Question

Please please help me out

Answer 1

Explanation:

`{ \sin(x) }^(2) + { \cos(x) }^(2) = 1 \\ { \cos(theta) }^(2) = (9)/(16) \\ \: { \sin(theta) }^(2) = (6)/(16) = (3)/(8) \\ \: \sin(theta) = ( √(3) )/( √(8) ) \\ \sin(theta ) = ( √(24) )/(8)</p><p> = ( 2√(6) )/(8) = ( √(6) )/(4) </p><p>`

Answer 2

`(√(7) )/(4)`

Explanation:

Using the Pythagorean identity

sin²x + cos²x = 1, then

sinx =
`√(1-cos^2x)`

sinΘ =
`√(1-(3/4)^2)`

=
`\sqrt{1-(9)/(16) }` =
`\sqrt{(7)/(16) }` =
`(√(7) )/(4)`