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) $\sec 285^\circ = \sec 75^\circ = \dfrac{1}{\cos 75^\circ} = \dfrac{1}{\cos(45^\circ + 30^\circ)} = \dfrac{4}{\sqrt{6} - \sqrt{2}}$ $\left(= \sqrt{6} + \sqrt{2}\right)$
Let’s deduce formulas to calculate the trigonometric functions of the sum $(a + b)$ and the difference $(a - b)$ of any two real numbers $a$ and $b$, known as the circular functions of $a$ and $b$. The story In Almagest, Ptolemy proved that, in an inscribed quadrilateral, the product of the diagonals is equal to the sum of the products of the opposite sides — Ptolemy’s theorem. Applied to a quadrilateral with a side equal to the diameter, it gives exactly the chord formulas for the sum and difference of two arcs, which Ptolemy used to assemble his half-by-half degree table. In the form of sines, the formulas appear with the Persian astronomer Abu al-Wafa (10th century), whom we already met in chapter II. The demonstration we will do — based on the distance between two points — uses Descartes’ analytical geometry and is much more recent.