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`        in triangle ABC, sin3A.cos(B-C)+sin3B.cos(C-A)+sin3C.cos(A-B)=`
5 years ago Arun Kumar
IIT Delhi
256 Points
```							Hellocorrect questionSin3A Cos(B-C)+Sin3B Cos(C-A)+Sin3C Cos(A-B)=3 SinA SinB SinC. = sin2A sin(B + C) cos(B - C) + sin2B sin(C + A) cos(C - A) + sin2C sin(A + B) cos(A - B)= 1/2sin2A (sin2B + sin2C) + 1/2sin2B (sin2C + sin2A) + 1/2sin2C (sin2A + sin2B)= sin2A (sinB cosB + sinC cosC) + sin2B (sinC cosC + sinA cosA) + sin2C (sinA cosA + sinB cosB)= sinA sinB (sinA cosB + cosA sinB) + sinB sinC (sinB cosC + cosB sinC) + sinC sinA (sinA cosC + cosA sinC)= sinA sinB sin(A + B) + sinB sinC sin(B + C) + sinC sinA sin(A + C)= 3sinA sinB sinC =Thanks & RegardsArun KumarBtech IIT delhiAskiitians Faculty
```
5 years ago
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