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Prove that cos theta.cos theta/2-cos 3 theta.cos9 theta/2= sin 7 theta.sin 8 theta |
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Answer» -step explanation:TEXT SolutionSolution : LHS = cosθcosθ2−cos3θcos9θ2 =12[2cosθ⋅cosθ2−2cos3θ⋅cos9θ2] =12[cos(θ+θ2)+cos(θ−θ2)−cos(3θ+9θ2)−cos(3θ−9θ2)] 12(cos3θ2+cosθ2−cos15θ2−cos3θ2 =12[cosθ2−cos15θ2] =−12[2sin(θ+15θ2)⋅sin(θ−15θ2)] [∵cosx−cosy=−2sinx+y2⋅sinx−y2] =+(sin8θ⋅sin7θ)=RHS∴ LHS=RHS HENCE proved. |
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