Question

find the exact values of the six trigonometric functions of the angle. - 315° sin(-315°)= (simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators.)

Explanation:

Step1: Rewrite the angle

We know that \(-315^{\circ}= - 360^{\circ}+45^{\circ}\), and trigonometric - function values of an angle \(\theta\) and \(\theta + 360^{\circ}n\) (\(n\in\mathbb{Z}\)) are the same. So, the trigonometric - function values of \(-315^{\circ}\) are the same as those of \(45^{\circ}\).

Step2: Calculate \(\sin(-315^{\circ})\)

Since \(\sin(-315^{\circ})=\sin(45^{\circ})\), and for an angle of \(45^{\circ}\) in a right - triangle with sides \(a = b = 1\) and hypotenuse \(c=\sqrt{1^{2}+1^{2}}=\sqrt{2}\), by the definition of sine function \(\sin\theta=\frac{y}{r}\), we have \(\sin(45^{\circ})=\frac{\sqrt{2}}{2}\).

Step3: Calculate \(\cos(-315^{\circ})\)

\(\cos(-315^{\circ})=\cos(45^{\circ})\), and by the definition of cosine function \(\cos\theta=\frac{x}{r}\), for \(45^{\circ}\), \(\cos(45^{\circ})=\frac{\sqrt{2}}{2}\).

Step4: Calculate \(\tan(-315^{\circ})\)

\(\tan\theta=\frac{\sin\theta}{\cos\theta}\), so \(\tan(-315^{\circ})=\tan(45^{\circ})=\frac{\sin(45^{\circ})}{\cos(45^{\circ})}=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1\).

Step5: Calculate \(\csc(-315^{\circ})\)

\(\csc\theta=\frac{1}{\sin\theta}\), so \(\csc(-315^{\circ})=\csc(45^{\circ})=\frac{1}{\sin(45^{\circ})}=\sqrt{2}\).

Step6: Calculate \(\sec(-315^{\circ})\)

\(\sec\theta=\frac{1}{\cos\theta}\), so \(\sec(-315^{\circ})=\sec(45^{\circ})=\frac{1}{\cos(45^{\circ})}=\sqrt{2}\).

Step7: Calculate \(\cot(-315^{\circ})\)

\(\cot\theta=\frac{1}{\tan\theta}\), so \(\cot(-315^{\circ})=\cot(45^{\circ})=\frac{1}{\tan(45^{\circ})}=1\).

Answer:

\(\sin(-315^{\circ})=\frac{\sqrt{2}}{2}\), \(\cos(-315^{\circ})=\frac{\sqrt{2}}{2}\), \(\tan(-315^{\circ}) = 1\), \(\csc(-315^{\circ})=\sqrt{2}\), \(\sec(-315^{\circ})=\sqrt{2}\), \(\cot(-315^{\circ}) = 1\)