Question

iii. determine the exact value of each trigonometric function. 6 pts
\\( \cos ( - 690 ^ { \circ } ) = \\)
\\( \sin ( - \frac { 29 \pi } { 6 } ) = \\)
\\( \tan ( - \frac { 2 \pi } { 3 } ) = \\)
\\( \sec ( 840 ^ { \circ } ) = \\)
\\( \csc ( - \frac { 7 \pi } { 6 } ) = \\)
\\( \cot ( 1500 ^ { \circ } ) = \\)

Explanation:

Step1: Calculate \(\cos(-690^{\circ})\)

Use the property \(\cos(-\alpha)=\cos\alpha\), so \(\cos(-690^{\circ})=\cos690^{\circ}\).
\(\cos690^{\circ}=\cos(720^{\circ} - 30^{\circ})\).
Since \(\cos(A - B)=\cos A\cos B+\sin A\sin B\) and \(\cos720^{\circ} = 1\), \(\sin720^{\circ}=0\), then \(\cos(720^{\circ}-30^{\circ})=\cos30^{\circ}=\frac{\sqrt{3}}{2}\).

Step2: Calculate \(\sin(-\frac{29\pi}{6})\)

Use the property \(\sin(-\alpha)=-\sin\alpha\), so \(\sin(-\frac{29\pi}{6})=-\sin\frac{29\pi}{6}\).
\(\sin\frac{29\pi}{6}=\sin(4\pi+\frac{5\pi}{6})\).
Since \(\sin(A + 2k\pi)=\sin A\) (\(k\in Z\)), then \(\sin(4\pi+\frac{5\pi}{6})=\sin\frac{5\pi}{6}=\frac{1}{2}\).
So \(\sin(-\frac{29\pi}{6})=-\frac{1}{2}\).

Step3: Calculate \(\tan(-\frac{2\pi}{3})\)

Use the property \(\tan(-\alpha)=-\tan\alpha\), so \(\tan(-\frac{2\pi}{3})=-\tan\frac{2\pi}{3}\).
\(\tan\frac{2\pi}{3}=\tan(\pi-\frac{\pi}{3})\).
Since \(\tan(\pi - A)=-\tan A\), then \(\tan(\pi-\frac{\pi}{3})=-\tan\frac{\pi}{3}=-\sqrt{3}\).
So \(\tan(-\frac{2\pi}{3})=\sqrt{3}\).

Step4: Calculate \(\sec(840^{\circ})\)

\(\sec(840^{\circ})=\frac{1}{\cos840^{\circ}}\).
\(\cos840^{\circ}=\cos(720^{\circ}+120^{\circ})\).
Since \(\cos(A + 2k\pi)=\cos A\) (\(k\in Z\)), then \(\cos(720^{\circ}+120^{\circ})=\cos120^{\circ}=-\frac{1}{2}\).
So \(\sec(840^{\circ})=-2\).

Step5: Calculate \(\csc(-\frac{7\pi}{6})\)

Use the property \(\csc(-\alpha)=-\csc\alpha\), so \(\csc(-\frac{7\pi}{6})=-\csc\frac{7\pi}{6}\).
\(\csc\frac{7\pi}{6}=\frac{1}{\sin\frac{7\pi}{6}}\).
\(\sin\frac{7\pi}{6}=\sin(\pi+\frac{\pi}{6})\).
Since \(\sin(\pi + A)=-\sin A\), then \(\sin(\pi+\frac{\pi}{6})=-\frac{1}{2}\).
So \(\csc\frac{7\pi}{6}=-2\), and \(\csc(-\frac{7\pi}{6}) = 2\).

Step6: Calculate \(\cot(1500^{\circ})\)

\(\cot(1500^{\circ})=\frac{\cos1500^{\circ}}{\sin1500^{\circ}}\).
\(\cos1500^{\circ}=\cos(1440^{\circ}+60^{\circ})=\cos60^{\circ}=\frac{1}{2}\).
\(\sin1500^{\circ}=\sin(1440^{\circ}+60^{\circ})=\sin60^{\circ}=\frac{\sqrt{3}}{2}\).
So \(\cot(1500^{\circ})=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{\sqrt{3}}{3}\).

Answer:

\(\cos(-690^{\circ})=\frac{\sqrt{3}}{2}\), \(\sin(-\frac{29\pi}{6})=-\frac{1}{2}\), \(\tan(-\frac{2\pi}{3})=\sqrt{3}\), \(\sec(840^{\circ})=-2\), \(\csc(-\frac{7\pi}{6}) = 2\), \(\cot(1500^{\circ})=\frac{\sqrt{3}}{3}\)