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Math Help - 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 \sin t = \frac{2}{11} and \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, t is in Quadrant 2.


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

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

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

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

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


    Back to the problem:

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



    c)\;\;\tan t
    \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}}


    d)\;\;\csc(t + 6\pi)
    \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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