Question

Se cos α= ¼ e 90º < α < 180º, calcule sen α e cotg α.

Answer 1

Sabe-se que:
sen²x + cos²x = 1
Então:
sen²$\alpha$ + cos²$\alpha$ = 1
sen²$\alpha$ + (1/4)² = 1
sen²$\alpha$ + 1/16 = 1
sen²$\alpha$ = 1- 1/16
sen²$\alpha$ = 15/16
sen$\alpha$ = $\frac{ \sqrt{15}}{4}$

cotg$\alpha$ = $\frac{1}{tg \alpha}$

tg$\alpha$ = $\frac{sen \alpha }{cos \alpha }$
cotg$\alpha$ = $\frac{cos \alpha }{sen \alpha }$
cotg$\alpha$ = $\frac{ \frac{1}{4} }{ \frac{ \sqrt{15} }{4} }$
cotg$\alpha$ = $\frac{1.4}{4. \sqrt{15}}$
$cotg \alpha$ = $\frac{1}{ \sqrt{15}}$
$cotg \alpha$ = $\frac{ \sqrt{15} }{15}$