Half Angle Formula

In any triangle A, B, C

\(\sin \frac{A}{2}=\sqrt{\frac{\left( s-b \right)\left( s-c \right)}{bc}}\)
\(\sin \frac{B}{2}=\sqrt{\frac{\left( s-c \right)\left( s-a \right)}{ac}}\)
\(\sin \frac{C}{2}=\sqrt{\frac{\left( s-a \right)\left( s-b \right)}{ab}}\)
\(\cos \frac{A}{2}=\sqrt{\frac{s\left( s-a \right)}{bc}}\)
\(\cos \frac{B}{2}\,=\sqrt{\frac{s\left( s-b \right)}{ac}}\)
\(\cos \frac{C}{2}=\sqrt{\frac{s\left( s-c \right)}{ab}}\)
\(\tan \frac{A}{2}=\sqrt{\frac{\left( s-b \right)\left( s-c \right)}{s\left( s-a \right)}}\)
\(\tan \frac{B}{2}=\sqrt{\frac{\left( s-c \right)\left( s-a \right)}{s\left( s-b \right)}}\)
\(\tan \frac{C}{2}=\sqrt{\frac{\left( s-a \right)\left( s-b \right)}{s\left( s-c \right)}}\)
\(\Delta =\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}\)

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.