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

Suppose asinA=bsinB=csinC=k . We need to prove that a #160;cos #160;C cos #160;B=2 #160;b c #160;cos2A2. Consider LHS=acosC cosB=ksinAcosC sinAcosB ...

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Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

In Δ A B C, p...

Question

In ∆ABC, prove that $a (\cos C - \cos B) = 2 (b - c) \cos^{2} \frac{A}{2}$ .

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Solution

Suppose $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$ .

We need to prove that $a (\cos C - \cos B) = 2 (b - c) \cos^{2} \frac{A}{2}$ .
Consider

$LHS = a (\cos C - \cos B) = k (\sin A \cos C - \sin A \cos B) = \frac{k}{2} (2 \sin A \cos C - 2 \sin A \cos B) = \frac{k}{2} [\sin (A + C) + \sin (A - C) - \sin (A + B) - \sin (A - B)] = \frac{k}{2} [\sin (π - B) + \sin (A - C) - \sin (π - C) - \sin (A - B)] (∵ A + B + C = π) = \frac{k}{2} [\sin B - \sin C + \sin (A - C) - \sin (A - B)] = \frac{k}{2} [2 \sin (\frac{B - C}{2}) \cos (\frac{B + C}{2}) + 2 \sin (\frac{A - C - A + B}{2}) \cos (\frac{A - C + A - B}{2})] = k \sin (\frac{B - C}{2}) [\cos (\frac{π - A}{2}) + \cos (\frac{2 A - (π - A)}{2})] = k \sin (\frac{B - C}{2}) [\cos (\frac{π - A}{2}) + \cos (\frac{(π - 3 A)}{2})] = k \sin (\frac{B - C}{2}) (\sin \frac{A}{2} + \sin \frac{3 A}{2}) = 2 k \sin (\frac{B - C}{2}) \sin (\frac{\frac{A}{2} + \frac{3 A}{2}}{2}) \cos (\frac{\frac{3 A}{2} - \frac{A}{2}}{2}) = 2 k \sin (\frac{B - C}{2}) \sin A \cos \frac{A}{2} = 4 k \sin (\frac{B - C}{2}) \sin \frac{A}{2} \cos^{2} \frac{A}{2} . . . (1) RHS = 2 (b - c) \cos^{2} \frac{A}{2} = 2 k (\sin B - \sin C) \cos^{2} \frac{A}{2} = 4 k \sin (\frac{B - C}{2}) \cos (\frac{B + C}{2}) \cos^{2} \frac{A}{2} = 4 k \sin (\frac{B - C}{2}) \cos (\frac{π}{2} - \frac{A}{2}) \cos^{2} \frac{A}{2} = 4 k \sin (\frac{B - C}{2}) \sin \frac{A}{2} \cos^{2} \frac{A}{2} = LHS (from equation 1)$

Hence, $a (\cos C - \cos B) = 2 (b - c) \cos^{2} \frac{A}{2}$

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In ∆ABC, prove that a cos C-cos B=2 b-c cos2A2.

In ∆ABC, prove that $a (\cos C - \cos B) = 2 (b - c) \cos^{2} \frac{A}{2}$ .