$sin\alpha=3cos\alpha$
$⇒\dfrac{sin\alpha}{cos\alpha}=3=tan\alpha$
$cot\alpha=\dfrac{1}{tan\alpha}=\dfrac{1}{3}$
$1+tan^2\alpha=\dfrac{1}{cos^2\alpha}$
$⇒cos\alpha=±\sqrt{\dfrac{1}{1+tan^2\alpha}}=±\sqrt{\dfrac{1}{1+3^2}}=±\dfrac{\sqrt{10}}{10}$
$⇒sin\alpha=±\dfrac{3\sqrt{10}}{10}$