Answer:
[tex]\cos \theta=-\frac{2\sqrt{5}}{5}[/tex]
Step-by-step explanation:
It was given that:
[tex]\tan(\theta)=-\frac{1}{2}[/tex]
and [tex]\frac{\pi}{2}\le \theta \le \pi[/tex].
We use the relation:
[tex]\sec^2 \theta=1+\tan^2\theta[/tex]
We substitute the value to get:
[tex]\sec^2 \theta=1+(\frac{1}{2})^2[/tex]
[tex]\sec^2 \theta=1+\frac{1}{4}[/tex]
[tex]\sec^2 \theta=\frac{5}{4}[/tex]
[tex]\sec \theta=\pm \frac{\sqrt{5}}{2}[/tex]
Reciprocate both sides to get
[tex]\cos \theta=\pm \frac{2}{\sqrt{5}}[/tex]
Rationalize the denominator:
[tex]\cos \theta=\pm \frac{2\sqrt{5}}{5}[/tex]
The given interval, [tex]\frac{\pi}{2}\le \theta \le \pi[/tex] is the same as the 2nd quadrant where the cosine ratio is negative.
[tex]\therefore \cos \theta=-\frac{2\sqrt{5}}{5}[/tex]
