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in triangle ABC, sin3A.cos(B-C)+sin3B.cos(C-A)+sin3C.cos(A-B)=
5 years ago

Answers : (1)

Arun Kumar
IIT Delhi
askIITians Faculty
256 Points
							
Hello
correct question
Sin3A 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 & Regards
Arun Kumar
Btech IIT delhi
Askiitians Faculty
5 years ago
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