Use the geometric series e^(ix) - (1/2)e^(3ix) + (1/4)e^(5ix) - ... to find the exact value sin1 -(1/2)sin3 + (1/4)sin5 - ...

S = eix - (1/2)e3ix + (1/4)e5ix - … is an infinite geometric series, equal to a/(1 - r). a = eix and r = (1/2)e2ix thus S = eix/(1+(1/2)e2ix = 2eix/(2+e2ix). Rationalising the denominator: S = 2eix(2+e-2ix)/(2+e2ix)(2+e-2ix) = (4eix + 2e-ix)/(4 + 2(e2ix + e-2ix) + 1) = (4(cosx + isinx) + 2(cosx - isinx))/(5 + 2(2cosx)) = 6cosx/(5 + 4cos2x) + i(2sinx/(5 + 4cos2x)). We know that S = cosx + isinx - (1/2)cos3x -(1/2)isin3x+ (1/4)cos5x + (1/4)isin5x - … Hence Im(S) = sinx - (1/2)sin3x + (1/4)sin5x - … = 2sinx/(5+4cosx). Hence sin1 - (1/2)sin3 + (1/4)sin5 - … = 2sin1/(5 + 4cos2)

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