Prove 1 sin alpha+sin beta+sin gama---sin(alpha+beta+gama)=4 sin(alpha+beta/2) sin(beta+gama/2) sin(gama+alpha/2) - Maths - Trigonometric Functions

Question

Prove 1 sin alpha+sin beta+sin gama---sin(alpha+beta+gama)=4
sin(alpha+beta/2) sin(beta+gama/2) sin(gama+alpha/2)
2 (sin3Acos4A--sinAcos2A) / sin4AsinA+cos6AcosA) = tan 2A
3 cos3 A +cos3(120+A) + cos3
(240+A) =3/4 cos 3A

Ankush Jain · Accepted Answer

1 To prove: $\sin α + \sin β + \sin γ - \sin (α + β + γ) = 4 \sin (\frac{α + β}{2}) \sin (\frac{β + γ}{2}) \sin (\frac{γ + α}{2})$ Proof: LHS $= \sin α + \sin β + \sin γ - \sin (α + β + γ) = 2 \sin (\frac{α + β}{2}) \cos (\frac{α - β}{2}) + 2 \sin (\frac{γ - (α + β + γ)}{2}) \cos (\frac{α + β + 2 γ}{2}) = 2 \sin (\frac{α + β}{2}) [\cos (\frac{α - β}{2}) - \cos (\frac{α + β + 2 γ}{2})] = 2 \sin (\frac{α + β}{2}) [(- 2) \sin (\frac{\frac{α - β + α + β + 2 γ}{2}}{2}) \sin (\frac{\frac{α - β - α - β - 2 γ}{2}}{2})] = 4 \sin (\frac{α + β}{2}) [\sin (\frac{α + γ}{2}) \sin (\frac{β + γ}{2})] = R H S [H e n c e p r o v e d]$ 3 To prove: $\cos^{3} A + \cos^{3} (120 + A) + \cos^{3} (240 + A) = \frac{3}{4} \cos 3 A$ Proof: We know that $\cos 3 A = 4 \cos^{3} A - 3 \cos A \Rightarrow \cos^{3} A = \frac{1}{4} (\cos 3 A + 3 \cos A)$ LHS $= \frac{1}{4} (\cos 3 A + 3 \cos A) + \frac{1}{4} [\cos (360 + 3 A) + 3 \cos (120 + A)] + \frac{1}{4} [\cos (720 + 3 A) + 3 \cos (240 + A)] = \frac{1}{4} (\cos 3 A + 3 \cos A) + \frac{1}{4} [\cos 3 A + 3 \cos (120 + A)] + \frac{1}{4} [\cos 3 A + 3 \cos (240 + A)] = \frac{3}{4} \cos 3 A + \frac{3}{4} [\cos A + \cos (120 + A) + \cos (240 + A)] = \frac{3}{4} \cos 3 A + \frac{3}{4} [\cos A + 2 \cos (180 + A) \cos 60] = \frac{3}{4} \cos 3 A + \frac{3}{4} [\cos A - 2 \cos A \frac{1}{2}] = \frac{3}{4} \cos 3 A + \frac{3}{4} [\cos A - \cos A] = \frac{3}{4} \cos 3 A = R H S$ [Hence proved] Due to paucity of time it would not be possible for us to provide the answers to all your queries We are providing solutions to two of your good queries Try solving the remaining one yourself and if you face any difficulty then do get back to us

Prove 1. sin alpha+sin beta+sin gama---sin(alpha+beta+gama)=4 sin(alpha+beta/2) . sin(beta+gama/2) . sin(gama+alpha/2) 2. (sin3Acos4A--sinAcos2A) / sin4AsinA+cos6AcosA) = tan 2A 3. cos3 A +cos3(120+A) + cos3 (240+A) =3/4 cos 3A

Prove 1. sin alpha+sin beta+sin gama---sin(alpha+beta+gama)=4 sin(alpha+beta/2) . sin(beta+gama/2) . sin(gama+alpha/2)

2. (sin3Acos4A--sinAcos2A) / sin4AsinA+cos6AcosA) = tan 2A

3. cos³A +cos³(120+A) + cos³ (240+A) =3/4 cos 3A