Question

Need help quick please!

Answer 1

cosΘ =
`(√(7) )/(4)`

Explanation:

using the identity

sin²x + cos²x = 1 ( subtract sin²x from both sides )

cos²x = 1 - sin²x ( take square root of both sides )

cosx = ±
`√(1-sin^2x)`

given

sinΘ =
`(3)/(4)` , then

cosΘ = ±
`\sqrt{1-((3)/(4))^2 }`

= ±
`\sqrt{1-(9)/(16) }`

= ±
`\sqrt{(16)/(16)-(9)/(16) }`

= ±
`\sqrt{(7)/(16) }`

= ±
`(√(7) )/(√(16) )`

since Θ is in quadrant 1 , then cosΘ > 0

cosΘ =
`(√(7) )/(4)`