You have that [tex]\cos \theta=- \frac{2}{5} [/tex] and [tex]\tan \theta\ \textgreater \ 0[/tex]. Since [tex]\tan \theta= \frac{\sin\theta}{\cos\theta} [/tex] you can conclude that [tex]\sin\theta\ \textless \ 0[/tex].
Let consider the right triangle with leg 2, hypotenuse 5 and an angle [tex]\theta[/tex], which lies between given leg and hypotenuse. Be Pythagorean theorem the second leg is equal to [tex] \sqrt{5^2-2^2} = \sqrt{21} [/tex]. Then [tex]\sin\theta=- \frac{ \sqrt{21} }{5} [/tex].
