Question

Prove that:- cos(A-B/2)=(A+B/C)sinC/2

Answer 1

Answer:

$\dfrac{A+B}{C}$ = $\dfrac{Cos \frac{(A-B)}{2} }{Sin\frac{C}{2} }$   proved

Step-by-step explanation:

Given as :

To prove :- $\dfrac{A+B}{C}$ = $\dfrac{Cos \frac{(A-B)}{2} }{Sin\frac{C}{2} }$

Let $\dfrac{A}{Sin A}$  = $\dfrac{B}{Sin B} =\dfrac{C}{Sin C}$  = k

So , A = k Sin A

     B = k Sin B

     C = k Sin C

From Left hand side

     $\dfrac{A+B}{C}$  =  $\dfrac{k Sin A + k Sin B}{k SinC}$

i.e              =  $\dfrac{Sin A + Sin B}{Sin C}$

                 = $\dfrac{2 Sin(\frac{A+B}{2}) Cos (\frac{A-B}{2}) }{2Sin(\frac{C}{2})Cos (\frac{C}{2}) }$

                 = $\dfrac{2 Sin(90- \frac{C}{2}) Cos (\frac{A-B}{2}) }{2Sin(\frac{C}{2})Cos (\frac{C}{2}) }$           ( ∵ A + B + C = 180° )

                 =   $\dfrac{2Cos (\frac{C}{2}) Cos (\frac{A-B}{2}) }{2Sin(\frac{C}{2})Cos (\frac{C}{2}) }$

                 =     $\dfrac{Cos \frac{(A-B)}{2} }{Sin\frac{C}{2} }$    Proved

So, LHS = RHS

Hence,  $\dfrac{A+B}{C}$ = $\dfrac{Cos \frac{(A-B)}{2} }{Sin\frac{C}{2} }$   proved   Answer