Question

Prove $(b+c)\cos \frac{B+C}{2}=a\cos \frac{B-C}{2}$

Answer 1

In any triangle with the sides $a,b,c$ and $A,B,C$ are the angles respectively,
the sine Formula is $\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=K$

$a=KsinA,b=KsinB,c=KsinC$
And sum of the angles of a triangle $={180}^0$
it means $A+B+C=180$,
$A=180-(B+C)$,
$B+C=180-A$
now, we have to prove,
$(b+c)cos\frac{B+C}{2}=cos\frac{B-C}{2}$
we can write this as,
$\frac{b+c}{a}=\frac{cos\frac{B-C}{2}}{cos\frac{B+C}{2}}$

So, L.H.S, $\frac{b+c}{a}=\frac{KsinB+KsinC}{KsinA}$

From,$sinX+sinY=2sin\frac{X+Y}{2}\ast cos\frac{X-Y}{2}$ and $sin2\theta =2sin\theta \ast cos\theta$

$\therefore \frac{b+c}{a}=\frac{2Ksin\frac{B+C}{2}\ast cos\frac{B-C}{2}}{KsinA}$

$=\frac{2sin\frac{180-A}{2}\ast cos\frac{B-C}{2}}{2sin\frac{A}{2}\ast cos\frac{A}{2}}$

$=\frac{cos\frac{A}{2}\ast cos\frac{B-C}{2}}{sin\frac{A}{2}\ast cos\frac{A}{2}}$

$=\frac{cos\frac{B-C}{2}}{sin\frac{180-(B+C)}{2}}$

$\frac{b+c}{a}=\frac{cos\frac{B-C}{2}}{cos\frac{B+C}{2}}$
HENCE,
$(b+c)cos\frac{B+C}{2}=acos\frac{B-C}{2}$