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Question
If cosθ + sinθ = `sqrt2` cosθ, show that cosθ - sinθ = `sqrt2` sinθ.
Solution 1
(cosθ + sinθ)2 = (`sqrt2`. cosθ)2
cos2θ + sin2θ + 2.cosθ.sinθ = 2cos2θ
1 + 2.cosθ.sinθ = 2cos2θ
2.cosθ.sinθ = 2cos2θ − 1
(cosθ.sinθ)2 = cos2θ + sin2θ − 2.cosθ.sinθ
= 1 − (2.cos2θ − 1)
= 1 − 2.cos2θ +1
= 2 − 2.cos2θ
= 2(1 − cos2θ)
cosθ − sinθ = `sqrt(2sin^2θ)`
= `sqrt2`sinθ
Hence proved
Solution 2
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