if cos Theta-sin theta =√2sin theta,prove that cos theta+sin theta=√2cosTheta
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Answered by
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Let theta be A
CosA - SinA = √2SinA
CosA = √2SinA + SinA
CosA = SinA (√2 + 1)
CosA / (√2 + 1) = SinA
On rationalizing we get
CosA(√2 - 1) / 2 - 1 = SinA
√2CosA - CosA = SinA
√2CosA = SinA + CosA
Hence Proved.
CosA - SinA = √2SinA
CosA = √2SinA + SinA
CosA = SinA (√2 + 1)
CosA / (√2 + 1) = SinA
On rationalizing we get
CosA(√2 - 1) / 2 - 1 = SinA
√2CosA - CosA = SinA
√2CosA = SinA + CosA
Hence Proved.
Answered by
1
sA+sinA=√2cosA=>cos2A+sin2A+2cosAsinA=2cos2A=>−cos2A+2cosAsinA−sin2a=−2sin2A=>−(cosA−sinA)2=−2sin2A=>cosA−sinA=√2sinA
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