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If cosθ + sinθ = √2cosθ Prove that: cosθ - sinθ = √2sinθ
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cosθ + sinθ = √2cosθ squaring both sides we get : cos^2θ + sin^2θ+2cosθsinθ = 2cos^2(θ) => 1+ 2cosθsinθ = 2cos^2(θ) => 2cosθsinθ = 2cos^2(θ)-1 now cosθ - sinθ ( squaring ) we get : 1-2cosθsinθ = 1-(2cos^2(θ)-1) = 1+1- 2cos^2(θ) = 2- 2cos^2(θ) = 2[(1-cos^2(θ)] = 2sin^2θ =>(cosθ - sinθ)^2=2sin^2θ ( cancelling square on L.H.S) we get cosθ - sinθ = √2sinθ ( hence proved )
thanx
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