Question

If `sinx=1/3`, how do you find `cos2x`?

Answer 1

I found: `cos(2x)=7/9`

Explanation:

Considering that:
`cos(2x)=cos^2(x)-sin^2(x)`
but
`1=sin^2(x)+cos^2(x)`
rearranging:
`cos^2(x)=1-sin^2(x)`
so:
`cos(2x)=cos^2(x)-sin^2(x)=`
`=1-sin^2(x)-sin^2(x)=`
`=1-2sin^2(x)`
Substituting:
`cos(2x)=1-2sin^2(x)=1-2(1/3)^2=1-2/9=7/9`