Question
If \triangle ABC has inradius r amd circumradius R, show that
r = 4R\sin{\dfrac{A}{2}} \sin{\dfrac{B}{2}} \sin{\dfrac{C}{2}}
r=4Rsin(A/2)sin(B/2)sin(C/2)
Answer
In the figure, the circle is inscribed in the \triangle ABC . Join the origin to the vertices of the triangle. The segments bisect the three internal angles , respectively.
According to the Law of Sines
\dfrac{a}{\sin A} = 2R
in which R is the radius of circumscribed circle of the triangle.
Multiplying both sides by \sin A, then we get
a = 2R\sin A
(1)
On the other hand,
a = BD+CD
=r\cot \dfrac{B}{2} +r\cot \dfrac{C}{2}
=r\Bigg( \dfrac{\cos\dfrac{B}{2} }{\sin\dfrac{B}{2} } +\dfrac{\cos\dfrac{C}{2} }{\sin\dfrac{C}{2} }\Bigg)
=r\dfrac{\cos\dfrac{B}{2} \sin\dfrac{C}{2}+\cos\dfrac{C}{2}\sin\dfrac{B}{2} }{\sin\dfrac{B}{2}\sin\dfrac{C}{2} }
Apply sum identity for sines function for numerator
\cos\dfrac{B}{2} \sin\dfrac{C}{2}+\cos\dfrac{C}{2}\sin\dfrac{B}{2}=\sin(\dfrac{B+C}{2} )
=\sin(\dfrac{\pi-A}{2} )
=\sin(\dfrac{\pi}{2} -\dfrac{A}{2} )
=\cos\dfrac{A}{2}
Then,
a = r\dfrac{\cos\dfrac{A}{2}}{\sin\dfrac{B}{2}\sin\dfrac{C}{2} }
(2)
Based on (1) and (2), using double angle identity for \sin A, we get
2R\sin A =4R\sin\dfrac{A}{2}\cancel{ \cos\dfrac{A}{2}} = r\dfrac{\cancel{\cos\dfrac{A}{2}}}{\sin\dfrac{B}{2}\sin\dfrac{C}{2} }
Then we have proved
r = 4R\sin{\dfrac{A}{2}} \sin{\dfrac{B}{2}} \sin{\dfrac{C}{2}}