prove that 2(bc cosA+ac cosB)=aa+bb+cc

Question

It is a problem of trigonometric function of 12 std

Answer 1

Test the proposed identity. Let A=B=C=60° so cosA=cosB=cosC=½, a=b=c (equilateral triangle).

2bccosA=a²=2accosB; a²+b²+c²=3a², so the proposition is false, because a²≠3a².

Cosine Rule:

a²=b²+c²-2bccosA; b²=a²+c²-2accosB; c²=a²+b²-2abcosC.

2bccosA=b²+c²-a²; 2accosB=a²+c²-b²,

2bccosA+2accosB=b²+c²-a²+a²+c²-b²=2c²,

bccosA+accosB=c² is a true identity.

Also:

a²+b²+c²=b²+c²-2bccosA+a²+c²-2accosB+a²+b²-2abcosC,

a²+b²+c²=2(a²+b²+c²)-2(bccosA+accosB+abcosC),

2(bccosA+accosB+abcosC)=a²+b²+c² is a true identity and is probably the one intended in the question.