Suppose that $\frac{\pi}{2} < t < \pi$ , and that $\displaystyle \csc t=2$ . Find the values of the trigonometric functions $\sin(t)$ , $\cos(t)$ , and $\tan(t)$ .
$\displaystyle \left[ \sin t={{\sqrt{3}}\over{2}} , \cos t={{1}\over{2}} , \tan t= \sqrt{3} \right]$
$\displaystyle \left[ \sin t={{1}\over{2}} , \cos t=-{{\sqrt{3}}\over{2}} , \tan t= -{{1}\over{\sqrt{3}}} \right]$
$\displaystyle \left[ \sin t=-{{\sqrt{3}}\over{2}} , \cos t={{1}\over{2}} , \tan t= -\sqrt{3} \right]$
$\displaystyle \left[ \sin t={{1}\over{2}} , \cos t={{\sqrt{3}}\over{2}} , \tan t= {{1}\over{\sqrt{3}}} \right]$
Choose the correct trigonometric identity.
$\displaystyle -\sin x\,\tan x-\cos x=2\,\cos x-\sec x$
$\displaystyle -\sin x\,\tan x-\cos x=\sec x$
$\displaystyle -\sin x\,\tan x-\cos x=-\sec x$
$\displaystyle -\sin x\,\tan x-\cos x={{2\,\sec x}\over{\left(\csc x\right)^2}}- \sec x$
Choose the correct trigonometric identity.
$\displaystyle \left(\sin ^2t+1\right)\,\left(1-\tan ^2t\right)=1$
$\displaystyle \left(\sin ^2t+1\right)\,\left(1-\tan ^2t\right)=-{{2\,\sin ^4t+ \sin ^2t-1}\over{\cos ^2t}}$
$\displaystyle \left(\sin ^2t+1\right)\,\left(1-\tan ^2t\right)={{\sin ^2t+1 }\over{\cos ^2t}}$
$\displaystyle \left(\sin ^2t+1\right)\,\left(1-\tan ^2t\right)=2\,\cos ^2t-1$
Find the derivative for $f(x) =\displaystyle -3\,\csc x-2\,\cos x$ .
$f'(x) =\displaystyle -2\,\sin x-3\,\sec x$ $f'(x) =\displaystyle 3\,\sec x\,\tan x+2\,\cos x$ $f'(x) =\displaystyle -2\,\sin x-3\,\cot x\,\csc x$ $f'(x) =\displaystyle -3\,\sec x\,\tan x-2\,\cos x$ $f'(x) =\displaystyle 2\,\sin x+3\,\cot x\,\csc x$
Suppose that $\displaystyle g\left( {{4\,\pi}\over{3}} \right) = 3$ and $\displaystyle g'\left( {{4\,\pi}\over{3}} \right) = 2$ . Then find the derivative $\displaystyle f'\left( {{4\,\pi}\over{3}} \right)$ for $f(x) = \displaystyle g\left(x\right)\,\sin x$ .
2 $\displaystyle {{3}\over{2}}-\sqrt{3}$ $\displaystyle -{{3^{{{3}\over{2}}}}\over{2}}$ $\displaystyle -\sqrt{3}-{{3}\over{2}}$ $\displaystyle \sqrt{3}-{{3}\over{2}}$
Suppose that $0 < t < \frac{\pi}{2}$ , and that $\displaystyle \csc t=4$ . Find the values of the trigonometric functions $\sin(t)$ , $\cos(t)$ , and $\tan(t)$ .
$\displaystyle \left[ \sin t={{\sqrt{15}}\over{4}} , \cos t={{1}\over{4}} , \tan t= \sqrt{15} \right]$
$\displaystyle \left[ \sin t=-{{1}\over{15}} , \cos t={{1}\over{4}} , \tan t=-{{4 }\over{15}} \right]$
$\displaystyle \left[ \sin t={{1}\over{4}} , \cos t=-{{1}\over{15}} , \tan t=-{{15 }\over{4}} \right]$
$\displaystyle \left[ \sin t={{1}\over{4}} , \cos t={{\sqrt{15}}\over{4}} , \tan t= {{1}\over{\sqrt{15}}} \right]$
Find the derivative for $f(x) =\displaystyle e^{x}\,\sec x$ .
$f'(x) =\displaystyle e^{x}\,\sec x\,\left(\tan x+1\right)$ $f'(x) =\displaystyle -e^{x}\,\left(\cot x-1\right)\,\csc x$ $f'(x) =\displaystyle -e^{x}\,\sec x\,\left(\tan x+1\right)$ $f'(x) =\displaystyle e^{x}\,\left(\cot x-1\right)\,\csc x$
Choose the correct trigonometric identity.
$\displaystyle -{{\sin \left(2\,t\right)}\over{\cos \left(2\,t\right)+1}}=-\tan t$
$\displaystyle -{{\sin \left(2\,t\right)}\over{\cos \left(2\,t\right)+1}}=\tan t$
$\displaystyle -{{\sin \left(2\,t\right)}\over{\cos \left(2\,t\right)+1}}=\cot t$
$\displaystyle -{{\sin \left(2\,t\right)}\over{\cos \left(2\,t\right)+1}}=-\cot t$
Find the values of $\displaystyle \sin t$ and $\displaystyle \cos t$ for $\displaystyle t={{5\,\pi}\over{4}}$ .
$\displaystyle \left[ \sin t={{1}\over{\sqrt{2}}} , \cos t={{1}\over{\sqrt{2}}} \right]$
$\displaystyle \left[ \sin t=-{{1}\over{\sqrt{2}}} , \cos t={{1}\over{\sqrt{2}}} \right]$
$\displaystyle \left[ \sin t={{1}\over{\sqrt{2}}} , \cos t=-{{1}\over{\sqrt{2}}} \right]$
$\displaystyle \left[ \sin t=-{{1}\over{\sqrt{2}}} , \cos t=-{{1}\over{\sqrt{2}}} \right]$
Choose the correct trigonometric identity.
$\sin( t-{{\pi}\over{2}}) = -\sin t$
$\sin( t-{{\pi}\over{2}}) = \sin t$
$\sin( t-{{\pi}\over{2}}) = \cos t$
$\sin( t-{{\pi}\over{2}}) = -\cos t$

Department of Mathematics
Last modified: 2025-08-29