4- bc -cos 2A-2-c a cos 2B-2-a b cos 2C-2-a-b-c2

4(bc cos^2 A/2+ca cos^2 B/2+ab cos^2 C/2 )=(a+b+c )^2 LHS, 4(bc cos^2 A/2+ca cos^2 B/2+ab cos^2 C/2 ) =2(bc.2cos^2 A/2+ca.2cos^2 B/2+ab.2cos^2 C/2 ) =2(bc.( ...

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Byju's Answer

Standard XII

Mathematics

Cosine Rule

4. bc .cos 2A...

Question

$4 (b c c o s^{2} \frac{A}{2} + c a c o s^{2} \frac{B}{2} + a b c o s^{2} \frac{C}{2}) = {(a + b + c)}^{2}$

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Solution

$4 (b c c o s^{2} \frac{A}{2} + c a c o s^{2} \frac{B}{2} + a b c o s^{2} \frac{C}{2}) = {(a + b + c)}^{2} L H S, 4 (b c c o s^{2} \frac{A}{2} + c a c o s^{2} \frac{B}{2} + a b c o s^{2} \frac{C}{2}) = 2 (b c .2 c o s^{2} \frac{A}{2} + c a .2 c o s^{2} \frac{B}{2} + a b .2 c o s^{2} \frac{C}{2}) = 2 (b c . (1 - c o s A) + c a . (1 - c o s B) + a b . (1 - c o s C)) = 2 b c - 2 b c c o s A + 2 c a - 2 c a c o s B + 2 a b - 2 a b c o s C = 2 b c - 2 b c \frac{b^{2} + c^{2} - a^{2}}{2 b c} + 2 c a - 2 c a \frac{a^{2} + c^{2} - b^{2}}{2 c a} + 2 a b - 2 a b (\frac{b^{2} + a^{2} - c^{2}}{2 a b}) [c o s r u l e] = 2 b c - b^{2} - c^{2} + a^{2} + 2 c a - a^{2} - c^{2} + b^{2} + 2 a b - b^{2} - a^{2} + c^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a = {(a + b + c)}^{2} = R H S .$

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4(bc cos2 A2+ca cos2 B2+ab cos2 C2)=(a+b+c)2

$4 (b c c o s^{2} \frac{A}{2} + c a c o s^{2} \frac{B}{2} + a b c o s^{2} \frac{C}{2}) = {(a + b + c)}^{2}$