Question

in a triangle ABC ,sigma cos (B-C)÷sinBsinC​

Answer 1

∑ Cos(B - C)/SinBSinC = (Sin2A + Sin2B  + Sin2C)/(SinASinBSinC)

Step-by-step explanation:

∑ Cos(B - C)/SinBSinC

= Cos(B - C)/SinBSinC  + Cos(C - A)/SinCSinA  +  Cos(A - B)/SinASinB

= (SinACos(B-C) + SinBCos(C-A) + SinCCos(A - B)) / (SinASinBSinC)

SinA = Sin(180 - (B + C)  = Sin(B + C)

SinB = Sin(A +C)

SinC = Sin(A + B)

= (Sin(B + C)Cos(B-C) + Sin(C + A)Cos(C-A) + Sin(A + B)Cos(A - B)) / (SinASinBSinC)

Now using Sin(x + y)Cos(x - y)  = 1/2 (Sin2x + sin2y)

=  (1/2) (Sin2B + sin2C  + Sin2C + Sin2A  + Sin2A + Sin2B) / (SinASinBSinC)

= (Sin2A + Sin2B  + Sin2C)/(SinASinBSinC)