Calculate the trigonometric ratios of the angle of $$15$$ degrees in terms of the angles of $$45$$ degrees and $$30$$ degrees.
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: $$$\sin(45^\circ)=\dfrac{\sqrt{2}}{2} \ \ \ \ \sin(30^\circ)=\dfrac{1}{2}$$$ $$$\cos(45^\circ)=\dfrac{\sqrt{2}}{2} \ \ \ \ \cos(30^\circ)=\dfrac{\sqrt{3}}{2}$$$ $$$\tan(45^\circ)=1 \ \ \ \ \tan(30^\circ)=\dfrac{\sqrt{3}}{3}$$$
From the following formulas, we have:
$$$\sin(15^\circ)=\sin(45^\circ-30^\circ)=\sin(45^\circ)\cdot \cos(30^\circ)-\cos(45^\circ)\cdot \sin(30^\circ)=$$$ $$$=\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt{2}\cdot\sqrt{3}}{4}-\dfrac{\sqrt{2}}{4}=\dfrac{\sqrt{2}}{4}(\sqrt{3}-1) \\ $$$
$$$\cos(15^\circ)=\cos(45^\circ-30^\circ)=\cos(45^\circ)\cdot \cos(30^\circ)+\sin(45^\circ)\cdot \sin(30^\circ)=$$$ $$$=\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt{2}\cdot\sqrt{3}}{4}+\dfrac{\sqrt{2}}{4}=\dfrac{\sqrt{2}}{4}(\sqrt{3}+1) \\ $$$
$$$\tan(15^\circ)=\tan(45^\circ-30^\circ)=\dfrac{\tan(45^\circ)-\tan(30^\circ)}{1+\tan(45^\circ)\cdot \tan(30^\circ)}=$$$ $$$\dfrac{1-\dfrac{\sqrt{3}}{3}}{1+\dfrac{\sqrt{3}}{3}}=\dfrac{\dfrac{3-\sqrt{3}}{3}}{\dfrac{3+\sqrt{3}}{3}}=\dfrac{3-\sqrt{3}}{3+\sqrt{3}}= \dfrac{3-\sqrt{3}}{3+\sqrt{3}}\cdot \dfrac{3-\sqrt{3}}{3-\sqrt{3}}=\dfrac{9+3-6\sqrt{3}}{9-3}=$$$ $$$=\dfrac{12-6\sqrt{3}}{6}=2-\sqrt{3}$$$
Solution:
$$\sin(15^\circ)=\dfrac{\sqrt{2}}{4}(\sqrt{3}-1)$$
$$\cos(15^\circ)=\dfrac{\sqrt{2}}{4}(\sqrt{3}+1)$$
$$\tan(15^\circ)=2-\sqrt{3}$$