In ABC- prove the following-4 b c cos 2 A 2-c a cos 2 B 2-a b cos 2 C 2-a-b-c 2

LHS=4bc #160;cos2A2+ca #160;cos2B2+ab #160;cos2 #160;C2=4bc1+cosA2+ca1+cosB2+ab1+cosC2=2bc+2bccosA+2ca+2cacosB+2ab+2abcosC=2ab+bc+ca+2bcb2+c2 a22bc+2cac ...

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Byju's Answer

Standard XI

Economics

Relationship between the Short Run Average and Marginal Cost Curves

In ABC, prove...

Question

In ∆ABC, prove the following:
$4 (b c \cos^{2} \frac{A}{2} + c a \cos^{2} \frac{B}{2} + a b \cos^{2} \frac{C}{2}) = {(a + b + c)}^{2}$

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Solution

$LHS = 4 (b c \cos^{2} \frac{A}{2} + c a \cos^{2} \frac{B}{2} + a b \cos^{2} \frac{C}{2}) = 4 [b c (\frac{1 + \cos A}{2}) + c a (\frac{1 + \cos B}{2}) + a b (\frac{1 + \cos C}{2})] = 2 b c + 2 b c \cos A + 2 c a + 2 c a \cos B + 2 a b + 2 a b \cos C = 2 (a b + b c + c a) + 2 b c (\frac{b^{2} + c^{2} - a^{2}}{2 b c}) + 2 c a (\frac{c^{2} + a^{2} - b^{2}}{2 c a}) + 2 a b (\frac{a^{2} + b^{2} - c^{2}}{2 a b})$

$= 2 (a b + b c + a c) + b^{2} + c^{2} - a^{2} + c^{2} + a^{2} - b^{2} + a^{2} + b^{2} - c^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c = {(a + b + c)}^{2} = RHS$

Hence proved.

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Similar questions

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In ∆ABC, prove the following: 4bc cos2A2+ca cos2B2+ab cos2 C2=a+b+c2

LHS=4bc cos2A2+ca cos2B2+ab cos2 C2=4bc1+cosA2+ca1+cosB2+ab1+cosC2=2bc+2bccosA+2ca+2cacosB+2ab+2abcosC=2ab+bc+ca+2bcb2+c2-a22bc+2cac2+a2-b22ca+2aba2+b2-c22ab =2ab+bc+ac+b2+c2-a2+c2+a2-b2+a2+b2-c2=a2+b2+c2+2ab+2bc+2ac=a+b+c2=RHS Hence proved.

In ∆ABC, prove the following:
$4 (b c \cos^{2} \frac{A}{2} + c a \cos^{2} \frac{B}{2} + a b \cos^{2} \frac{C}{2}) = {(a + b + c)}^{2}$