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

Development:

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: $$\sin ({45}^{\circ })={\displaystyle \frac{\sqrt{2}}{2}}\sin ({30}^{\circ })={\displaystyle \frac{1}{2}}$$ $$\cos ({45}^{\circ })={\displaystyle \frac{\sqrt{2}}{2}}\cos ({30}^{\circ })={\displaystyle \frac{\sqrt{3}}{2}}$$ $$\tan ({45}^{\circ })=1\tan ({30}^{\circ })={\displaystyle \frac{\sqrt{3}}{3}}$$

From the following formulas, we have:

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

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

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

Solution:

$\sin ({15}^{\circ })={\displaystyle \frac{\sqrt{2}}{4}}(\sqrt{3}-1)$

$\cos ({15}^{\circ })={\displaystyle \frac{\sqrt{2}}{4}}(\sqrt{3}+1)$

$\tan ({15}^{\circ })=2-\sqrt{3}$

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