Question

If sin θ + cos θ = x, prove that  `sin^6 theta + cos^6 theta = (4- 3(x^2 - 1)^2)/4`

Answer 1

Given `sin theta + cos theta = x`

Squaring the given equation, we have

`(sin theta + cos theta)^2 = x^2`

`=> sin^2 theta + 2 sin theta cos theta = cos^2 theta = x^2`                                   

`=> (sin^2 theta + cos^2 theta) + 2sin theta cos theta = x^2`

`=> 1 + 2 sin theta cos theta = x^2`

`=> 2 sin theta cos theta = x^2  -1`

`=> sin theta cos theta = (x^2- 1)/2`

Squaring the last equation, we have

`(sin theta cos theta)^2 = (x^2  - 1)^2/4` 

`=> sin^2 theta cos^2 theta = (x^2 - 1)^2/4`

`=> sin^2 theta cos^2 theta = (s^2 -1)/4`

Therefore, we have

`sin^6 theta + cos^6 theta = (sin^2 theta)^3 + (cos^2 theta)^3`

`= (sin^2 theta + cos^2 theta)^3 -  3sin^3 theta cos^2 theta (sin^2 theta + cos^2 theta)`

`= (1)^3 - 3 ((x^2 - 1)^2)/4 (1)`

`= 1 - 3 (x^2 - 1)^2/4 (1)`

`x = 1 - 3 (x^2 - 1)^2/4`

`= (4- 3(x^2 - 1)^2)/4`

hence Proved