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

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jeew-m's profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

`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^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 `


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


Top Answer

llltkl's profile pic

llltkl | College Teacher | (Level 3) Valedictorian

Posted on


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


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


Now, simplify and rearrange







` 2sinAcosA=sin2A`




Hence, proved.


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