Question

In AABC, C= then prove that
cos^2A + cos^2 B - cosAcosB = 3/4​

Answer 1

Answer:

given

C = 2π/3

now

A + B + C = π

A + B = π - C = π - 2π/3 = π/3

A = π/3 - B

applying cos

cosA = cos(π/3 - B)

cosA = cosπ/3*cosB + sinπ/3*sinB

cosA = (cosB)/2 + (√3/2)sinB

2cosA - cosB = √3sinB

squaring both side

4cos^2A + cos^2B - 4 cosAcosB = 3sin^2B

4cos^2A + cos^2B - 4cosAcosB = 3 -3cos^2B

4cos^2A + 4cos^2B - 4cosAcosB = 3

4(cos^2A + cos^2B - cosAcosB) = 3

cos^2A + cos^2B - cosAcosB = 3/4