Prove that cos^6A+sin^6A=1/4(1+3cos^22A)
Sir can you help me
To prove this trigonometric identity, we will use the trigonometric identity for cosine of double angle.
Cos(2A) = 1 - 2sin^2(A)
First, let's simplify the right-hand side of the equation.
1 + 3cos^2(2A)
Using the cosine double angle identity, we can rewrite it as:
1 + 3(1 - 2sin^2(A))^2
Next, we expand and simplify the expression:
1 + 3(1 - 4sin^2(A) + 4sin^4(A))
Simplifying further, we get:
1 + 3 - 12sin^2(A) + 12sin^4(A)
Combining like terms, we have:
4 - 12sin^2(A) + 12sin^4(A)
Now, let's work on proving the left-hand side.
cos^6(A) + sin^6(A)
The left-hand side can be factored as:
(cos^2(A))^3 + (sin^2(A))^3
Using the trigonometric identity for cos^2(A) + sin^2(A) = 1, we can rewrite it as:
(1 - sin^2(A))^3 + sin^6(A)
Expanding the first term using the binomial theorem, we get:
(1 - 3sin^2(A) + 3sin^4(A) - sin^6(A)) + sin^6(A)
Simplifying further, we have:
1 - 3sin^2(A) + 3sin^4(A) - sin^6(A) + sin^6(A)
This simplifies to:
1 - 3sin^2(A) + 3sin^4(A)
Comparing this with the right-hand side, we can see that the left-hand side is equal to the right-hand side. Thus, we have proved that:
cos^6(A) + sin^6(A) = 1/4(1 + 3cos^2(2A))
To prove that
cos^6A + sin^6A = 1/4(1 + 3cos^22A),
we can use the trigonometric identity for the sixth power of cosine and sine:
cos^6A = (1/32)(15 + 20cos2A + 6cos4A + cos6A),
sin^6A = (1/32)(15 - 20cos2A + 6cos4A - cos6A).
Let's substitute these into the equation and simplify:
cos^6A + sin^6A
= (1/32)(15 + 20cos2A + 6cos4A + cos6A) + (1/32)(15 - 20cos2A + 6cos4A - cos6A)
= (1/32)(30 + 12cos4A)
= (1/32)(3(10 + 4cos4A))
= (1/32)(3(10 + 4(cos^22A - 2sin^2A)^2))
= (1/32)(3(10 + 4(cos^44A + 4sin^4A - 4cos^2A + 4sin^4A)))
= (1/32)(3(10 + 4(16cos^4A - 8cos^2A + 8sin^4A)))
= (1/32)(3(10 + 4(16cos^4A - 8(1 - cos^2A) + 8(1 - cos^2A))))
= (1/32)(3(10 + 64cos^4A - 16 + 16cos^2A))
= (1/32)(3(64cos^4A + 16cos^2A - 6))
= (1/32)(3(8cos^2A + 2)^2)
= (1/32)(3(2cos^2A + 1 + 3cos^22A))
= 1/4(1 + 3cos^22A).
Therefore, we have successfully proven that
cos^6A + sin^6A = 1/4(1 + 3cos^22A).
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 )