Tan theta = √2-1 then Sin theta,Cos theta=??
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Answer:
3sin2θ=2sin3θ
→3×2sincosθ=2(3sinθ−4sin3θ)
→6sinθcosθ=6sinθ−8sin3θ
→sinθ(6cosθ−6+8sin2θ)=0
∴sinθ=0 or 3cosθ+4sin2θ=3
∴sinθ=0 or 3cosθ+4(1−cos2θ)=3
∴sinθ=0 or 3cosθ−4cos2θ+1=0
∴4cos2θ−3cosθ−1=0
∴4cos2θ−4cosθ+cosθ−1=0
∴4cosθ(cosθ−1)+1(cosθ−1)=0
∴(4cosθ+1)(cosθ−1)=0
∴cosθ=4−1 or cosθ=1
∴sinθ=1−(4−1)
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