Question

PLEASE ANSWER! Given that cos(Θ)=3/8 and the angle Θ is in the first quadrant, find the value for sin(Θ).

Answer 1

sinΘ =
`(√(55) )/(8)`

Explanation:

using the identity

sin²Θ + cos²Θ = 1 ( subtract cos²Θ from both sides )

sin²Θ = 1 - cos²Θ ( take square root of both sides )

sinΘ = ±
`√(1 -cos^20)`

given

cosΘ =
`(3)/(8)` , then

sinΘ = ±
`\sqrt{1-((3)/(8))^2 }`

= ±
`\sqrt{1-(9)/(64) }`

= ±
`\sqrt{(64)/(64)-(9)/(64) }`

= ±
`\sqrt{(55)/(64) }`

= ±
`(√(55) )/(√(64) )`

= ±
`(√(55) )/(8)`

since Θ is in the first quadrant then sinΘ > 0 , so

sinΘ =
`(√(55) )/(8)`