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

First of all, we observe that $15=45-30$. Therefore, knowing the ratios of $45$ degrees and $30$ degrees, we will be able to obtain those of the $15$ degree angle. Let's calculate: $$\sin(45^\circ)=\dfrac{\sqrt{2}}{2} \ \ \ \ \sin(30^\circ)=\dfrac{1}{2}$$ $$\cos(45^\circ)=\dfrac{\sqrt{2}}{2} \ \ \ \ \cos(30^\circ)=\dfrac{\sqrt{3}}{2}$$ $$\tan(45^\circ)=1 \ \ \ \ \tan(30^\circ)=\dfrac{\sqrt{3}}{3}$$

From the following formulas, we have:

$$\sin(15^\circ)=\sin(45^\circ-30^\circ)=\sin(45^\circ)\cdot \cos(30^\circ)-\cos(45^\circ)\cdot \sin(30^\circ)=$$ $$=\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt{2}\cdot\sqrt{3}}{4}-\dfrac{\sqrt{2}}{4}=\dfrac{\sqrt{2}}{4}(\sqrt{3}-1) \\ $$

$$\cos(15^\circ)=\cos(45^\circ-30^\circ)=\cos(45^\circ)\cdot \cos(30^\circ)+\sin(45^\circ)\cdot \sin(30^\circ)=$$ $$=\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt{2}\cdot\sqrt{3}}{4}+\dfrac{\sqrt{2}}{4}=\dfrac{\sqrt{2}}{4}(\sqrt{3}+1) \\ $$

$$\tan(15^\circ)=\tan(45^\circ-30^\circ)=\dfrac{\tan(45^\circ)-\tan(30^\circ)}{1+\tan(45^\circ)\cdot \tan(30^\circ)}=$$ $$\dfrac{1-\dfrac{\sqrt{3}}{3}}{1+\dfrac{\sqrt{3}}{3}}=\dfrac{\dfrac{3-\sqrt{3}}{3}}{\dfrac{3+\sqrt{3}}{3}}=\dfrac{3-\sqrt{3}}{3+\sqrt{3}}= \dfrac{3-\sqrt{3}}{3+\sqrt{3}}\cdot \dfrac{3-\sqrt{3}}{3-\sqrt{3}}=\dfrac{9+3-6\sqrt{3}}{9-3}=$$ $$=\dfrac{12-6\sqrt{3}}{6}=2-\sqrt{3}$$