Question

3 - Calcular, por redução ao primeiro quadrante:
a) sen 150°
b) sen 225°
c) sen 330°
d) sen 3π/4
e) cos 11π/6
f) tg 5π/3
g) cos 5π/4
h) sen 11π/6
i) cos 5π/6
j) tg 35π/4
k) tg 15 π/4

Para entender melhor tem a imagem.

Answer 1

$\boxed{\begin{array}{l}\tt a)~\sf sen(150^\circ)=sen (180^\circ-150^\circ)=sen(30^\circ)\\\sf sen(150^\circ)=\dfrac{1}{2}\\\tt b)~\sf sen(225^\circ)=-sen(225^\circ-180^\circ)=-sen(45^\circ)\\\sf sen(225^\circ)=-\dfrac{\sqrt{2}}{2}\\\tt c)~\sf sen(330^\circ)=-sen(360^\circ-330^\circ)=-sen(30^\circ)\\\sf sen(330^\circ)=-\dfrac{1}{2}\\\tt d)~\sf sen\bigg(\dfrac{3\pi}{4}\bigg)=sen\bigg(\pi-\dfrac{3\pi}{4}\bigg)=sen\bigg(\dfrac{\pi}{4}\bigg)\\\\\sf sen\bigg(\dfrac{3\pi}{4}\bigg)=\dfrac{\sqrt{2}}{2}\end{array}}$$\large\boxed{\begin{array}{l}\tt e)~\sf cos\bigg(\dfrac{11\pi}{6}\bigg)=cos\bigg(2\pi-\dfrac{11\pi}{6}\bigg)=cos\bigg(\dfrac{\pi}{6}\bigg)\\\\\sf cos\bigg(\dfrac{11\pi}{6}\bigg)=\dfrac{\sqrt{3}}{2}\\\tt f)~\sf tg\bigg(\dfrac{5\pi}{3}\bigg)=-tg\bigg(2\pi-\dfrac{5\pi}{3}\bigg)=-tg\bigg(\dfrac{\pi}{3}\bigg)\\\\\sf tg\bigg(\dfrac{5\pi}{3}\bigg)=-\sqrt{3}\end{array}}$

$\large\boxed{\begin{array}{l}\tt g)~\sf cos\bigg(\dfrac{5\pi}{4}\bigg)=-cos\bigg(\dfrac{5\pi}{4}-\pi\bigg)=-cos\bigg(\dfrac{\pi}{4}\bigg)\\\\\sf cos\bigg(\dfrac{5\pi}{4}\bigg)=-\dfrac{\sqrt{2}}{2}\\\tt h)~\sf sen\bigg(\dfrac{11\pi}{6}\bigg)=-sen\bigg(2\pi-\dfrac{11\pi}{6}\bigg)=-sen\bigg(\dfrac{\pi}{6}\bigg)\\\\\sf sen\bigg(\dfrac{11\pi}{6}\bigg)=-\dfrac{1}{2}\end{array}}$

$\Large\boxed{\begin{array}{l}\tt i)~\sf cos\bigg(\dfrac{5\pi}{6}\bigg)=-cos\bigg(\pi-\dfrac{5\pi}{6}\bigg)=-cos\bigg(\dfrac{\pi}{6}\bigg)\\\\\sf cos\bigg(\dfrac{5\pi}{6}\bigg)=-\dfrac{1}{2}\end{array}}$

$\large\boxed{\begin{array}{l}\tt j)~\sf \dfrac{35\pi}{4}=\dfrac{32\pi}{4}+\dfrac{3\pi}{4}=8\pi+\dfrac{3\pi}{4}\\\sf \dfrac{35\pi}{4}=4\cdot 2\pi+\dfrac{3\pi}{4}\\\sf tg\bigg(\dfrac{35\pi}{4}\bigg)=tg\bigg(\dfrac{3\pi}{4}\bigg)=-tg\bigg(\pi-\dfrac{3\pi}{4}\bigg)=-tg\bigg(\dfrac{\pi}{4}\bigg)\\\sf tg\bigg(\dfrac{35\pi}{4}\bigg)=-1\end{array}}$

$\Large\boxed{\begin{array}{l}\tt k)~\sf\dfrac{15\pi}{4}=\dfrac{8\pi}{4}+\dfrac{7\pi}{4}=2\pi+\dfrac{7\pi}{4}\\\\\sf tg\bigg(\dfrac{15\pi}{4}\bigg)=tg\bigg(\dfrac{7\pi}{4}\bigg)=-tg\bigg(2\pi-\dfrac{7\pi}{4}\bigg)\\\\\sf tg\bigg(\dfrac{15\pi}{4}\bigg)=-tg\bigg(\dfrac{\pi}{4}\bigg)=-1\end{array}}$