/2)*sin x + sin(x/2)*sin 2x + ...... + sin(x/2)*sin nx`

You need to convert each product into a difference, using the following formula, such that:

`sin a*sin b = (1/2)*(cos(a - b) - cos(a + b))`

Reasoning by analogy yields:

`sin(x/2)*sin x = (1/2)*(cos(x/2 - x) - cos(x/2 + x))`

`sin(x/2)*sin x = (1/2)*(cos(-x/2) - cos((3x)/2))`

`sin(x/2)*sin 2x = (1/2)*(cos(x/2 - 2x) - cos(x/2 + 2x))`

`sin(x/2)*sin 2x = (1/2)*(cos((3x)/2) - cos((5x)/2))`

........

`sin(x/2)*sin nx = (1/2)*(cos(x/2 - nx) - cos(x/2 + nx))`

`sin(x/2)*sin nx = (1/2)*(cos((2n-1)x/2) - cos((2n+1)x/2))`

Evaluating the summation yields:

`S*sin(x/2) = (1/2)(cos(x/2) - cos((3x)/2) + cos((3x)/2) - cos((5x)/2) + cos((5x)/2) - ...... + cos((2n-1)x/2) - cos((2n+1)x/2))`

Reducing duplicate terms yields:

`S*sin(x/2) = (1/2)(cos(x/2) - cos((2n+1)x/2))`

Converting the difference into a product yields:

`S*sin(x/2) = (1/2)(sin((x + 2nx + x)/4)sin((2nx + x - x)/4))`

`S = (sin((nx)/2)*sin(((n+1)x)/2))/(2sin(x/2))`

**Hence, converting the given summation into a product yields **`S = (sin((nx)/2)*sin(((n+1)x)/2))/(2sin(x/2)).`