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

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`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)...

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`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)`

Approved by eNotes Editorial Team