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Math Help - Trig Identities

  1. #1
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    Exclamation Trig Identities

    Hey guy, I need help with another problem,

    Suppose that 3\pi /2 < \Theta < 2\pi and  \sec \Theta = 3 Evaluate the following

     a. \sin \Theta
     b. \cos (-\Theta)
     c. \tan (\pi /2 - \Theta)
     d. \cot ( \pi + \Theta )
     e. \csc (2\pi + \Theta )
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by goliath View Post
    Hey guy, I need help with another problem,

    Suppose that 3\pi /2 < \Theta < 2\pi and  \sec \Theta = 3 Evaluate the following

     a. \sin \Theta
     b. \cos (-\Theta)
     c. \tan (\pi /2 - \Theta)
     d. \cot ( \pi + \Theta )
     e. \csc (2\pi + \Theta )
    The angle is in the third quadrant, where the sine if negative and the cosine is positive. The rest of the tri ratios have signs that depend on the sine and cosine.

    Secant is 1/cosine, so cosine(theta) is 1/3.

    Use the Trig Identity \sin^2x + \cos^2 x = 1 and definitions/properties to find the remaining measures.

    Good luck!
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  3. #3
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    Hello, goliath!

    Suppose that \tfrac{3\pi}{2} \,<\, \theta \,<\, 2\pi and  \sec\theta \,=\, 3.
    \sec\theta \:=\:\frac{3}{1} \:=\:\frac{hup}{adj}

    We have: . adj = 1,\;hyp = 3
    And Pythagorus says: . opp = \pm2\sqrt{2}

    Since \theta is in Quadrant 4,  opp = -2\sqrt{2}

    And we have this list: . \begin{Bmatrix}\sin\theta &=&-\dfrac{2\sqrt{2}}{3} \\ \\[-3mm] \cos\theta &=& \dfrac{1}{3} \\ \\[-3mm] \tan\theta &=& -2\sqrt{2} \end{Bmatrix}


    Evaluate the following.

    a.\;\sin\theta
    From our list: . \sin\theta \:=\:-\frac{2\sqrt{2}}{3}


    b.\;\cos (-\theta)
    \cos(-\theta) \;=\;\cos\theta \;=\;\frac{1}{3}


    c.\;\tan \left(\tfrac{\pi}{2} - \theta\right)
    \tan\left(\tfrac{\pi}{2}-\theta\right) \;=\;\cot\theta \;=\;\frac{1}{\tan\theta} \;=\;-\frac{1}{2\sqrt{2}}


    d.\;\cot(\pi + \theta)
    \cot(\pi + \theta) \;=\;\cot\theta \;=\;-\frac{1}{2\sqrt{2}}


    e.\;\csc(2\pi + \theta )
    \csc(2\pi + \theta) \;=\;\csc\theta \;=\;\frac{1}{\sin\theta} \;=\;\frac{1}{\text{-}\frac{2\sqrt{2}}{3}} \;=\;-\frac{3}{2\sqrt{2}}

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  4. #4
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    Thank you!, kudos
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