What is the value of cos [(180 – θ)/2] cos [(180 – 9θ)/2] + sin [(180 – 3θ)/2] sin [(180 – 13θ)/2]?
Sin θ/ 2 . sin9 θ / 2 + cos3 θ / 2 . cos13 θ / 2 Use these two formula :- 2 sin A sin B = cos(A – B) – cos(A + B) 2 cos A cos B = cos(A + B) + cos(A – B) = [cos4 θ – cos5 θ + cos8 θ + cos5 θ] / 2 = (cos4 θ + cos8 θ) / 2 = 2 cos6 θ . cos2 θ / 2 = cos6 θ . cos2 θ Put θ = 00 0 + 1 = 1 Cos6 θ . cos2 θ = 1
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