Vivierra
Answered

Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Get detailed and accurate answers to your questions from a community of experts on our comprehensive Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

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

Sagot :

[tex]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}[/tex]


[tex]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}[/tex]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.