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If sinx + sin 2 x + sin 3 x = 1,then find the value of cos 6 x - 4cos 4 x + 8cos 2 x.
Write sinx + sin2x + sin3x = 1, as:sinx + sin3x = 1 – sin2xNow squaring both sides,U would get,sin^6x + sin^4x + 3sin^2x = 1and write each sin^2x to (1-cos^2x) => (1-cos^2x) ^3 + (1-cos^2x)^2 + 3(1-cos^2x) = 1 ould get further cos6x - 4cos4x + 8cos2x. value as 1.
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