# Half Angle Formula

## Half Angle Formula

In any triangle A, B, C

1. $$\sin \frac{A}{2}=\sqrt{\frac{\left( s-b \right)\left( s-c \right)}{bc}}$$
2. $$\sin \frac{B}{2}=\sqrt{\frac{\left( s-c \right)\left( s-a \right)}{ac}}$$
3. $$\sin \frac{C}{2}=\sqrt{\frac{\left( s-a \right)\left( s-b \right)}{ab}}$$
4. $$\cos \frac{A}{2}=\sqrt{\frac{s\left( s-a \right)}{bc}}$$
5. $$\cos \frac{B}{2}\,=\sqrt{\frac{s\left( s-b \right)}{ac}}$$
6. $$\cos \frac{C}{2}=\sqrt{\frac{s\left( s-c \right)}{ab}}$$
7. $$\tan \frac{A}{2}=\sqrt{\frac{\left( s-b \right)\left( s-c \right)}{s\left( s-a \right)}}$$
8. $$\tan \frac{B}{2}=\sqrt{\frac{\left( s-c \right)\left( s-a \right)}{s\left( s-b \right)}}$$
9. $$\tan \frac{C}{2}=\sqrt{\frac{\left( s-a \right)\left( s-b \right)}{s\left( s-c \right)}}$$
10. $$\Delta =\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$$

Example: If P₁ , P₂  and P₃ are the altitudes of a triangle ABC from the vertical A, B and C respectively and Δ denotes its area then prove that $$\frac{1}{{{P}_{1}}}+\frac{1}{{{P}_{2}}}+\frac{1}{{{P}_{3}}}=\frac{2ab}{\left( a+b+c \right)}{{\cos }^{2}}\frac{C}{2}$$

Solution: We have $$\Delta =\frac{1}{2}a{{P}_{1}}=\frac{1}{2}b{{P}_{2}}=\frac{1}{2}c{{P}_{3}}$$

Now,

L H S = $$\frac{1}{{{P}_{1}}}+\frac{1}{{{P}_{2}}}+\frac{1}{{{P}_{3}}}$$,

$$=\frac{a}{2\Delta }+\frac{b}{2\Delta }-\frac{c}{2\Delta }$$,

$$=\frac{a+b-c}{2\Delta }$$,

$$=\frac{2\left( s-c \right)}{2\Delta }$$,

$$=\frac{s-c}{\Delta }$$,

And

$$R\,H\,S=\frac{2ab}{\left( a+b+c \right)\Delta }{{\cos }^{2}}\frac{c}{2}$$,

$$=\frac{2ab}{2s\Delta }\times \frac{s\left( s-c \right)}{ab}$$,

$$\,=\frac{s-c}{\Delta }$$,

∴ $$\frac{1}{{{P}_{1}}}+\frac{1}{{{P}_{2}}}+\frac{1}{{{P}_{3}}}=\frac{2ab}{\left( a+b+c \right)\Delta }{{\cos }^{2}}\frac{c}{2}$$.

Example: In a triangle ABC, if $$a{{\cos }^{2}}\frac{C}{2}+c{{\cos }^{2}}\frac{A}{2}=\frac{3b}{2}$$ prove that a, b, c are in A.P.

Solution: We have,

$$a{{\cos }^{2}}\frac{c}{2}+c{{\cos }^{2}}\frac{A}{2}=\frac{3b}{2}$$,

⇒ $$a.\frac{s\left( s-c \right)}{ab}+c.\frac{s\left( s-c \right)}{bc}=\frac{3b}{2}$$

⇒ $$\frac{s}{b}\left( s-c+s-a \right)=\frac{3b}{2}$$

⇒ 2s (2s – a – c) = 3b²

⇒ (a + b + c) b = 3b²

⇒ a + c = 2b

⇒ a, b, c are in A.P.