Question

Someone please help! ​

Answer 1

Given:

`(\sin x)^2+(\cos x)^2=1`

`\cos \theta = (4√(2))/(7)`

To find:

The value of
`\sin \theta`.

Solution:

We have,

`\cos \theta = (4√(2))/(7)`

Putting this value in the Pythagorean identity, we get

`(\sin \theta)^2+(\cos \theta)^2=1`

`(\sin \theta)^2+\left((4√(2))/(7)\right)^2=1`

`(\sin \theta)^2+(16(2))/(49)=1`

`(\sin \theta)^2+(32)/(49)=1`

`(\sin \theta)^2=1-(32)/(49)`

`(\sin \theta)^2=(49-32)/(49)`

`(\sin \theta)^2=(17)/(49)`

Taking square root on both sides, we get

`\sin \theta=\pm \sqrt{(17)/(49)}`

`\sin \theta=\pm (√(17))/(7)`

The given value of
`\sin \theta` is positive. So,
`\sin \theta= (√(17))/(7)`.

Therefore, the value of value
`\sin \theta` is
`\sin \theta= (√(17))/(7)`.