In a - ABC- prove that a cos C -cos B -2 b - c cos 2 A -2-

Let a=k sin A a(cos C cos B )=2(b c )cos^2A/2 LHS =a(cos C cos B ) =a2.sinC+B/2.sinB C/2 =2k sin A sinπ A/2.sinB C/2 =2k2 sinA/2.cosA/2.cosA/2.sinB C/2 ...

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Standard XII

Mathematics

Sine Rule

In a Δ ABC, p...

Question

$I n a Δ A B C, prove that a (c o s C - c o s B) = 2 (b - c) c o s^{2} \frac{A}{2} .$

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Solution

$L e t a = k s i n A a (c o s C - c o s B) = 2 (b - c) c o s^{2} \frac{A}{2} L H S = a (c o s C - c o s B) = a 2. s i n \frac{C + B}{2} . s i n \frac{B - C}{2} = 2 k s i n A s i n \frac{π - A}{2} . s i n \frac{B - C}{2} = 2 k 2 s i n \frac{A}{2} . c o s \frac{A}{2} . c o s \frac{A}{2} . s i n \frac{B - C}{2} = 2 K 2 c o s^{2} \frac{A}{2} (2 s i n \frac{B - C}{2} . s i n \frac{A}{2}) = 2 k c o s^{2} \frac{A}{2} (2 s i n \frac{B - C}{2} . s i n \frac{π - (B - C)}{2}) = 2 K c o s^{2} \frac{A}{2} (s i n B - s i n C) = 2 c o s^{2} \frac{A}{2} (k s i n B - k s i n C) = 2 c o s^{2} \frac{A}{2} (b - c) = R H S .$

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

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In a ΔABC, prove that a(cos C−cos B)=2(b−c)cos2A2.

Let a=k sin Aa(cos C−cos B)=2(b−c)cos2A2LHS=a(cos C−cos B)=a2.sinC+B2.sinB−C2=2k sin A sinπ−A2.sinB−C2=2k2 sinA2.cosA2.cosA2.sinB−C2=2K2 cos2A2(2sinB−C2.sinA2)=2k cos2A2(2sinB−C2.sinπ−(B−C)2)=2K cos2A2(sin B−sin C)=2 cos2A2 (k sin B−k sin C)=2 cos2A2 (b−c)=RHS.

$I n a Δ A B C, prove that a (c o s C - c o s B) = 2 (b - c) c o s^{2} \frac{A}{2} .$