91Ó°ÊÓ

We have the angle \(\theta\) in the range of \(-\frac{\pi}{2}<\theta<0\). This means \(\theta\) belongs to fourth quadrant where \(\cos{\theta}\) is positive and \(\sin{\theta}\) is negative. Also, we already know \(\tan \theta = -2\). Using the Pythagorean identity, \(\sin^2 \theta+\cos^2\theta=1\), we can find the values of \(\sin \theta\) and \(\cos \theta\).

Since we know the value of \(\tan\theta\), which is \(-2\), we can express \(\sin\theta\) and \(\cos\theta\) in terms of \(\tan \theta\): \[\tan\theta = \frac{\sin\theta}{\cos\theta}\] From this, we can express \(\sin\theta\) and \(\cos\theta\) as follows: \[\sin\theta = \tan\theta \cdot \cos\theta\] Since \(\theta\) is in the fourth quadrant, \(\cos\theta\) is positive. Thus, disregarding the '\(-\)' in front of the 2 gives us \(\tan\theta = 2\). Now, we can find \(\cos\theta\) using Pythagorean identity: \[\cos^2\theta = 1 - \sin^2\theta = 1 - (\tan\theta \cdot \cos\theta)^2 = 1-(2\cos\theta)^2\] Let \(x=\cos\theta\), then we have a quadratic equation in terms of \(x\): \[1 - (2x)^2 = 1-4x^2=0\] Now, we can solve for \(x=\cos\theta\).

We will solve the following quadratic equation for \(x=\cos\theta\): \[4x^2=1\] \[x^2=\frac{1}{4}\] \[x=\pm\frac{1}{2}\] As \(\theta\) is in the 4th quadrant, \(\cos\theta\) is positive, so we can select the positive value of \(x\): \[\cos\theta=\frac{1}{2}\]

Since we now know the value of \(\cos\theta\), we can find \(\sin\theta\): \[\sin\theta = \tan\theta \cdot \cos\theta = (-2)\left(\frac{1}{2}\right) = -1\]

With the steps above, we conclude the following values: (a) \(\cos\theta = \frac{1}{2}\) (b) \(\sin\theta = -1\)