Knowing that [tex]\sec x=\dfrac{1}{\cos x}[/tex]:
[tex]\sec\theta=4\iff\dfrac{1}{\cos\theta}=4\iff\cos\theta=\dfrac{1}{4}[/tex]
Using that [tex]\sin^2x+\cos^2x=1[/tex]:
[tex]\sin^2\theta+(\dfrac{1}{4})^2=1\iff \sin^2\theta=\dfrac{15}{16}\iff\\\\\boxed{\sin\theta=\pm\dfrac{\sqrt{15}}{4}}[/tex]
