Question

Show that (cos theata + isin theata)square = (cos3 theata + isin 3 theata)

Answer 1

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cos3x - sin3x = (cosx + sinx)(1 - 2sin2x)

Step-by-step explanation:

cos3x - sin3x = (cosx + sinx)(1 - 2sin2x)

cos3x = 4cos^3x - 3cosx

sin3x = 3sinx - 4 sin^3x

LHS

= cos3x - sin3x

= 4cos^3x - 3cosx - (3sinx - 4 sin^3x)

= 4(cos^3x + sin^3x) -3(cosx + sinx)

= 4(cosx + sinx)(cos^2x + sin^2x - cosxsinx) - 3(cosx + sinx)

= 4(cosx + sinx)(1 - cosxsinx) - 3(cosx + sinx)

= (cosx + sinx)(4 - 4cosxsinx) - 3(cosx + sinx)

= (cosx + sinx)(4 - 2× 2 cosxsinx) - 3(cosx + sinx)

= (cosx + sinx)(4 - 2sin2x) - 3(cosx + sinx)

= (cosx + sinx)(4 - 2sin2x - 3)

= (cosx + sinx)(1 - 2sin2x)

= rhs

qed

provef

cos3x - sin3x = (cosx + sinx)(1 - 2sin2x)

Answer 2

cos3x - sin3x = (cosx + sinx)(1 - 2sin2x)

Step-by-step explanation:

cos3x - sin3x = (cosx + sinx)(1 - 2sin2x)

cos3x = 4cos^3x - 3cosx

sin3x = 3sinx - 4 sin^3x

LHS

= cos3x - sin3x

= 4cos^3x - 3cosx - (3sinx - 4 sin^3x)

= 4(cos^3x + sin^3x) -3(cosx + sinx)

= 4(cosx + sinx)(cos^2x + sin^2x - cosxsinx) - 3(cosx + sinx)

= 4(cosx + sinx)(1 - cosxsinx) - 3(cosx + sinx)

= (cosx + sinx)(4 - 4cosxsinx) - 3(cosx + sinx)

= (cosx + sinx)(4 - 2× 2 cosxsinx) - 3(cosx + sinx)

= (cosx + sinx)(4 - 2sin2x) - 3(cosx + sinx)

(cosx + sinx)(4 - 2sin2x - 3)

= (cosx + sinx)(1 - 2sin2x)

= rhs

qed

provef

cos3x - sin3x = (cosx + sinx)(1 - 2sin2x)

Mark ❤️