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Question

7)Prove that Cos 2π/3 cos π/3 – Sin 2π/3 sin π/3 = -(√3+1)/2√2

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Solution

Note: In your question R.H.S. should be -1 instead of -(√3+1)/2√2 .
L.H.S.=cos2π3cosπ3-sin2π3sinπ3 =cos2π3+π3 Using the formula, cosAcosB-sinAsinB=cosA+B =cos3π3 =cos π =-1 =R.H.S.Hence Proved.

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