Question

Determine:
tg 285° =

Cos 105° =

Answer 1

Tangente da soma entre dois arcos:

$\mathrm{tg}(a+ b)=\dfrac{\mathrm{tg\,}a+ \mathrm{tg\,}b}{1- \mathrm{tg\,}a\cdot \mathrm{tg\,}b}~~~~~~(a+ b\ne \dfrac{\pi}{2}+k\pi)$


$\bullet\;\;\mathrm{tg\,}285^\circ\\\\ =\mathrm{tg}(135^\circ+150^\circ)\\\\ =\dfrac{\mathrm{tg\,}135^\circ+\mathrm{tg\,}150^\circ}{1- \mathrm{tg\,}135^\circ\cdot \mathrm{tg\,}150^\circ}\\\\\\ =\dfrac{-1+\left(-\frac{\sqrt{3}}{3} \right )}{1-(-1)\cdot \left(-\frac{\sqrt{3}}{3} \right )}\\\\\\ =\dfrac{-1-\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3}}\\\\\\ =\dfrac{-3-\sqrt{3}}{3-\sqrt{3}}$

$=\dfrac{(-3-\sqrt{3})\cdot (3+\sqrt{3})}{(3-\sqrt{3})\cdot (3+\sqrt{3})}\\\\\\ =\dfrac{-9-3\sqrt{3}-3\sqrt{3}-3}{9-3}\\\\\\ =\dfrac{-12-6\sqrt{3}}{6}\\\\\\ =\dfrac{\diagup\!\!\!\! 6\cdot (-2-\sqrt{3})}{\diagup\!\!\!\! 6}\\\\\\ =-2-\sqrt{3}$

______________________

Cosseno da soma de dois arcos:

$\cos(a+b)=\cos a\cos b-\mathrm{sen\,}a\,\mathrm{sen\,}b$


$\bullet\;\;\cos 105^\circ\\\\ =\cos(60^\circ+45^\circ)\\\\ =\cos 60^\circ\cdot \cos 45^\circ-\mathrm{sen\,}60^\circ\cdot \mathrm{sen\,}45^\circ\\\\ =\dfrac{1}{2}\cdot \dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{3}}{2}\cdot \dfrac{\sqrt{3}}{2}\\\\\\ =\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{6}}{4}\\\\\\ =\dfrac{\sqrt{2}-\sqrt{6}}{4}$


Bons estudos! :-)