Question

in any trianangle ABC prove that :- a(Cos C - Cos B) = 2 (b-c)cos^2 A/2​

Answer 1

Answer:

To Prove a(cosC−cosB) = 2(b−c) $\frac{cos ^{2}A }{2}$ , in a given triangle with sides a,b and c also angles are A,B and C

soln:

We know that in a triangle have sides a,b&c and angles A,B&C,

 $\frac{a}{sin A}$ = $\frac{b}{sin B}$ = $\frac{c}{sin C}$ = k (some constant)   ---------------------- (1)

also we have A+B+C = π (ie, 180°)  ---------------------------- (2)

from the statement take the LHS

LHS  = a(cosC−cosB)

        = k sinA [ - 2 sin($\frac{C+B}{2}$) sin($\frac{C-B}{2}$) ]     [ by using (1) & CosA - CosB half angle formula ]

       = k sinA [ - 2 sin($\frac{C+B}{2}$) sin($\frac{C-B}{2}$) ]

       = -2k * 2 sin($\frac{A}{2}$) * cos ($\frac{A}{2}$) [ sin($\frac{\pi-A }{2}$) sin($\frac{C-B}{2}$) ]   [ using (2) and sin 2A formula ]

       = -4k sin($\frac{\pi-(B+C)}{2}$) . cos ($\frac{A}{2}$). [ cos($\frac{A}{2}$) . sin($\frac{C-B}{2}$) ]  [ using sin(90-α) & (2) ]

       = 4k cos² ($\frac{A}{2}$).cos($\frac{B+C}{2}$). sin($\frac{B-C}{2}$)   [ using - sin A = sin (-A) ]

       = 4k cos² ($\frac{A}{2}$).cos($\frac{B+C}{2}$). sin($\frac{B-C}{2}$)

       = 2k cos² ($\frac{A}{2}$). [ 2.cos($\frac{B+C}{2}$). sin($\frac{B-C}{2}$)  ]

       = 2k cos² ($\frac{A}{2}$). [ sin($\frac{B+C}{2}$ + $\frac{B-C}{2}$) - sin($\frac{B+C}{2}$ - $\frac{B-C}{2}$)  ]        [ Using formula, 2cosA.sinB = sin(A+B)-sin(A-B) ]

       = 2k cos² ($\frac{A}{2}$). [ sinB- sinC ]

       = 2cos² ($\frac{A}{2}$). [ k sinB- k sinC ]

       = 2cos² ($\frac{A}{2}$). [ b - c ]    [ using (1) ]

       = 2(b-c)cos² ($\frac{A}{2}$)

       = RHS

ie, LHS = RHS

=> a(cosC−cosB)  = 2(b-c)cos² ($\frac{A}{2}$)

Hence proved