Prove that cos^6A+sin^6A=1/4(1+3cos^22A)

Answer 1

cos ^ 2 A = 1 - sin ^ 2 A

cos ^ 6 A + sin ^ 6 A =

( 1 - sin ^ 2 A ) ^ 3 + ( sin ^ 2 A ) ^ 3 =

_________________________________Remark :

( a - b ) ^ 3 = a ^ 3 - 3 a ^ 2 b + 3 a b ^ 2 - b ^ 3

In case ( 1 - sin ^ 2 A ) ^ 3

a = 1, b = sin ^ 2 A so :
_________________________________

( 1 - sin ^ 2 A ) ^ 3 + ( sin ^ 2 A ) ^ 3 =

1 ^ 3 - 3 * 1 ^ 2 * sin ^ 2 A + 3 * 1 * [ sin ^ 2 A ] ^ 2 - [ sin ^ 2 A ] ^ 3 + ( sin ^ 2 A ) ^ 3 =

1 - 3 sin ^ 2 A + 3 sin ^ 4 A - sin ^ 6 A + sin ^ 6 A =

1 - 3 sin ^ 2 A + 3 sin ^ 4 A =

1 + 3 sin ^ 2 A ( - 1 + sin ^ 2 A ) =

______________
Remark :

cos ^ 2 A = 1 - sin ^ 2 A

so :

( - 1 + sin ^ 2 A ) =

( - 1 ) * ( 1 - sin ^ 2 A ) =

- cos ^ 2 A
____________

1 + 3 sin ^ 2 A ( - 1 + sin ^ 2 A ) =

1 + 3 sin ^ 2 A ( - cos ^ 2 A ) =

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

1 - 3 [ ( sin A * cos A ) ] ^ 2 =

____________________________
Remark :

2 sin A cos A = sin 2A

so :

sin A cos A = ( 1 / 2 ) sin 2A
_____________________________

1 - 3 [ ( sin A * cos A ) ] ^ 2 =

1 - 3 [ ( 1 / 2 ) sin 2 A ] ^ 2 =

1 - 3 * ( 1 / 2 ) ^ 2 * [ sin 2A ] ^ 2 =

1 - 3 * ( 1 / 4 ) * [ sin 2A ] ^ 2 =

1 - ( 3 / 4 ) * sin ^ 2 2A =

1 - ( 3 / 4 ) * ( 1 - cos ^ 2 2A ) =

1 - ( 3 / 4 ) * 1 - ( 3 / 4 ) * ( - cos ^ 2 2A ) =

1 - 3 / 4 + ( 3 / 4 ) cos ^ 2 2A =

4 / 4 - 3 / 4 + ( 3 / 4 ) cos ^ 2 2A =

1 / 4 + ( 3 / 4 ) cos ^ 2 2A =

( 1 / 4 ) ( 1 + 3 cos ^ 2 2A )

Answer 2

Sir can you help me