If A + B + C = \(3\pi\over 2\), then cos2A + cos2B + cos2C is equal to

Solution :

cos2A + cos2B + cos2C = 2cos(A+B)cos(A-B) + cos2C

= 2cos(\(3\pi\over 2\) – C)cos(A-B) + cos2C  \(\because\)  A + B + C = \(3\pi\over 2\)

= -2sinC cos(A-B) + 1 – 2\(sin^2C\) = 1 – 2sinC[cos(A-B)+sinC]

= 1 – 2sinC[cos(A-B) + sin(\(3\pi\over 2\)-(A+B))]

= 1 – 2sinC[cos(A-B)-cos(A+B)]

= 1 – 4sinA sinB sinC


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