Generating...                               s20quiz10_n0

  1. 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={{1}\over{2}} , \cos t={{\sqrt{3}}\over{2}} , \tan t= {{1}\over{\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={{\sqrt{3}}\over{2}} , \cos t={{1}\over{2}} , \tan t= \sqrt{3} \right] $

  2. Suppose that $\displaystyle g\left( -{{\pi}\over{4}} \right) = 1 $ and $\displaystyle g'\left( -{{\pi}\over{4}} \right) = 2 $. Then find the derivative $\displaystyle f'\left( -{{\pi}\over{4}} \right)$ for $f(x) = \displaystyle g\left(x\right)\,\sin x $ .

    2 $\displaystyle -{{3}\over{\sqrt{2}}}$ $\displaystyle -{{1}\over{\sqrt{2}}}$ $\displaystyle {{3}\over{\sqrt{2}}}$ $\displaystyle -{{1}\over{\sqrt{2}}}$

  3. Find the derivative $f'(x)$ for $f(x) =\displaystyle \cot x\,\sec x $ .

    $f'(x) =\displaystyle -\sec x\,\tan x $ $f'(x) =\displaystyle \cot x\,\csc x $ $f'(x) =\displaystyle \sec x\,\tan x $ $f'(x) =\displaystyle \sec x $ $f'(x) =\displaystyle -\cot x\,\csc x $

  4. Suppose that $0 < t < \frac{\pi}{2}$, and that $\displaystyle \csc t=3$ . Find the values of the trigonometric functions $\sin(t)$, $\cos(t)$, and $\tan(t)$.

    $\displaystyle \left[ \sin t={{1}\over{3}} , \cos t={{2^{{{3}\over{2}}}}\over{3}} , \tan t={{1}\over{2^{{{3}\over{2}}}}} \right] $

    $\displaystyle \left[ \sin t=-{{1}\over{8}} , \cos t={{1}\over{3}} , \tan t=-{{3 }\over{8}} \right] $

    $\displaystyle \left[ \sin t={{2^{{{3}\over{2}}}}\over{3}} , \cos t={{1}\over{3}} , \tan t=2^{{{3}\over{2}}} \right] $

    $\displaystyle \left[ \sin t={{1}\over{3}} , \cos t=-{{1}\over{8}} , \tan t=-{{8 }\over{3}} \right] $

  5. Find the derivative $f'(x)$ for $f(x) =\displaystyle {{\cos x}\over{x}} $ .

    $f'(x) =\displaystyle -{{x\,\sin x+\cos x}\over{x^2}} $ $f'(x) =\displaystyle {{x\,\sin x+\cos x}\over{x^2}} $ $f'(x) =\displaystyle -{{\sin x-x\,\cos x}\over{x^2}} $ $f'(x) =\displaystyle {{\sin x-x\,\cos x}\over{x^2}} $

  6. Choose the correct trigonometric identity.

    $\sin( t+{{3\,\pi}\over{2}}) = \cos t$

    $\sin( t+{{3\,\pi}\over{2}}) = -\sin t$

    $\sin( t+{{3\,\pi}\over{2}}) = \sin t$

    $\sin( t+{{3\,\pi}\over{2}}) = -\cos t$

  7. Choose the correct trigonometric identity.

    $\displaystyle \left(\csc t\right)^2+\left(\cot t\right)^2={{\cos ^2t+1}\over{ \sin ^2t}}$

    $\displaystyle \left(\csc t\right)^2+\left(\cot t\right)^2=-1$

    $\displaystyle \left(\csc t\right)^2+\left(\cot t\right)^2=-{{\cos ^2t+1}\over{ \sin ^2t}}$

    $\displaystyle \left(\csc t\right)^2+\left(\cot t\right)^2=1$

  8. Suppose that $\frac{3\pi}{2} < t < 2\pi$, and that $\displaystyle \csc t=-2$ . Find the values of the trigonometric functions $\sin(t)$, $\cos(t)$, and $\tan(t)$.

    $\displaystyle \left[ \sin t=-{{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=-{{1}\over{3}} , \tan t={{3 }\over{2}} \right] $

    $\displaystyle \left[ \sin t=-{{1}\over{3}} , \cos t=-{{1}\over{2}} , \tan t={{2 }\over{3}} \right] $

  9. Find the derivative $f'(x)$ for $f(x) =\displaystyle e^{x}\,\sin x $ .

    $f'(x) =\displaystyle e^{x}\,\left(\sin x-\cos x\right) $ $f'(x) =\displaystyle -e^{x}\,\left(\sin x+\cos x\right) $ $f'(x) =\displaystyle -e^{x}\,\left(\sin x-\cos x\right) $ $f'(x) =\displaystyle e^{x}\,\left(\sin x+\cos x\right) $

  10. Choose the correct trigonometric identity.

    $\displaystyle \left(\sin ^2t+1\right)\,\left(1-\tan ^2t\right)=2\,\cos ^2t-1$

    $\displaystyle \left(\sin ^2t+1\right)\,\left(1-\tan ^2t\right)=1$

    $\displaystyle \left(\sin ^2t+1\right)\,\left(1-\tan ^2t\right)=-{{2\,\sin ^4t+ \sin ^2t-1}\over{\cos ^2t}}$

    $\displaystyle \left(\sin ^2t+1\right)\,\left(1-\tan ^2t\right)={{\sin ^2t+1 }\over{\cos ^2t}}$


Department of Mathematics
Last modified: 2025-06-19