Question

in a triangle ABC
Σ
cos (B-C) /
sin B sin C=​

Answer 1

Answer:

∑ 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)

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Answer 2

Step-by-step explanation:

hope this helps you dear ♥

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