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# How does cos^2A+cos^2B+cos^2C+2cosAcosBcosC=1,where A,B,C are angles of triangles?

69 Points
10 years ago

Dear Aparna Tiwari,

2cosAcosBcosC =(2cosAcosB)cosC                               here A+B+C=pi=∏       thereforen  C=pi-(A+B)
=(cos(A + B) + cos(A - B))cosC
=cos(pi - C)cosC + cos(A - B)cos(pi - (A + B))
=-2cos^2C - 2cos(A + B)cos(A - B)

=-cos^2C - cos^2A + sin^2B

cos^2A+cos^2B+cos^2C+2cosAcosBcosC=cos^2A+cos^2B+cos^2C-cos^2C - cos^2A + sin^2B

=   cos^2B+ sin^2B =1

All the best.

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IIT Bombay

Lab Bhattacharjee
121 Points
3 years ago
As cos(A+B)=cos(pi-C)=-cos Cπ2

cos^2A+cos^2B+cos^2C=1+\cos^2A-\sin^2B+\cos^2C=1+cos(A+B)cos(A-B)+cos^2C=1-cos C cos(A-B)-cos C cos(A+B)

=1-cos C[cos(A+B)+cos(A-B)]=1-cos C(2ocs Acos B)