# Prove the identity sin^6 A +cos ^6 A =1-3/4 sin^2 2A

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### 2 Answers

`sin^6A = (sin^2A)^3`

`cos^6A = (cos^2A)^3`

We know that;

`x^3+y^3 = (x+y)(x^2-xy+y^2)`

`sin^6A+cos^6A = (sin^2A+cos^A)(sin^4A-sin^2Acos^2A+cos^4A)`

We can write;

`x^2+y^2 = (x+y)^2-2xy`

`sin^4A+cos^4A = (sin^2A+cos^2A)^2-2sin^2Acos^2A`

We know that `sin^2A+cos^A = 1`

`sin^6A+cos^6A`

`= (sin^2A+cos^A)(sin^4A-sin^2Acos^2A+cos^4A)`

`= 1xx((sin^2A+cos^2A)^2-2sin^2Acos^2A-sin^2Acos^2A)`

`= 1xx(1-3sin^2Acos^2A)`

`sin2A = 2sinAcosA`

`(sin2A)^2 = 4sin^2Acos^2A`

`3/4sin^2(2A) = 3sin^2Acos^2A`

So this gives the answer as;

`sin^6A+cos^6A `

`=(1-3sin^2Acos^2A)`

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

**Sources:**

L.H.S.=`sin^(6)A+cos^(6)A`

Using the identity `sin^(2)x=1-cos^(2)x`

`L.H.S.={1-cos^(2)A}^3+cos^(6)A`

now expand using `(a-b)^(3)=a^3-b^3-3a^2b+3ab^2`

`=1-cos^6A-3cos^2A+3cos^4A+cos^6A`

Now, simplify and rearrange

`1+3cos^4A-3cos^2A`

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

`=1+3cos^2A*-sin^2A`

`=1-3sin^2Acos^2A`

`=1-3/4.(4sin^2Acos^2A)`

`=1-3/4*(2sinAcosA)^2`

` 2sinAcosA=sin2A`

`1-3/4(sin2A)^2`

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

=R.H.S.

**Hence, proved.**

**Sources:**