Question

If cos ⁡θ=−8/17 and 180°θ < 270°, what is sin θ?

Answer 1

Here is your answer


`[I am using ☆ instead of theta]`


`cos☆= -8/17`


`{cos}^(2)☆= {(-8/17)}^(2)= 64/289`

Now,


`sin☆= \sqrt{1-{cos}^(2)☆}`


`sin☆= √(1- (64/289))`


`sin☆= √((289-64)/289)`


`sin☆= √(225/289)`


`sin☆= 15/17`

Since, 180<theta<270

so, it must lie in 3rd quadrant.

In third quadrant


`sin☆ is -ve`


`Hence, sin☆= -15/17`

HOPE IT IS USEFUL

Answer 2

sinΘ = -
`(15)/(17)`

Explanation:

Using the trigonometric identity

sin²x + cos²x = 1

⇒ sinx = ±
`√(1-cos^2x)`

Since 180 < Θ < 270 then sinΘ < 0

sinΘ = -
`√(1-(-8/17)^2)`

= -
`\sqrt{1-(64)/(289) }`

= -
`\sqrt{(225)/(289) }` = -
`(15)/(17)`