find the exact values of the six trigonometric functions of the given angle. do not use a calculator.\n210°\n…

Explanation:

Step1: Find the reference angle

The reference angle $\theta'$ for an angle $\theta = 210^{\circ}$ (which is in the third - quadrant, since (180^{\circ}<210^{\circ}<270^{\circ})) is given by (\theta'=\theta - 180^{\circ}). So, (\theta'=210^{\circ}-180^{\circ}=30^{\circ})

Step2: Determine the sign of the trigonometric functions

In the third - quadrant, (\sin\theta<0), (\cos\theta<0), and (\tan\theta>0)

Step3: Calculate (\sin210^{\circ})

We know that (\sin\theta=-\sin\theta') (because (\sin) is negative in the third - quadrant). Since (\sin30^{\circ}=\frac{1}{2}), then (\sin210^{\circ}=-\frac{1}{2})

Step4: Calculate (\cos210^{\circ})

We know that (\cos\theta=-\cos\theta') (because (\cos) is negative in the third - quadrant). Since (\cos30^{\circ}=\frac{\sqrt{3}}{2}), then (\cos210^{\circ}=-\frac{\sqrt{3}}{2})

Step5: Calculate (\tan210^{\circ})

We know that (\tan\theta=\tan\theta') (because (\tan=\frac{\sin}{\cos}), and both (\sin) and (\cos) are negative in the third - quadrant, so their ratio is positive). Since (\tan30^{\circ}=\frac{\sqrt{3}}{3}), then (\tan210^{\circ}=\frac{\sqrt{3}}{3})

Answer:

For (\sin210^{\circ}): A. (\sin210^{\circ}=-\frac{1}{2}) For (\cos210^{\circ}): A. (\cos210^{\circ}=-\frac{\sqrt{3}}{2}) For (\tan210^{\circ}): A. (\tan210^{\circ}=\frac{\sqrt{3}}{3})