Question

find the value of the trigonometric function sin (t) if sec t = -4/3 and the terminal side of angle t lies in quadrant II

Answer 1

`sin(t) =(√(7))/(4)`

Explanation:

By definition we know that

`sec(t) = (1)/(cos(t))`

and

`cos ^ 2(t) = 1-sin ^ 2(t)`

As
`sec(t) = -(4)/(3)`

Then

`sec(t) = -(4)/(3)\\\\(1)/(cos(t)) =-(4)/(3)\\\\cos(t) = -(3)/(4)`

Now square both sides of the equation:

`cos^2(t) = (-(3)/(4))^2`

`cos^2(t) = (9)/(16)\\\\`

`1-sin^2(t) =(9)/(16)\\\\sin^2(t) =1-(9)/(16)\\\\sin^2(t) =(7)/(16)\\\\sin(t) =\±\sqrt{(7)/(16)}`

In the second quadrant sin (t) is positive. Then we take the positive root

`sin(t) =\sqrt{(7)/(16)}`

`sin(t) =(√(7))/(4)`