If A+B+C=180 then sinA+sinB+sinC?
Answers
\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
)
Step-by-step explanation:
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