Given and is in quadrant II, find exact values of the six trigonometric functions.

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I know I need to set $\displaystyle cos(2\alpha)=1-2\sin^2(\alpha)$ which is then $\displaystyle \frac{31}{49}=1-2\sin^2(\alpha)$ which is then $\displaystyle 2\sin^2(\alpha)=1-\frac{31}{49}$ now I think I get $\displaystyle 2\sin^2(\alpha)= \frac{49}{49}-\frac{39}{49}=\frac{18}{49}$ from here I need some help ?$\displaystyle 2\sin^2(\alpha)=\frac{18}{49}$