Question

14.0s arcos da forma k.180°(-1)ᵏ .30°, k ᗴ Z, têm extremidades em que quadrantes?

15.Determine os valores de:
a) y = 3 cos 540°-2sen90°+tg 180°

Answer 1

14)

k=0  ==>0+30° =30°

k=1  ==>180°-30=150°

k=2  ==>360 +30  ==> 30°  

k=3  ==>540-30  ==>510 ==>360 + 150°

Se k for par 30°

Se k for ímpar 150°

15)

a)

540=360 + 180 ==>vamos usar 180°

cos(540)=cos(180) =-1

sen(90)=1

tg(180)=sen(180)/cos(180)=0/(-1)=0

y = 3 * (-1) - 2 * 1 + 0  

y= -5

Answer 2

$\boxed{\begin{array}{l}\rm14)~k\cdot180^\circ+(-1)^k\cdot30^\circ,k\in\mathbb{Z}\\\sf quando~k=0\\\sf 0\cdot180^\circ+(-1)^0\cdot30^\circ=30^\circ\\\sf quando~k=1\\\sf 1\cdot180^\circ+(-1)^1\cdot30^\circ=180^\circ-30^\circ=150^\circ\\\sf portanto\\\sf k\cdot180^\circ+(-1)^k\cdot30^\circ\begin{cases}\sf \in1^o~quadrante~quando~k~\acute e~par\\\sf\in 2^o~quadrante~quando~k~\acute e~\acute impar\end{cases}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm15)\\\sf y=3cos(540^\circ)-2sen(90^\circ)+tg(180^\circ)\\\sf 540^\circ=1\cdot360^\circ+180^\circ\\\sf cos(540^\circ)=cos(180^\circ)=-1\\\sf y=3\cdot(-1)-2\cdot1+0\\\sf y=-3-2\\\sf y=-5\end{array}}$