Prove that sin alpha + sin beta + sin gamma - sin (alpha + beta + gamma)=4sin(alpha+beta/2) sin(beta+gamma/2) sin(gamma + alpha /2)
Answers
Answer:
Step-by-step explanation:
Answer:
sinC+sinD=2sin(
2
C+D
)cos(
2
C−D
)
sinC-sinD=2\:cos(\frac{C+D}{2})\:sin(\frac{C-D}{2})sinC−sinD=2cos(
2
C+D
)sin(
2
C−D
)
cosC-cosD=-2\:sin(\frac{C+D}{2})\:sin(\frac{C-D}{2})cosC−cosD=−2sin(
2
C+D
)sin(
2
C−D
)
\text{consider,}consider,
sin\alpha+sin\beta+sin\gamma-sin(\alpha+\beta+\gamma)sinα+sinβ+sinγ−sin(α+β+γ)
=2\:sin(\frac{\alpha+\beta}{2})\:cos(\frac{\alpha-\beta}{2})+2\:cos(\frac{\gamma+\alpha+\beta+\gamma}{2})\:sin(\frac{\gamma-(\alpha+\beta+\gamma)}{2})=2sin(
2
α+β
)cos(
2
α−β
)+2cos(
2
γ+α+β+γ
)sin(
2
γ−(α+β+γ)
)
=2\:sin(\frac{\alpha+\beta}{2})\:cos(\frac{\alpha-\beta}{2})+2\:cos(\frac{\alpha+\beta+2\gamma}{2})\:sin(\frac{-(\alpha+\beta)}{2})=2sin(
2
α+β
)cos(
2
α−β
)+2cos(
2
α+β+2γ
)sin(
2
−(α+β)
)
=2\:sin(\frac{\alpha+\beta}{2})\:cos(\frac{\alpha-\beta}{2})-2\:cos(\frac{\alpha+\beta+2\gamma}{2})\:sin(\frac{\alpha+\beta}{2})=2sin(
2
α+β
)cos(
2
α−β
)−2cos(
2
α+β+2γ
)sin(
2
α+β
)
=2\:sin(\frac{\alpha+\beta}{2})[cos(\frac{\alpha-\beta}{2})-cos(\frac{\alpha+\beta+2\gamma}{2})]=2sin(
2
α+β
)[cos(
2
α−β
)−cos(
2
α+β+2γ
)]
=2\:sin(\frac{\alpha+\beta}{2})[-2\:sin(\frac{(\frac{\alpha-\beta}{2})+(\frac{\alpha+\beta+2\gamma}{2})}{2}\:sin(\frac{(\frac{\alpha-\beta}{2})-(\frac{\alpha+\beta+2\gamma}{2})}{2})]=2sin(
2
α+β
)[−2sin(
2
(
2
α−β
)+(
2
α+β+2γ
)
sin(
2
(
2
α−β
)−(
2
α+β+2γ
)
)]
=2\:sin(\frac{\alpha+\beta}{2})[-2\:sin(\frac{(\frac{2\alpha+2\gamma}{2})}{2})\:sin(\frac{(\frac{-2\beta-2\gamma}{2})}{2})=2sin(
2
α+β
)[−2sin(
2
(
2
2α+2γ
)
)sin(
2
(
2
−2β−2γ
)
)
=2\:sin(\frac{\alpha+\beta}{2})[-2\:sin(\frac{\alpha+\gamma}{2})\:sin(\frac{-(\beta+\gamma)}{2})=2sin(
2
α+β
)[−2sin(
2
α+γ
)sin(
2
−(β+γ)
)
=2\:sin(\frac{\alpha+\beta}{2})[-2\:sin(\frac{\alpha+\gamma}{2})\:(-sin(\frac{\beta+\gamma}{2}))=2sin(
2
α+β
)[−2sin(
2
α+γ
)(−sin(
2
β+γ
))
=4\:sin(\frac{\alpha+\beta}{2})\:sin(\frac{\alpha+\gamma}{2})\:sin(\frac{\beta+\gamma}{2})=4sin(
2
α+β
)sin(
2
α+γ
)sin(
2
β+γ
)
\implies\:\bf{sin\alpha+sin\beta+sin\gamma-sin(\alpha+\beta+\gamma)=4\:sin(\frac{\alpha+\beta}{2})\:sin(\frac{\beta+\gamma}{2})\:sin(\frac{\gamma+\alpha}{2})}⟹sinα+sinβ+sinγ−sin(α+β+γ)=4sin(
2
α+β
)sin(
2
β+γ
)sin(
2
γ+α
)
Step-by-step explanation: