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