We have an angle θ in the II quadrant for which sin(θ) = 12/13.
We have to find the cosine of θ.
We can start by drawing a possible angle:
We can relate the cosine and sine of an angle with the identity:
[tex]\sin ^2(\theta)+\cos ^2(\theta)=1[/tex]We then can calculate cos(θ) as:
[tex]\begin{gathered} \cos ^2(\theta)=1-\sin ^2(\theta) \\ \cos ^2(\theta)=1-(\frac{12}{13})^2 \\ \cos ^2(\theta)=1-\frac{144}{169} \\ \cos ^2(\theta)=\frac{169-144}{169} \\ \cos ^2(\theta)=\frac{25}{169} \\ \cos ^2(\theta)=\frac{5^2}{13^2} \\ \cos ^2(\theta)=(\frac{5}{13})^2 \\ \frac{\pi}{2}<\theta<\pi\Rightarrow\cos (\theta)<0\Rightarrow\cos (\theta)=-\frac{5}{13} \end{gathered}[/tex]NOTE: the last step is because we know that the angle θ is in the second quadrant. Therefore, the cosine of θ is negative (see picture above). The identity is still valid, because the square of (-5/13) is still 25/169.
Now, we can calculate the tangent of θ as:
[tex]\tan (\theta)=\frac{\sin (\theta)}{\cos (\theta)}=\frac{\frac{12}{13}}{-\frac{5}{13}}=\frac{12}{-5}=-\frac{12}{5}[/tex]Answer:
cos(θ) = -5/13
tan(θ) = -12/5
