Question

prove cos6A+ sin6A=1-3 sin2A*cos2A

Answer 1

sin6A+cos6A = (sin2A)3 + (cos2A)3
                     = (sin2A + cos2A)3 − 3 sin2A cos2A(sin2A + cos2A)  [Since, (a+b)3 = a3 + b3 +3ab(a+b)]
                     = (1)3 − 3 sin2A cos2A(1)  [Since, sin2A + cos2A = 1]
                     =   1 − 3 sin2A cos2A
                     = RHS

Answer 2

Answer:

To prove:

$cos^6A+ sin^6A=1-3 sin^2A.cos^2A$

L.H.S.

$cos^6A+ sin^6A$

$=(cos^2A)^3 + (sin^2 A)^3$

$=(cos^2A+sin^2A)[ (cos^2A)^2-cos^2A\times sin^2A +(sin^2A)^2]$

( Because, a³ + b³ = (a+b) (a²-ab+b²) )

$=(cos^2A)^2-cos^2A\times sin^2A +(sin^2A)^2$

( Because, sin² x + cos²x = 1 )

$=(cos^2A)^2-cos^2A\times sin^2A +(sin^2A)^2+ 2 cos^2A .sin^2A - 2 cos^2A .sin^2A$

$=(cos^2A+sin^2A)^2-3 cos^2A .sin^2A$

( Because, a² + 2ab + b² = (a+b)² )

$=1-3 cos^2A. sin^2A$

= R.H.S.

Hence, proved.