Question

if cosx=1/12 and sinx>0 find tan2x

Answer 1

$cosx=\frac{1}{12}\\\\sin^2x+cos^2x=1\\\\sin^2x+\left(\frac{1}{12}\right)^2=1\\\\sin^2x+\frac{1}{144}=1\\\\sin^2x=1-\frac{1}{144}\\\\sin^2x=\frac{144}{144}-\frac{1}{144}\\\\sin^2x=\frac{143}{144}\\\\sinx=\sqrt\frac{143}{144}\\\\sinx=\frac{\sqrt{143}}{12}$


$tan2x=\frac{sin2x}{cos2x}\\\\sin2x=2sinxcosx;\ cos2x=cos^2x-sin^2x\\\\sin2x=2\cdot\frac{\sqrt{143}}{12}\cdot\frac{1}{12}=\frac{2\sqrt{143}}{144}\\\\cos2x=\left(\frac{1}{12}\right)^2-\left(\frac{\sqrt{143}}{12}\right)^2=\frac{1}{144}-\frac{143}{144}=-\frac{142}{144}\\\\tan2x=\frac{2\sqrt{143}}{144}:\left(-\frac{142}{144}\right)=-\frac{2\sqrt{143}}{144}\cdot\frac{144}{142}=-\frac{\sqrt{143}}{71}$