# Half Angle formulas

Half Angle formulas?:

The Half angle formulas are stated below:

If ABC is a triangle , A , B and C are the three angles of the triangle and a , b , c are the sides opposite to the corresponding angles and

“s” is the semi perimeter or , $s = \dfrac{a + b+ c}{2}$  , Then:

$\sin \frac{A}{2} = \sqrt{\dfrac{(s-b)(s-c)}{bc}} \\ \\ \sin \frac{B}{2} = \sqrt{\dfrac{(s-a)(s-c)}{ac}} \\ \\ \sin \frac{C}{2} = \sqrt{\dfrac{(s-a)(s-b)}{ab}} \\ \\ \\ \cos \frac{A}{2} = \sqrt{\dfrac{s(s-a)}{bc}} \\ \\ \cos \frac{B}{2} = \sqrt{\dfrac{s(s-b)}{ac}} \\ \\ \cos \frac{C}{2} = \sqrt{\dfrac{s(s-c)}{ab}} \\ \\ \\ \tan \frac{A}{2} = \sqrt{\dfrac{(s-b)(s-c)}{s(s-a)}} \\ \\ \tan \frac{B}{2} = \sqrt{\dfrac{(s-a)(s-c)}{s(s-b)}} \\ \\ \tan \frac{C}{2} = \sqrt{\dfrac{(s-a)(s-b)}{s(s-c)}}$

Proof of Half angle formula:

First of all let’s prove the half angle formula for $\cos \frac{A}{2}$

Using the cosine law:

$2bc \cos A = b^2 + c^2 - a^2 \\ \\ or, 2bc + 2bc \cos A = 2bc + b^2 + c^2 - a^2 \\ \\ or, 2bc (1 + \cos A) = (b+c)^2 - a^2$

Now using the trigonometric sub-multiple angle formula:

$2bc . 2 \cos ^2 \frac{A}{2} = (b+c+a)(b+c-a) \\ \\ or , 4bc \cos ^2 \frac{A}{2} = (2s - 2a) . 2s \, \, \, \, ( because : a+b+c = 2s ) \\ So , \cos \frac{A}{2} = \sqrt{\dfrac{s(s-a)}{bc}}$

Now , let us prove the half angle formula for $\sin \frac{A}{2}$

Using the cosine law:

$- 2bc \cos A = a^2 - (b^2 + c^2) \\ \\ or, 2bc - 2bc \cos A = 2bc + a^2 - (b^2 + c^2 ) \\ \\ or, 2bc (1 - \cos A) = a^2 - (b-c)^2 \\ \\ or, 2bc . 2 \sin ^2 \frac{A}{2} = (a - b + c)( a + b - c) \\ \\ or, 2bc . 2 \sin^2 \frac{A}{2} = (2s - 2b)(2s - 2c) \\ \\ \\ So , \sin \frac{A}{2} = \sqrt{\dfrac{(s-b)(s-c)}{bc}}$

Lastly , Dividing $\sin \frac{A}{2}$ by $\cos \frac{A}{2}$ we get :

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

Similarly we can also prove the half angle formula of angle B and C.

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