If (cos θ + sin θ) = √2sin θ, prove that (sin θ – cos θ) = √2cos θ.

Given: cos θ + sin θ = √2sin θ


Consider (sin θ + cos θ)2 + (sin θ – cos θ)2 = sin2 θ + cos2 θ + 2sin θ cos θ + sin2 θ + cos2 θ –


 


2 sin θ cos θ


 


(sin θ + cos θ)2 + (sin θ – cos θ)2 = 2sin2 θ + 2 cos2 θ


 


(sin θ + cos θ)2 + (sin θ – cos θ)2 = 2(sin2 θ + cos2 θ)


 


(sin θ + cos θ)2 + (sin θ – cos θ)2 = 2


 


(√2sin θ)2 + (sin θ – cos θ)2 = 2


 


(sin θ – cos θ)2 = 2 – 2sin2 θ


 


(sin θ – cos θ)2 = 2(1 – sin2 θ)


 


(sin θ – cos θ)2 = 2(cos2 θ)


 


(sin θ – cos θ) = ± √cos θ


 


Hence, proved.


 

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