If A+B+C = 180 then prove that cos^2A + cos^2B - cos^2C = 1 - 2 sinA sinB sinC
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given: A+B+C=180 .............(1)
LHS of the given equation:
cos2A+cos2B−cos2C=12.[2cos2A+2cos2B−2cos2C]=12.[1+cos2A+1+cos2B−2cos2C] [since cos2θ=2cos2θ−1=12.[2+cos2A+cos2B−2cos2C]=12.[2+2cos(A+B).cos(A−B)−2cos2C] [since cosC+cosD=2cosC+D2.cosC−D2
=12.[2+2cos(180−C).cos(A−B)−2cos2C]=12.[2−2cosC.cos(A−B)−2cos2C]=1−cosC.[cos(A−B)+cosC]=1−cosC.[cos(A−B)−cos(A+B)] [from eq(1)=1−cosC.[2sinA.sinB]=1−2sinAsinB.cosC=RHS
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LHS of the given equation:
cos2A+cos2B−cos2C=12.[2cos2A+2cos2B−2cos2C]=12.[1+cos2A+1+cos2B−2cos2C] [since cos2θ=2cos2θ−1=12.[2+cos2A+cos2B−2cos2C]=12.[2+2cos(A+B).cos(A−B)−2cos2C] [since cosC+cosD=2cosC+D2.cosC−D2
=12.[2+2cos(180−C).cos(A−B)−2cos2C]=12.[2−2cosC.cos(A−B)−2cos2C]=1−cosC.[cos(A−B)+cosC]=1−cosC.[cos(A−B)−cos(A+B)] [from eq(1)=1−cosC.[2sinA.sinB]=1−2sinAsinB.cosC=RHS
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