Olá, Marlize.
[tex]P = \cos{(\frac{k\pi}3)}, Q = \sin{(\frac{k\pi}6)}, k=1,2,3\\\\\\ \text{a) Se }k=1:P+2Q= \cos{(\frac{\pi}3)}+2\sin{(\frac{\pi}6)}=\frac12+2\cdot \frac12=\frac32[/tex]
[tex]\text{b) Se }k=2:P+2Q= \cos{\underbrace{(\frac{2\pi}3)}_{=\pi-\frac{\pi}3}+2\sin{(\frac{2\pi}6)}=[/tex]
[tex]-\cos{(\frac{\pi}3)}+2\sin{(\frac{\pi}3)}=-\frac12+2\frac{\sqrt3}2=-\frac12+\sqrt3\\[/tex]
[tex]\text{c) Se }k=3:P+2Q= \cos{(\frac{3\pi}3)}+2\sin{(\frac{3\pi}6)}=\cos{\pi}}+2\sin{(\frac{\pi}2)}=\\=-1+2\cdot 1=1[/tex]
