Question

A,B,C are angles of triangle, prove that

cos 2a cos 2b cos 2c=-4cosa cosb cosc- 1​

Answer 1

Given: A+B+C=π

L.H.S=cos2A+cos2B+cos2C

=[2cos(A+B)cos(A−B+(2cos2C−1).....cosC+cosD

=2cos 2C+D cos 2C−D

=2cos(180−C)cos(A−B)+2cos 2 (C−1)

=−2cosC(cos(A−B)−cosC)−1

=−2cosC(cos(A−B)+cos(A+B))−1

=−2cosC(2cosAcosB)−1

=−4cosAcosBcosC−1