# Prove that: (cosx)^3*(sinx)^2=1/16*(2cosx-cos3x-cos5x)

beckden | High School Teacher | (Level 1) Educator

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`cos2x = cos^2x - sin^2x `

`sin2x = 2sinxcosx`

`cos3x = cosxcos2x - sinxsin2x = cosx(cos^2x-sin^2x) - sinx(2sinxcosx) `

`cos3x = cos^3x - cosxsin^2x - 2sin^2xcosx = cos^3x - 3sin^2xcosx `

`cos3x = cos^3x - 3(1-cos^2x)cosx = cos^3x - 3cosx + 3cos^3x `

`cos3x = 4cos^3x - 3cosx `

`sin3x = sin2xcosx + cos2xsinx = 2cosxsinxcosx + (2cos^2x - 1)sinx `

`sin3x = 2cos^2xsinx + 2cos^2xsinx - sinx = 4cos^2xsinx - sinx `

`sin3x = 4sinx - 4sin^3x - sinx = -4sin^3x + 3sinx `

`cos5x = cos3xcos2x - sin3xsin2x `

`cos5x = (4cos^3x - 3cosx)(2cos^2x - 1) - (-4sin^3x + 3sinx)(2sinxcosx) `

`cos5x = 8cos^5x - 10cos^3x + 8cosxsin^4x - 6cosxsin^2x + 3cosx `

`cos5x = 8cos^5x - 10cos^3x + 3cosx + 8cosx(1-cos^2x)^2 - 6cosx(1-cos^2x)`

`cos5x = 8cos^5x - 10cos^3x + 3cosx + 8cosx - 16cos^3x + 8cos^5x - 6cosx + 6cos^3x `

`cos5x = 16cos^5x - 20cos^3x + 5cosx `

So

`2cosx - cos3x - cos5x = 2cosx - 4cos^3x + ``3cosx - 16cos^5x + 20cos^3x - 5cosx `

`2cosx - cos3x - cos5x = -16cos^5x + 16cos^3x `

`2cosx - cos3x - cos5x = 16cos^3x(-1 + cos^2x)`

` = 16cos^3x(1-cos^2x) `

Finally

`2cosx - cos3x - cos5x = 16cos^3xsin^2x` so

`1/16(2cosx - cos3x - cos5x) = cos^3xsin^2x`