\textbf{Given:}Given:

\displaystyle\,A+B+C=180^{\circ}A+B+C=180

∘

\displaystyle\,sin\,A-sin\,B+sin\,CsinA−sinB+sinC

\text{Using}Using

\boxed{\bf\sin\,C-sin\,D=2\,COS(\frac{C+D}{2})\,sin(\frac{C-D}{2})}

sinC−sinD=2COS(

2

C+D

)sin(

2

C−D

)

=\displaystyle\,2\,cos(\frac{A+B}{2})\,sin(\frac{A-B}{2})+sin\,C=2cos(

2

A+B

)sin(

2

A−B

)+sinC

=\displaystyle2\,cos(\frac{180^{\circ}-C}{2})\,sin(\frac{A-B}{2})+sin\,C=2cos(

2

180

∘

−C

)sin(

2

A−B

)+sinC

\text{using}using

\boxed{\bf\sin\,A=2\,sin(\frac{A}{2})\,cos(\frac{A}{2})}

sinA=2SIN(

2

A

)cos(

2

A

)

=\displaystyle\,2\,cos(90^{\circ}-(\frac{C}{2}))\,sin(\frac{A-B}{2})+2\,sin\,(\frac{C}{2})\,cos(\frac{c}{2})=2cos(90

∘

−(

2

C

))sin(

2

A−B

)+2sin(

2

C

)cos(

2

c

)

=\displaystyle\,2\,sin(\frac{C}{2})\,sin\,(\frac{A-B}{2})+2\,sin(\frac{C}{2})\,cos(\frac{C}{2})=2sin(

2

C

)sin(

2

A−B

)+2sin(

2

C

)cos(

2

C

)

=\displaystyle\,2\,sin(\frac{C}{2})\,(sin(\frac{A-B}{2})+cos(\frac{C}{2}))=2sin(

2

C

)(sin(

2

A−B

)+cos(

2

C

))

=\displaystyle\,2\,sin(\frac{C}{2})[sin(\frac{A-B}{2})+cos(\frac{180^{\circ}-(A+B)}{2})]=2sin(

2

C

)[sin(

2

A−B

)+cos(

2

180

∘

−(A+B)

)]

=\displaystyle\,2\,sin(\frac{C}{2})\,[sin(\frac{A-B}{2})+cos\,(90^{\circ}-(\frac{A+B}{2}))]=2sin(

2

C

)[sin(

2

A−B

)+cos(90

∘

−(

2

A+B

))]

\boxed{\bf\,sin(A+B)+sin(A-B)=2\,sin\,A\,cos\,B}

sin(A+B)+sin(A−B)=2sinAcosB

=\displaystyle\,2\,sin(\frac{C}{2})\,[sin(\frac{A-B}{2})+sin(\frac{A+B}{2})]=2sin(

2

C

)[sin(

2

A−B

)+sin(

2

A+B

)]

=\displaystyle\,2\,sin(\frac{C}{2})[2\,sin(\frac{A}{2})\,cos(\frac{B}{2})]=2sin(

2

C

)[2sin(

2

A

)cos(

2

B

)]

=\displaystyle\,4\,sin(\frac{A}{2})\,cos(\frac{B}{2})\,sin(\frac{C}{2})=4sin(

2

A

)cos(

2

B

)sin(

2

C

)

\implies\boxed{\bf\,sin\,A-sin\,B+sin\,C=4\,sin(\frac{A}{2})\,cos(\frac{B}{2})\,sin(\frac{C}{2})}⟹

sinA−sinB+sinC=4sin(

2

A

)cos(

2

B

)sin(

2

C

)