Math, asked by gurugavale14, 20 days ago

sin (theta -π/6) + cos(theta-π/3) =√3 sin theta​

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Answered by ITZURADITYAKING
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3sin2θ=2sin3θ

3sin2θ=2sin3θ→3×2sincosθ=2(3sinθ−4sin3θ)

3sin2θ=2sin3θ→3×2sincosθ=2(3sinθ−4sin3θ)→6sinθcosθ=6sinθ−8sin3θ

3sin2θ=2sin3θ→3×2sincosθ=2(3sinθ−4sin3θ)→6sinθcosθ=6sinθ−8sin3θ→sinθ(6cosθ−6+8sin2θ)=0

3sin2θ=2sin3θ→3×2sincosθ=2(3sinθ−4sin3θ)→6sinθcosθ=6sinθ−8sin3θ→sinθ(6cosθ−6+8sin2θ)=0∴sinθ=0 or 3cosθ+4sin2θ=3

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

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

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

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 

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

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

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

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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