In calculation exercises, the fundamental relationship gives the module and the quadrant gives the sign. In the identities, indicate which of the three methods you used. Calculation of functions from one of them 73 Knowing that $\operatorname{sen} x = \dfrac{4}{5}$ and $\dfrac{\pi}{2} < x < \pi$, calculate the other circular functions of $x$. See the resolutionAs $\dfrac{\pi}{2} < x < \pi \Rightarrow \cos x < 0$, we have: $$\cos x = -\sqrt{1 - \operatorname{sen}^2 x} = -\sqrt{1 - \frac{16}{25}} = -\sqrt{\frac{9}{25}} = -\frac{3}{5}$$$$\operatorname{tg} x = \frac{\operatorname{sen} x}{\cos x} = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3} \qquad \operatorname{cotg} x = \frac{\cos x}{\operatorname{sen} x} = \frac{-\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4}$$$$\sec x = \frac{1}{\cos x} = \frac{1}{-\frac{3}{5}} = -\frac{5}{3} \qquad \operatorname{cossec} x = \frac{1}{\operatorname{sen} x} = \frac{1}{\frac{4}{5}} = \frac{5}{4}.$$ 74 Knowing that $\operatorname{cossec} x = -\dfrac{25}{24}$ and $\pi < x < \dfrac{3\pi}{2}$, calculate the other circular functions of $x$. Ver resposta$\operatorname{sen} x = -\dfrac{24}{25}$, $\cos x = -\dfrac{7}{25}$, $\operatorname{tg} x = \dfrac{24}{7}$, $\operatorname{cotg} x = \dfrac{7}{24}$, $\sec x = -\dfrac{25}{7}$
All relations in this topic come from two plane geometry tools: the Pythagorean theorem and the triangle similarity. The figure below shows the three triangles we will use. Drag $P$ and choose the relationship. The panel numerically checks each equality. Ative o JavaScript para ver esta figura interativa. The story In Almagest, Ptolemy already used the equivalent of $\operatorname{sen}^2 x + \cos^2 x = 1$: if he knew the chord of a bow, he obtained the additional bowstring through Pythagoras’ theorem in the triangle inscribed in the semicircle. The relations with tangent and secant appear in the works of Abu al-Wafa (10th century) and Nasir al-Din al-Tusi (13th century), whose Treatise on the quadrilateral made trigonometry a discipline independent of astronomy.