Question

In ΔABC, prove that $(b + c)\ cos(\frac{B + C}{2})\ =\ a\ cos(\frac{B - C}{2})$.

Answer 1

Answer:

Step-by-step explanation:

Formula used:

In triangle ABC,

(a/SinA)=(b/sinB)=(c/sinC)=2R

In triangle ABC, A+B+C=180

sin2A= 2 sinA cosA

$=(b+c)\:cos(\frac{B+C}{2})$

$=(2R.sinB+2R.sinC)\:cos(\frac{B+C}{2})$

$=2R(sinB+sinC)\:cos(\frac{B+C}{2})$

$=2R[2\:sin(\frac{B+C}{2}).cos(\frac{B-C}{2})]\:cos(\frac{B+C}{2})$

$=2R[2\:sin(\frac{B+C}{2}).cos(\frac{B+C}{2})]\:cos(\frac{B-C}{2})$

$=2R[\:sin(2\frac{(B+C)}{2})]\:cos(\frac{B-C}{2})$

$=2R[\:sin(B+C)]\:cos(\frac{B-C}{2})$

$=2R[\:sin(180-A)]\:cos(\frac{B-C}{2})$

$=[2RsinA]\:cos(\frac{B-C}{2})$

$=a\:cos(\frac{B-C}{2})$