Question

If sinθ =4/7, what is cosθ?

Answer 1

`cosθ= +- 33^(1/2)/7`

Explanation:

Since

`Sin^2θ xx cos^2θ =1`

you have

`(4/7)^2 xx cos^2θ=1`

So

`Cos^2θ= 1- 16/49`

`Cosθ= +- 33^(1/2)/7`

Answer 2

`cos(theta)=+-sqrt(33)/7`

Explanation:

We want to find `cos(theta)`, when `sin(theta)=4/7`

By the pythagorean trig identity

`color(blue)(cos^2(theta)+sin^2(theta)=1`

`=>cos^2(theta)=1-sin^2(theta)`

`=>cos(theta)=+-sqrt(1-sin^2(theta))`

Substitute `color(red)(sin(theta)=4/7`

`cos(theta)=+-sqrt(1-(4/7)^2)`

`color(white)(cos(theta))=+-sqrt(1-(16/49)`

`color(white)(cos(theta))=+-sqrt((49-16)/49)`

`color(white)(cos(theta))=+-sqrt(33)/7`

Answer 3

`cos(theta)=pmsqrt(33)/7`

Explanation:

Using the relation
`sin^2(theta)+cos^2(theta)=1` then we get
`cos(theta)=pmsqrt(1-16/49)`