The Half Angle Formulae

In any triangle $ABC$, the half angle formulae are;

Sine: \[\sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}}\] \[\sin\frac{B}{2}=\sqrt{\frac{(s-c)(s-a)}{ca}}\] \[\sin\frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{ab}}\]

Cosine: \[\cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}}\] \[\cos\frac{B}{2}=\sqrt{\frac{s(s-b)}{ca}}\] \[\cos\frac{C}{2}=\sqrt{\frac{s(s-c)}{ab}}\]

Tangent: \[\tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\] \[\tan\frac{B}{2}=\sqrt{\frac{(s-c)(s-a)}{s(s-b)}}\] \[\tan\frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}\]

Let $ABC$ be a triangle. Let us denote the angle $BAC$, $CBA$, $ACB$ of the triangle $ABC$ by $A$, $B$, $C$ respectively; and the lengths of the sides $BC$, $CA$, $AB$ by $a$, $b$, $c$ respectively. Let $s$ be the semi perimeter of the triangle $ABC$. \[\therefore s=\frac{a+b+c}{2}\]

Now, we shall prove the above half angle formulae.

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

We know, from the cosine law, \[\cos A=\frac{b^2+c^2-a^2}{2bc}\] \[2bc\cos A=b^2+c^2-a^2\] \[2bc\left(1-2\sin^2\frac{A}{2}\right)=b^2+c^2-a^2\] \[2bc-4bc\sin^2\frac{A}{2}=b^2+c^2-a^2\] \[4bc\sin^2\frac{A}{2} =2bc-b^2-c^2+a^2\] \[4bc\sin^2\frac{A}{2}=a^2-(b^2-2bc+c^2)\] \[4bc\sin^2\frac{A}{2}=a^2-(b-c)^2\] \[4bc\sin^2\frac{A}{2}=(a-b+c)(a+b-c)\] \[4bc\sin^2\frac{A}{2}=2(s-b)\cdot 2(s-c)\] \[[\because a+b+c=2s]\] \[\sin^2\frac{A}{2}=\frac{(s-b)(s-c)}{bc}\] \[\therefore\sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}}\]

Similarly, we can prove that, \[\sin\frac{B}{2}=\sqrt{\frac{(s-c)(s-a)}{ca}}\] \[\sin\frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{ab}}\]

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Proof: $\cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}}$

We know, from the cosine law, \[\cos A=\frac{b^2+c^2-a^2}{2bc}\] \[2bc\cos A=b^2+c^2-a^2\] \[2bc\left(2\cos^2\frac{A}{2}-1\right)=b^2+c^2-a^2\] \[4bc\cos^2\frac{A}{2}-2bc=b^2+c^2-a^2\] \[4bc\cos^2\frac{A}{2}=b^2+2bc+c^2-a^2\] \[4bc\cos^2\frac{A}{2}=(b+c)^2-a^2\] \[4bc\cos^2\frac{A}{2}=(b+c+a)(b+c-a)\] \[4bc\cos^2\frac{A}{2}=2s\cdot 2(s-a)\] \[[\because a+b+c=2s]\] \[\cos^2\frac{A}{2}=\frac{s(s-a)}{bc}\] \[\therefore\cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}}\]

Similarly, we can prove that, \[\cos\frac{B}{2}=\sqrt{\frac{s(s-b)}{ca}}\] \[\cos\frac{C}{2}=\sqrt{\frac{s(s-c)}{ab}}\]

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Proof: $\tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$

We know, \[\tan\frac{A}{2}=\frac{\sin\frac{A}{2}}{\cos\frac{A}{2}}\] \[=\frac{\sqrt{\frac{(s-b)(s-c)}{bc}}}{\sqrt{\frac{s(s-a)}{bc}}}\] \[\therefore\tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}\]

Similarly, we can prove that, \[\tan\frac{B}{2}=\sqrt{\frac{(s-c)(s-a)}{s(s-b)}}\] \[\tan\frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}\]

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