iii. determine the exact value of each trigonometric function. 6 pts \n\\( \\cos ( - 690 ^ { \\circ } ) =…

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

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

Answer

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})