- a3Sin (B-C)+b3Sin(C-A)+c3Sin ( A-B)= 0
- b2Sin2C+c2Sin2B=2bc SinA
- a(cos B+cos C)=2(b+c)Sin 2A/2
- a3Sin (B-C)+b3Sin(C-A)+c3Sin ( A-B)= 0
- b2Sin2C+c2Sin2B=2bc SinA
- a(cos B+cos C)=2(b+c)Sin 2A/2
Sahdev Singh , 7 Years ago
Grade 11
Trigonometry> a 3 Sin (B-C)+b 3 Sin(C-A)+c 3 Sin ( A-B...
1 Answers
Arun
1.
This is equivalent to proving that ∑8sin³(B + C)sin(B – C) = 0 because a = 2RsinA and A + B + C = π.
8sin³(B + C)sin(B – C) = [6sin(B + C) – 2sin3(B + C)]sin(B – C)
= 3(cos2C – cos2B) – [cos(2B + 4C) – cos(4B + 2C)]
= 3(cos2C – cos2B) – [cos(2C – 2A) – cos(2B – 2A), using 2A + 2B + 2C = 2π.
∑8sin³(B + C)sin(B – C) = [3(cos2C – cos2B) + 3(cos2B – cos2A) + 3(cos2A – cos2C)] – [cos(2C – 2A) – cos(2B – 2A) + cos(2A – 2B) – cos(2C – 2B) + cos(2B – 2C) – cos(2A – 2C)] = 0
→∑8R³sin³Asin(B – C) = ∑a³sin(B – C) = 0.
8sin³(B + C)sin(B – C) = [6sin(B + C) – 2sin3(B + C)]sin(B – C)
= 3(cos2C – cos2B) – [cos(2B + 4C) – cos(4B + 2C)]
= 3(cos2C – cos2B) – [cos(2C – 2A) – cos(2B – 2A), using 2A + 2B + 2C = 2π.
∑8sin³(B + C)sin(B – C) = [3(cos2C – cos2B) + 3(cos2B – cos2A) + 3(cos2A – cos2C)] – [cos(2C – 2A) – cos(2B – 2A) + cos(2A – 2B) – cos(2C – 2B) + cos(2B – 2C) – cos(2A – 2C)] = 0
→∑8R³sin³Asin(B – C) = ∑a³sin(B – C) = 0.
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