sinA-sinB-sinC=4cosA/2cosB/2sinC/2
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Answer:
sinA+sinB+sinC=4cos
2
A
cos
2
B
cos
2
C
Step-by-step explanation:
In ΔABC , Show that
sinA+sinB+sinC=4\:cos\frac{A}{2}\:cos\frac{B}{2}\:cos\frac{C}{2}sinA+sinB+sinC=4cos
2
A
cos
2
B
cos
2
C
\begin{gathered}Formula\:used:\\\\sinC+sinD=2\:sin(\frac{C+D}{2})\:cos(\frac{C-D}{2})\\\\sinA=2\:sin\frac{A}{2}\:cos\frac{A}{2}\end{gathered}
Formulaused:
sinC+sinD=2sin(
2
C+D
)cos(
2
C−D
)
sinA=2sin
2
A
cos
2
A
In ΔABC , A+B+C=\piA+B+C=π
\begin{gathered}sinA+sinB+sinC\\\\=2\:sin(\frac{A+B}{2})cos(\frac{A-B}{2})+2\:sin\frac{C}{2}\:cos\frac{C}{2}\\\\=2\:sin(\frac{\pi-C}{2})cos(\frac{A-B}{2})+2\:sin\frac{C}{2}\:cos\frac{C}{2}\end{gathered}
sinA+sinB+sinC
=2sin(
2
A+B
)cos(
2
A−B
)+2sin
2
C
cos
2
C
=2sin(
2
π−C
)cos(
2
A−B
)+2sin
2
C
cos
2
C
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