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प्रश्न
If cos θ + cos2 θ = 1, prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1
उत्तर
Given `cos theta + cos^2 theta = 1`
We have to prove sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1
From the given equation, we have
`cos theta + cos^2 theta = 1`
`=> cos theta = 1 - cos^2 theta`
`=> ccos theta = sin^2 theta`
`=> sin^2 theta = cos theta`
Therefore, we have
sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2
`= (sin^12 theta + 3 sin^10 theta + 3 sin^8 theta + sin^6 theta) + (2 sin^4 theta + 2 sin^2 theta) - 2`
`= {(sin^4 theta)^3 + 3(sin^4 theta)^2 sin^2 theta + 3 sin^4 theta(sin^2 theta)^2 + (sin^2 theta)^3} + 2(sin^4 theta + sin^2 theta) - 2`
`= (sin^4 theta + sin^2 theta)^3 + 2 (sin^4 theta + sin^2 theta) - 2`
`= (cos^2 theta + cos theta)^3 + 2 (cos^2 theta + cos theta) - 2`
= (1)^3 + 2(1) - 2
= 1
hence proved
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