Question

a) Sen π/5 rad =

b) Sen 36º =

c) Cos 11π/12 rad =

d) Cos 0 =​

Answer 1

Explicação passo-a-passo:

a)

$\sf sen~\dfrac{\pi}{5}~rad=sen~36^{\circ}$

$\sf \red{sen~\dfrac{\pi}{5}~rad=\dfrac{\sqrt{10-2\sqrt{5}}}{4}}$

b)

$\sf \red{sen~36^{\circ}=\dfrac{\sqrt{10-2\sqrt{5}}}{4}}$

c)

$\sf cos~\dfrac{11\pi}{12}~rad=cos~165^{\circ}$

$\sf cos~\dfrac{11\pi}{12}~rad=-cos~(180^{\circ}-165^{\circ})$

$\sf cos~\dfrac{11\pi}{12}~rad=-cos~15^{\circ}$

$\sf cos~\dfrac{11\pi}{12}~rad=-cos~(60^{\circ}-45^{\circ})$

$\sf cos~\dfrac{11\pi}{12}~rad=-(cos~60^{\circ}\cdot cos~45^{\circ}+sen~60^{\circ}\cdot sen~45^{\circ})$

$\sf cos~\dfrac{11\pi}{12}~rad=-\left(\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}\right)$

$\sf cos~\dfrac{11\pi}{12}~rad=-\left(\dfrac{\sqrt{2}}{4}+\dfrac{\sqrt{6}}{4}\right)$

$\sf \red{cos~\dfrac{11\pi}{12}~rad=\dfrac{-\sqrt{2}-\sqrt{6}}{4}}$

d) $\sf \red{cos~0=1}$