Question

Evalute costheta if sintheta = (sqrt5)/3

Answer 1

`\large\boxed{\cos\theta=\pm(2)/(3)}`

Explanation:

Use
`\sin^2x+\cos^2x=1`.

We have

`\sin\theta=(\sqrt5)/(3)`

Substitute:

`\left((\sqrt5)/(3)\right)^2+\cos^2\theta=1\qquad\text{use}\ \left((a)/(b)\right)^n=(a^n)/(b^n)\\\\((\sqrt5)^2)/(3^2)+\cos^2\theta=1\qquad\text{use}\ (√(a))^2=a\\\\(5)/(9)+\cos^2\theta=1\qquad\text{subtract}\ (5)/(9)\ \text{from both sides}\\\\\cos^2\theta=(9)/(9)-(5)/(9)\\\\\cos^2\theta=(4)/(9)\to \cos\theta=\pm\sqrt{(4)/(9)}\\\\\cos\theta=\pm(\sqrt4)/(\sqrt9)\\\\\cos\theta=\pm(2)/(3)`

Answer 2

Answer:
`cos(\theta)=\±(2)/(3)`

Explanation:

In this case we know that:

`sin(\theta) = (√(5))/(3)`

To find the value of
`cos(\theta)` we use the following trigonometric identity

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

So

`sin^2(\theta) = ((√(5))/(3))^2`

Therefore

`cos^2(\theta)=1-((√(5))/(3))^2`

`cos^2(\theta)=1-(5)/(9)`

`cos^2(\theta)=(4)/(9)`

`cos(\theta)=\±\sqrt{(4)/(9)}`

`cos(\theta)=\±(2)/(3)`