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Using the cosine double angle formula,
[tex]\cos 2\theta=2\cos^2 \theta-1=\frac{12}{13}\\\\2\cos^{2} \theta=\frac{25}{13}\\\\\cos^{2} \theta=\frac{25}{26}\\\\\boxed{\cos \theta=\frac{5}{\sqrt{26}}}[/tex]
(Note I took the positive case since [tex]\theta[/tex] terminates in the first quadrant)
Using the Pythagorean identity,
[tex]\sin^2 \theta+\cos^2 \theta=1\\\\\sin^2 \theta+\frac{25}{26}=1\\\\sin^2 \theta=\frac{1}{26}\\\\\boxed{\sin \theta=\frac{1}{\sqrt{26}}}[/tex]
(Note I took the positive case since [tex]\theta[/tex] terminates in the first quadrant)