Question

If the terminal ray of β lies in the fourth quadrant and sin (β) = -√3/3 determine cos(β) in simplest form.

Answer 1

`cos(\beta)=(√(6))/(3)}`

Explanation:

we know that

If the terminal ray of β lies in the fourth quadrant

then

sin (β) is negative and cos (β) is positive

Remember the trigonometric identity

`sin^2(\beta)+cos^2(\beta)=1`

we have

`sin(\beta)=-(√(3))/(3)`

substitute

`(-(√(3))/(3))^2+cos^2(\beta)=1`

`(3)/(9)+cos^2(\beta)=1`

`cos^2(\beta)=1-(3)/(9)`

`cos^2(\beta)=(6)/(9)`

apply root square both sides

`cos(\beta)=(√(6))/(3)}`