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I have read these formulae in my book but i could not understand how these are proved.

$\sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}};\sin\frac{B}{2}=\sqrt{\frac{(s-c)(s-a)}{ca}};\sin\frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{ab}}$ and

$\cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}};\cos\frac{B}{2}=\sqrt{\frac{s(s-b)}{ca}};\cos\frac{C}{2}=\sqrt{\frac{s(s-c)}{ab}}$
where $A,B,C$ are the angles of a triangle $ABC$ and $a,b,c$ are the sides opposite to the angles $A,B,C$ respectively.


I know that area of triangle$\Delta=\frac{1}{2}bc\sin A$

But area of triangle by Heron's formula is $\Delta=\sqrt{s(s-a)(s-b)(s-c)}$

$\sin A=\frac{2\Delta}{bc}$

$2\sin\frac{A}{2}\cos\frac{A}{2}=\frac{2\Delta}{bc}$

$\sin\frac{A}{2}\cos\frac{A}{2}=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{bc}$

I do not know how to prove it further or there is some other method to prove it.

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1 Answer 1

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\begin{aligned} \cos A &= \frac{b^2+c^2-a^2}{2bc} \\ \cos A &= 2\cos^2\frac{A}{2} - 1 \\ 2\cos^2\frac{A}{2} - 1 &= \frac{b^2+c^2-a^2}{2bc}\\ \cos^2 \frac{A}{2} &= \frac{b^2+c^2-a^2+2bc}{4bc} = \frac{(b+c-a)(b+c+a)}{4bc}\\ \cos \frac{A}{2} &= \sqrt{\frac{(2s-2a)(2s)}{4bc}} = \sqrt{\frac{s(s-a)}{bc}} \end{aligned}

$\sin \frac{A}{2}$ can be solved in the same way

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