Respuesta :
Use the identity,
[tex] \sin^2 \theta+\cos^2 \theta=1\\
\cos \theta=\pm\sqrt{1-\sin^2 \theta} [/tex].
Here [tex] \sin \theta=\frac{12}{13} \\
1-\sin^2 \theta=1-(\frac{12}{13} )^2\\
\cos^2 \theta=\frac{25}{169} \\
\cos \theta=\pm \sqrt{\frac{25}{169} } \\
\cos \theta=\pm \frac{5}{13} [/tex]
When [tex] \sin \theta [/tex] is positive [tex] \cos \theta [/tex] can be positive or negative.
