Question

Given tan0 = -1 and cos0 > 0, find sin0 Given tan0 = -1 and cos0 > 0, find sin0 @Mathematics

Answer 1

This is Unit Circle

Answer 2

\[\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}=-1 {\rightarrow} \frac{\sin^{2}{\theta}}{\cos^{2}{\theta}}=1\]after squaring both sides. Then, \[\sin^{2}{\theta}=\cos^{2}{\theta}\]\[\sin^{2}{\theta}=1-\sin^{2}{\theta}\]\[2\sin^{2}{\theta}=1 {\rightarrow} \sin^{2}{\theta}=\frac{1}{2}\]Hence,\[\sin{\theta}={\pm} \frac{1}{\sqrt{2}}\]Since tan(theta) is negative, but cos(theta) positive, taking the positive result for sin(theta) would lead to a contradiction; i.e.\[\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}} {\iff}(negative)=\frac{(positive)}{(negative)}\]which is false. We take the negative solution:\[\sin{\theta}=-\frac{1}{\sqrt{2}}\]