Question

if A+B+C =pi/2 then prove that cos2A+ cos2B+cos2C = 1+ 4sinAsinBsinC

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Answer 1

Step-by-step explanation:

LHS = cos2A + cos2B + cos2C

==> 2cos(2A + 2B/2) cos(2A - 2B/2) + cos2C

==> 2cos(A + B) cos(A - B) + 2cos²C - 1

==> 2cos(π/2 - C) cos(A - B) 2(1 - sin²C) - 1

==> 2sinC cos(A - B) + 2 - 2sin²C - 1

==> 2sinC cos(A - B) - 2sin²C + 1

==> 2sinC [cos(A - B) - sinC] + 1

==> 2sinC [cos(A - B) - sin(π/2 - A - B)] + 1

==> 2sinC [cos(A - B) - cos(A + B)] + 1

==> 2sinC [-2 sin(A - B + A + B/2)(A - B - A - B/2)] + 1

==> 2sinC [-2 sinA (-sinB)] + 1

==> 1 + 4 sinA sinB sinC ...