Question

ex 4: find the six trig functions of 315°.

Explanation:

Step1: Determine the reference angle

Since \(315^{\circ}=360^{\circ} - 45^{\circ}\), the reference angle \(\theta'=45^{\circ}\). And \(315^{\circ}\) is in the fourth - quadrant. In the fourth - quadrant, \(\sin\theta\lt0\), \(\cos\theta\gt0\), \(\tan\theta\lt0\), \(\csc\theta\lt0\), \(\sec\theta\gt0\), \(\cot\theta\lt0\).

Step2: Calculate \(\sin315^{\circ}\)

We know that \(\sin\theta=-\sin\theta'\) (because of the fourth - quadrant). Since \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\), then \(\sin315^{\circ}=-\frac{\sqrt{2}}{2}\)

Step3: Calculate \(\cos315^{\circ}\)

We know that \(\cos\theta=\cos\theta'\) (because of the fourth - quadrant). Since \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\), then \(\cos315^{\circ}=\frac{\sqrt{2}}{2}\)

Step4: Calculate \(\tan315^{\circ}\)

We know that \(\tan\theta=-\tan\theta'\) (because of the fourth - quadrant). Since \(\tan45^{\circ}=1\), then \(\tan315^{\circ}=- 1\)

Step5: Calculate \(\csc315^{\circ}\)

Since \(\csc\theta=\frac{1}{\sin\theta}\), and \(\sin315^{\circ}=-\frac{\sqrt{2}}{2}\), then \(\csc315^{\circ}=-\sqrt{2}\)

Step6: Calculate \(\sec315^{\circ}\)

Since \(\sec\theta=\frac{1}{\cos\theta}\), and \(\cos315^{\circ}=\frac{\sqrt{2}}{2}\), then \(\sec315^{\circ}=\sqrt{2}\)

Step7: Calculate \(\cot315^{\circ}\)

Since \(\cot\theta=\frac{1}{\tan\theta}\), and \(\tan315^{\circ}=-1\), then \(\cot315^{\circ}=-1\)

Answer:

\(\sin315^{\circ}=-\frac{\sqrt{2}}{2}\), \(\cos315^{\circ}=\frac{\sqrt{2}}{2}\), \(\tan315^{\circ}=-1\), \(\csc315^{\circ}=-\sqrt{2}\), \(\sec315^{\circ}=\sqrt{2}\), \(\cot315^{\circ}=-1\)