Trigonometric angle sum formula

Question

I got: $$\cos \left(\alpha -\frac{\pi }{4}\right),\:\cos \left(\alpha \right)=-\frac{1}{3},\:\frac{\pi }{2}<\alpha <\pi $$ $$\cos \left(\alpha -\frac{\pi }{4}\right)=?$$

I know that: $$\cos \left(\alpha -\beta \right)=\cos \left(\alpha \right)\cos \left(\beta \right)+\sin \left(\beta \right)\sin \left(\alpha \right)$$ $$cos\left(α−\frac{\pi }{4}\right)=\cos \left(\alpha \right)\cos \left(\frac{\pi }{4}\right)+\sin \left(\frac{\pi }{4}\right)\sin \left(\alpha \right)$$ $$-\frac{1}{3}\cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\cdot \sin \left(\alpha \right)$$

But I can't find $sin(a)$

Answer 1

$\cos^2{\alpha}=\frac{1}{9}$

$1-\sin^2{\alpha}=\frac{1}{9}$

$\implies \sin(\alpha)=\frac{2\sqrt2}{3}$

Answer 2

$\sin\alpha =\pm\sqrt{1-\cos^2\alpha}=\pm\sqrt{1-({-\frac13})^2}=\pm\frac{\sqrt8}3=\pm\frac{2\sqrt2}3$.

Since we are in the $2$nd quadrant, $\sin\alpha =\frac{2\sqrt2}3$.