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}}


Collected in the board: Trigonometry

Steven Zheng posted 1 day ago


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

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} )



a = r\dfrac{\cos\dfrac{A}{2}}{\sin\dfrac{B}{2}\sin\dfrac{C}{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}}

Steven Zheng posted 1 day ago

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