Question

find the exact values of the six trigonometric functions of the angle. - 495°

sin(-495°)=□
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)

Explanation:

Step1: Find a coterminal angle

Add \(360^{\circ}\) to \(- 495^{\circ}\) until we get an angle between \(0^{\circ}\) and \(360^{\circ}\). \(-495^{\circ}+360^{\circ}=-135^{\circ}\), \(-135^{\circ}+360^{\circ}=225^{\circ}\).

Step2: Recall the unit - circle definitions of trigonometric functions

The angle \(225^{\circ}\) is in the third quadrant. The reference angle \(\theta_{r}=225^{\circ}-180^{\circ} = 45^{\circ}\).
For \(\sin(-495^{\circ})=\sin(225^{\circ})\), and in the third quadrant, \(\sin\) is negative. Since \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\), then \(\sin(225^{\circ})=-\frac{\sqrt{2}}{2}\).

Step3: Find \(\cos(-495^{\circ})\)

\(\cos(-495^{\circ})=\cos(225^{\circ})\). In the third quadrant, \(\cos\) is negative. Since \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\), then \(\cos(225^{\circ})=-\frac{\sqrt{2}}{2}\).

Step4: Find \(\tan(-495^{\circ})\)

\(\tan(-495^{\circ})=\tan(225^{\circ})=\frac{\sin(225^{\circ})}{\cos(225^{\circ})}=\frac{-\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = 1\).

Step5: Find \(\csc(-495^{\circ})\)

\(\csc(-495^{\circ})=\frac{1}{\sin(-495^{\circ})}=\frac{1}{-\frac{\sqrt{2}}{2}}=-\sqrt{2}\).

Step6: Find \(\sec(-495^{\circ})\)

\(\sec(-495^{\circ})=\frac{1}{\cos(-495^{\circ})}=\frac{1}{-\frac{\sqrt{2}}{2}}=-\sqrt{2}\).

Step7: Find \(\cot(-495^{\circ})\)

\(\cot(-495^{\circ})=\frac{1}{\tan(-495^{\circ})}=\frac{1}{1}=1\).

Answer:

\(\sin(-495^{\circ})=-\frac{\sqrt{2}}{2}\), \(\cos(-495^{\circ})=-\frac{\sqrt{2}}{2}\), \(\tan(-495^{\circ}) = 1\), \(\csc(-495^{\circ})=-\sqrt{2}\), \(\sec(-495^{\circ})=-\sqrt{2}\), \(\cot(-495^{\circ}) = 1\)