Y= 1-cosx-i sinsx / 1+cosx+i sinsx
To simplify the given expression, we use the formula of complexification that involves the cosine and sine functions. This formula is given by: cos(x) + i sin(x) = e^(ix).\n\nAt first, we set Y1 = e^(ix). This means that cos(x) + i sin(x) = Y1.\n\nSecondly, we know that cos(-x) = cos(x) and sin(-x) = -sin(x), so we then set Y2 = cos(-x) + i sin(-x) = e^(-ix).\n\nSubstitute Y1 and Y2 into the given equation:\n\nY = [1 - Y1] / [1 + Y2]\n\nor:\n\nY = [1 - e^(ix)] / [1 + e^(-ix)].\n\nBy multiplying the numerator and denominator of the right-hand side by e^(ix) results in Y = [e^(ix) - 1] / [e^(ix) + e^(ix)] = [e^(ix) - 1] / 2e^(ix)\n\nHence,\n\nY = 1/2 - (1 / 2e^(ix)) = 1/2 - i / (2* (cos(x) + i sin(x))),\n\nwhich is a simplified form of the given equation.