Prove that cos (60 + x).cos(60-x) - sin(60 + x) sin(60-x) = -1/2

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Solution

$LHS = \cos (60 ° + x) . \cos (60 ° - x) - \sin (60 ° + x) . \sin (60 ° - x) = \cos [(60 ° + x) + (60 ° - x)] [as, \cos x . \cos y - \sin x . \sin y = \cos (x + y)] = \cos 120 ° = \cos (90 ° + 30 °) = - \sin 30 ° [as, \cos (90 ° + θ) = - \sin θ] = - \frac{1}{2} = RHS$

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