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Question
If sin θ + cos θ = m, show that cos6θ + sin6θ = `(4 - 3("m"^2 - 1)^2)/4`, where m2 ≤ 2
Solution
sin θ + cos θ = m
(sin θ + cos θ)2 = m2
sin2θ + cos2θ + 2 sin θ cos θ = m2
1 + 2 sin θ cos θ = m2
2 sin θ cos θ = m2 – 1
sin θ cos θ = `("m"^2 - 1)/2` ......(1)
cos6θ + sin6θ = (cos2θ)3 + (sin2θ)3
= (cos2θ) + sin2θ) ....`[(cos^2θ)^2 - cos^2θ*sin^2θ + (sin^2θ)^2]`
= (cos2θ)2 – cos2θ sin2θ + (sin2θ)2
= (cos2θ + sin2θ)2 – 2 cos2θ sin2θ – cos2θ sin2θ
= (cos2θ + sin2θ)2 – 3 cos2θ sin2θ
= 1 – 3 cos2θ sin2θ
= `1 - 3(("m"^2 - 1)/2)^2`
= `1 - 3 ("m"^2 - 1)^2/4`
cos6θ + sin6θ = `(4 - 3("m"^2 - 1)^2)/4`
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