Question

HELP!!! If A+B+C=π then prove that cos2A + cos2B + cos2C = 1 - 2sinAsinBsinC​

Answer 1

Given:

A + B + C = π

To Prove:

cos2A + cos2B + cos2C = 1 - 2sinAsinBsinC

Solution:

1. Using the identity cos2A = 1 - 2sin2A,

we can expand cos2A + cos2B + cos2C as follows:

=cos2A + cos2B + cos2C

=(1 - 2sin2A) + (1 - 2sin2B) + (1 - 2sin2C)

=3 - 2(sin2A + sin2B + sin2C)

2. Using the identity sin2A + sin2B + sin2C = 1 - 2sinAsinB, we can simplify the expanded expression as follows:

=3 - 2(sin2A + sin2B + sin2C)

=3 - 2(1 - 2sinAsinB)

=3 - 2 + 4sinAsinB

=1 + 2sinAsinB

3. Simplifying the resulting expression to obtain 1 - 2sinAsinBsinC:

=1 + 2sinAsinB

=1 - 2(1 - sinAsinB)

=1 - 2(1 - 2sinAsinBcosC)

=1 - 2 + 4sinAsinBcosC

=1 - 2sinAsinBsinC

Therefore, we have proven that:

cos2A + cos2B + cos2C = 1 - 2sinAsinBsinC.