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Thread: Trigonometric Identities

  1. #1
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    Trigonometric Identities


    Hi, I was wondering if anyone could help me with this problem? It's the last once I need to complete in order to get an A on the assignment. I'm pretty sure it has to do with trigonometric identities, I just don't know how to apply it out properly. I'd really appreciate the help, thank you in advance.
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  2. #2
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    Hello, DesiKid89!

    Suppose $\displaystyle \sin t = \frac{2}{11}$ and $\displaystyle \tan t < 0$

    Find each of the following . . .

    Sine is positive in Quadrants 1 and 2.
    Tangent is negative in Quadrants 2 and 4.
    . . Hence, $\displaystyle t$ is in Quadrant 2.


    $\displaystyle a)\;\;\sin(-t)$
    $\displaystyle \sin(-t) \:=\:-\sin(t) \:=\:\boxed{-\frac{2}{11}}$

    $\displaystyle b)\;\;\cos(6\pi-t)$
    We have the identity: .$\displaystyle \sin^2t + \cos^2t \:=\:1\quad\Rightarrow\quad\cos t \:=\:\pm\sqrt{1-\sin^2t}$

    We are given: .$\displaystyle \sin t \:=\:\frac{2}{11}$

    Hence: .$\displaystyle \cos t \;=\;\pm\sqrt{1 - \left(\frac{2}{11}\right)^2} \;=\;\pm\sqrt{\frac{117}{121}} \;=\;\pm\frac{3\sqrt{13}}{11} $

    Since $\displaystyle t$ is in Quadrant 2: .$\displaystyle \cos t \;=\;-\frac{3\sqrt{13}}{11}$


    Back to the problem:

    $\displaystyle \cos(6\pi-t) \;=\;\cos(-t) \;=\;\cos(t) \;=\;\boxed{-\frac{3\sqrt{13}}{11}}$



    $\displaystyle c)\;\;\tan t$
    $\displaystyle \tan t \;=\;\frac{\sin t}{\cos t} \;=\;\frac{\frac{2}{11}}{-\frac{3\sqrt{13}}{11}} \;=\;-\frac{2}{3\sqrt{13}} \;=\;\boxed{-\frac{2\sqrt{13}}{39}}$


    $\displaystyle d)\;\;\csc(t + 6\pi)$
    $\displaystyle \csc(t + 6\pi) \;=\;\csc t \;=\;\frac{1}{\sin t} \;=\;\frac{1}{\frac{2}{11}} \;=\;\frac{11}{2}$


    Are you getting the idea?
    Can you do the rest?

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