Show the inequality sin x+sin2x+sin3x+--+sin( nx)=(sin(nx)/2sin(n+1)x/2)/sin x/2
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You need to multiply the expression by `sin(x/2)` both sides, such that:
`sin(x/2)*(sin x + sin(2x) + ... + sin(nx)) = sin(x/2)*(sin((nx)/2)*sin((n+1)*x/2))/(sin(x/2))`
Reducing duplicate factors to the right side, yields:
`sin(x/2)*(sin x + sin(2x) + ... + sin(nx)) = (sin((nx)/2)*sin((n+1)*x/2))`
You need to convert the products `sin(x/2)*sin x + sin(x/2)*sin (2x) + ..` . into summations, such that:
`(1/2)(cos(x/2 - x) - cos(x/2 + x) + cos(x/2 - 2x) - cos(x/2 + 2x) + cos(x/2 - 3x) - cos(x/2 + 3x) + .... cos(x/2 - (n-1)x) - cos(x/2 + (n-1)x)+ cos(x/2 - nx) - cos(x/2 + nx)) = (sin((nx)/2)*sin((n+1)*x/2))`
`(1/2)(cos(-x/2) - cos((3x)/2) + cos((-3x)/2) - cos((5x)/2) + cos((-5x)/2) - cos((7x)/2) + .... cos((3x)/2 - nx) - cos(-(x/2 - nx)) + cos(x/2 - nx) - cos(x/2 + nx)) = (sin((nx)/2)*sin((n+1)*x/2))`
Since `cos(-a) = cos a` , you may reduce duplicate factors, such that:
`(1/2)(cos(-x/2) - cos(x/2 + nx)) = (sin((nx)/2)*sin((n+1)*x/2))`
Converting `cos(-x/2) - cos(x/2 + nx)` into a product yields:
`2*(1/2)*sin((-x/2 + x/2 + nx)/2)*sin((x/2 + nx + x/2)/2) = (sin((nx)/2)*sin((n+1)*x/2))`
Reducing duplicate factors yields:
`sin((nx)/2)*sin((x + nx)/2) = (sin((nx)/2)*sin((n+1)*x/2))`
`(sin((nx)/2)*sin((n+1)*x/2)) = (sin((nx)/2)*sin((n+1)*x/2))`
Hence, evaluating the left side of expression yields that `(sin((nx)/2)*sin((n+1)*x/2)) = (sin((nx)/2)*sin((n+1)*x/2)) ` hence, the expression holds.
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