Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 71146 by sadimuhmud 136 last updated on 12/Oct/19

$$\sin \alpha =\frac{12}{13}\mathrm{and}\alpha \mathrm{is}\mathrm{in}\mathrm{the}2\mathrm{nd}\mathrm{quadrent}.$$$$\mathrm{prove}\cos \alpha =-\frac{5}{13}$$

Commented by Rio Michael last updated on 12/Oct/19

$$sin\alpha =\frac{12}{13}$$$$from{sin}^2\alpha +{cos}^2\alpha =1$$$$\Rightarrow {cos}^2\alpha =1-{sin}^2\alpha$$$${cos}^2\alpha =1-{\left(\frac{12}{13}\right)}^2$$$$cos\alpha =\pm \sqrt{\frac{25}{169}}$$$$cos\alpha =-\frac{5}{13}$$$$reasonbeingthatthecosineratioisnegativeinthesecondquadrant.$$

Answered by Kunal12588 last updated on 12/Oct/19

$$sin\alpha =\frac{12}{13}$$$$usingidentity:-{\sin }^2\theta +{\cos }^2\theta =1$$$$\Rightarrow cos\alpha =-\frac{\sqrt{{13}^2-{12}^2}}{13}[\because \frac{\pi }{2}<\alpha ⩽\pi ]$$$$\Rightarrow cos\alpha =-\frac{\sqrt{169-144}}{13}=-\frac{\sqrt{25}}{13}=-\frac{5}{13}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com