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)