# Formula and Proof of Half Angle Identities

## Formulas

$$\sin (\frac{u}{2}) = \pm \sqrt{\dfrac{1 - \cos u}{2}}$$ $$\cos (\frac{u}{2}) = \pm \sqrt{\dfrac{1 + \cos u}{2}}$$ $$\tan (\frac{u}{2}) = \pm \sqrt{\dfrac{1 - \cos u}{1 + \cos u}}$$

The half-angle formulas allow the expression of trigonometric functions to determine the trigonometric values for another angle u/2 in terms of u. The half-angle formulas are useful in finding the values of unknown trigonometric functions. The half-angle formulas for any angle u can be stated as follows:

sin(u/2) = + or - √ (1 – cos u) / 2.

In the case of u is in first and second quadrant then function is positive and in the case of u is in third and forth quadrant then function is negative.

cos(u/2) = + or - √ (1 + cos u) / 2.

In the case of u is in first and forth quadrant then function is positive and in the case of u is in second and third quadrant then function is negative.

### Proof of Half Angle Identities

The Half angle formulas can be derived from the double-angle formula. the double-angle formulas are as follows:

cos 2u = 1 - 2sin2 u

cos 2u = 2cos2 u - 1

the above equations are true for any value of the variable u. Therefore lets substitute u with ½ u in the double-angle equations.

$$\cos 2(\frac{1}{2}u) = 1 - 2\sin^2 \frac{1}{2}u$$ $$\cos u = 1 - 2\sin^2 \frac{1}{2}u$$ $$2\sin^2 \frac{1}{2}u = 1 - \cos u$$ $$\sin \frac{1}{2}u = \sqrt{\dfrac{1 - \cos u}{2}}$$ Now derive for cos from the double-angle formula as follows:
$$\cos 2(\frac{1}{2}u) = 2\cos^2 \frac{1}{2}u - 1$$ $$\cos u = 2\cos^2 \frac{1}{2}u - 1$$ $$2\cos^2 \frac{1}{2}u = 1 + \cos u$$ $$\cos \frac{1}{2}u = \sqrt{\dfrac{1 + \cos u}{2}}$$ We can also derive for tan which is follows: $$\tan \frac{1}{2}u = \dfrac{\sin \frac{1}{2}u}{\cos \frac{1}{2}u}$$ in this equation we can substritute the derived value of sin ½ u and cos ½ u:
$$\tan \frac{1}{2}u = \dfrac{\sqrt{\dfrac{1 - \cos u}{2}}}{\sqrt{\dfrac{1 + \cos u}{2}}}$$ $$\tan \frac{1}{2}u = \sqrt{\dfrac{1 - \cos u}{1 + \cos u}}$$