Trigonometric identities: half an angle, double an angle, sum and difference of two angles

Trigonometric relationships of double-angle and half-angle

Known all the ratios of an angle, we can find all the ratios of the double of that angle and its half using the following identities:

$\sin (2 α) = 2 \cdot \sin α \cdot \cos α$
$\cos (2 α) = \cos^{2} α - \sin^{2} α$
$\tan (2 α) = \frac{2 \cdot \tan α}{1 - \tan^{2} α}$
$\sin (\frac{α}{2}) = \pm \sqrt{\frac{1 - \cos α}{2}}$
$\cos (\frac{α}{2}) = \pm \sqrt{\frac{1 + \cos α}{2}}$
$\tan (\frac{α}{2}) = \pm \sqrt{\frac{1 - \cos α}{1 + \cos α}}$

Example

Given $α$ , of which we know its trigonometric ratios, now we will be able to calculate the ratios of the double-angle and the half-angle. Bearing in mind that $α = 30^{\circ}$ , we will compute the ratios of $2 α = 60^{\circ}$ and $\frac{α}{2} = 15^{\circ}$ .

We have:

$\sin 2 α = 2 \cdot \sin α \cdot \cos α = 2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$

$\cos (2 α) = \cos^{2} α - \sin^{2} α = \frac{3}{4} - \frac{1}{4} = \frac{1}{2}$

$\tan (2 α) = \frac{2 \cdot \tan α}{1 - \tan^{2} α} = \frac{2 \cdot \frac{\sqrt{3}}{3}}{1 - \frac{1}{3}} = \sqrt{3}$

$\sin \frac{α}{2} = \sqrt{\frac{1 - \cos α}{2}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 - \sqrt{3}}{4}} = \frac{\sqrt{2 - \sqrt{3}}}{2}$

$\cos \frac{α}{2} = \sqrt{\frac{1 + \cos α}{2}} = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{2 + \sqrt{3}}{4}} = \frac{\sqrt{2 + \sqrt{3}}}{2}$

$\tan \frac{α}{2} = \sqrt{\frac{1 - \cos α}{1 + \cos α}} = \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{1 + \frac{\sqrt{3}}{2}}} = \sqrt{\frac{2 - \sqrt{3}}{2 + \sqrt{3}}} =$

$= \sqrt{\frac{2 - \sqrt{3}}{2 + \sqrt{3}}} \cdot \sqrt{\frac{2 + \sqrt{3}}{2 + \sqrt{3}}} = \frac{\sqrt{4 - 3}}{2 + \sqrt{3}} = \frac{1}{2 + \sqrt{3}} \cdot \frac{2 - \sqrt{3}}{2 - \sqrt{3}} =$

$= \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$

Trigonometric relationships of the sum and the difference of two angles

$\sin (A + B) = \sin A \cdot \cos B + \cos A \cdot \sin B$
$\sin (A - B) = \sin A \cdot \cos B - \cos A \cdot \sin B$
$\cos (A + B) = \cos A \cdot \cos B - \sin A \cdot \sin B$
$\cos (A - B) = \cos A \cdot \cos B + \sin A \cdot \sin B$
$\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}$
$\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \cdot \tan B}$

Example

We can calculate the trigonometrical ratios of $45^{\circ} = 60^{\circ} - 15^{\circ}$ .

$\sin (60 - 15) = \sin 60 \cdot \cos 15 - \cos 60 \cdot \sin 15 = \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2 + \sqrt{3}}}{2} - \frac{1}{2} \cdot \frac{\sqrt{2 - \sqrt{3}}}{2} =$

$= \frac{1}{4} (\sqrt{6 + 3 \sqrt{3}} - \sqrt{2 - \sqrt{3}}) = \frac{\sqrt{2}}{2}$

$\cos (60 - 15) = \cos 60 \cdot \cos 15 + \sin 60 \cdot \sin 15 = \frac{1}{2} \cdot \frac{\sqrt{2 + \sqrt{3}}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2 - \sqrt{3}}}{2} =$

$= \frac{1}{4} (\sqrt{2 + \sqrt{3}} + \sqrt{6 - 3 \sqrt{3}}) = \frac{\sqrt{2}}{2}$

$\tan (60 - 15) = \frac{\tan 60 - \tan 15}{1 + \tan 60 \cdot \tan 15} = \frac{\sqrt{3} - (2 - \sqrt{3})}{1 + \sqrt{3} \cdot (2 - \sqrt{3}} = \frac{2 \sqrt{3} - 2}{1 + 2 \sqrt{3} - 3} = \frac{2 \sqrt{3} - 2}{2 \sqrt{3} - 2} = 1$