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Thread: Trig Ratios

  1. #1
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    Trig Ratios

    Find the remaining trigonometric ratios.

    tan (a) = 4, 0<a<pi/2


    sin(α)
    =
    cos(α)
    =
    cot(α)
    =
    sec(α)
    =
    csc(α)
    =
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  2. #2
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    Re: Trig Ratios

    I would use the Pythagorean identities, keeping in mind all roots will be positive since the angle given is in the first quadrant. For example:

    $\displaystyle \tan(a)=4$

    $\displaystyle \frac{\sin(a)}{\cos(a)}=4$

    $\displaystyle \sin(a)=4\cos(a)$

    $\displaystyle \sin^2(a)=16\cos^2(a)$

    $\displaystyle \sin^2(a)=16(1-\sin^2(a))$

    $\displaystyle 17\sin^2(a)=16$

    $\displaystyle \sin^2(a)=\frac{16}{17}$

    $\displaystyle \sin(a)=\frac{4}{\sqrt{17}}$

    You can now easily find $\displaystyle \cos(a)$ from $\displaystyle \sin(a)=4\cos(a)$. Now that you have the three primary functions, finding the rest is straightforward...
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    Re: Trig Ratios

    So would cos(a) be the same as sin (a), and the rest would be just reciprocals? Sorry I haven't done this in a long time :S
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  4. #4
    MHF Contributor MarkFL's Avatar
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    Re: Trig Ratios

    No, from the equation relating sin(a) and cos(a), we see that cos(a) is 1/4 the value of sin(a). Then yes, the secondary functions would be the reciprocals of the corresponding primary functions.
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  5. #5
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    Re: Trig Ratios

    Hello, Oldspice1212!

    Given: .$\displaystyle \tan\theta \,=\,4,\;0 < \theta < \tfrac{\pi}{2}$
    Find the remaining trigonometric ratios.

    We are given: .$\displaystyle \tan\theta\:=\:\frac{4}{1}\:=\:\frac{opp}{adj}$

    $\displaystyle \theta$ is in a right triangle with: $\displaystyle opp = 4,\:adj = 1$
    Pythagorus tells us that: $\displaystyle hyp \:=\:\sqrt{4^2+1^2} \:=\:\sqrt{17}$

    So we have: .$\displaystyle \begin{Bmatrix}opp &=& 4 \\ adj &=& 1 \\ hyp &=& \sqrt{17} \end{Bmatrix}$


    And we can write all six trig ratios:

    . . $\displaystyle \begin{array}{cccccccccc}\sin\theta &=& \frac{4}{\sqrt{17}} &\;\;& \csc\theta &=& \frac{\sqrt{17}}{4} \\ \\[-3mm] \cos\theta &=& \frac{1}{\sqrt{17}} && \sec\theta &=& \sqrt{17} \\ \\[-3mm] \tan\theta &=& 4 && \cot\theta &=& \frac{1}{4} \end{array}$
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