This is the chapter with the most demonstration exercises. Before opening a solution, decide which family of formulas to use: addition, multiplication, division or transformation into a product. Addition formulas 131 Calculate the values of: a) $\cos 15^\circ$; b) $\operatorname{sen} 105^\circ$; c) $\operatorname{tg} 75^\circ$; d) $\sec 285^\circ$. See the resolutiona) $\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \operatorname{sen} 45^\circ \operatorname{sen} 30^\circ = \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$ b) $\operatorname{sen} 105^\circ = \operatorname{sen}(60^\circ + 45^\circ) = \operatorname{sen} 60^\circ \cos 45^\circ + \operatorname{sen} 45^\circ \cos 60^\circ = \dfrac{\sqrt{3}}{2} \cdot \dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$ c) $\operatorname{tg} 75^\circ = \operatorname{tg}(45^\circ + 30^\circ) = \dfrac{\operatorname{tg} 45^\circ + \operatorname{tg} 30^\circ}{1 - \operatorname{tg} 45^\circ \operatorname{tg} 30^\circ} = \dfrac{1 + \frac{\sqrt{3}}{3}}{1 - \frac{\sqrt{3}}{3}} = \dfrac{3 + \sqrt{3}}{3 - \sqrt{3}}$ $\left(= 2 + \sqrt{3}\right)$ d) $\sec 285^\circ = \sec 75^\circ = \dfrac{1}{\cos 75^\circ} = \dfrac{1}{\cos(45^\circ + 30^\circ)} = \dfrac{4}{\sqrt{6} - \sqrt{2}}$ $\left(= \sqrt{6} + \sqrt{2}\right)$
By doing $b = a$ in the addition formulas, we obtain the double arc functions; reversing these formulas, those of the arch half. These are the tools that allow, for example, calculating $\operatorname{sen}\frac{\pi}{8}$ from $\cos\frac{\pi}{4}$. The story In 1593, Belgian mathematician Adriaan van Roomen challenged European mathematicians to solve an equation of degree 45. François Viète, lawyer and advisor to the French king Henry IV, realized that it was the expression of $\operatorname{sen} 45\theta$ in function of $\operatorname{sen}\theta$ and solved it with the formulas of multiple arcs, which he himself had systematized. The half bow formulas are even older: Ptolemy used them to pass from the string of $12^\circ$ to those of $6^\circ$, $3^\circ$, $1{,}5^\circ$ and $0{,}75^\circ$ in the construction of his tablet.