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Math Help - Trig problem help

  1. #1
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    Trig problem help

    I'm having a little trouble with this homework problem.

    If tan(a) = -3/2 and sin(a) > 0, find the following:
    sin(a)
    cos(a)
    sec(a)
    csc(a)
    cot(a)

    I've already gotten that cot(a) = -2/3 but I have no idea how to do any of the others.
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by john11235 View Post
    I'm having a little trouble with this homework problem.

    If tan(a) = -3/2 and sin(a) > 0, find the following:
    sin(a)
    cos(a)
    sec(a)
    csc(a)
    cot(a)

    I've already gotten that cot(a) = -2/3 but I have no idea how to do any of the others.
    First of all, translate the information you're given :

    \tan(a)=-\frac 32=\frac{\sin(a)}{\cos(a)}

    Since -3/2 < 0 and sin(a) > 0, we have that cos(a) < 0.

    Next step, use the most common formula : \sin^2(x)+\cos^2(x)=1

    \implies \cos(a)=-\sqrt{1-\sin^2(a)} (it could have been + sqrt, but since it is negative...)
    So substitute this formula of cos(a) into tan(a) and get sin(a) !

    Then get cos(a) etc..
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    First of all, translate the information you're given :

    \tan(a)=-\frac 32=\frac{\sin(a)}{\cos(a)}

    Since -3/2 < 0 and sin(a) > 0, we have that cos(a) < 0.

    Next step, use the most common formula : \sin^2(x)+\cos^2(x)=1

    \implies \cos(a)=-\sqrt{1-\sin^2(a)} (it could have been + sqrt, but since it is negative...)
    So substitute this formula of cos(a) into tan(a) and get sin(a) !

    Then get cos(a) etc..
    So put the -sqrt(1-((sin^2)(a)) into the sin(a)/cos(a) equation?
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  4. #4
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    Quote Originally Posted by john11235 View Post
    I'm having a little trouble with this homework problem.

    If tan(a) = -3/2 and sin(a) > 0, find the following:
    sin(a)
    cos(a)
    sec(a)
    csc(a)
    cot(a)

    I've already gotten that cot(a) = -2/3 but I have no idea how to do any of the others.
    tan(a) < 0 and sin(a) > 0 ... "a" is in quad II

    sketch a reference triangle in quad II

    opposite side, y = 3
    adjacent side x = -2
    hypotenuse r = \sqrt{13}

    now use your definitions for the six trig ratios.

    sin(a) = y/r
    cos(a) = x/r
    tan(a) = y/x
    cot(a) = x/y
    sec(a) = r/x
    csc(a) = r/y
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  5. #5
    Senior Member pankaj's Avatar
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    \tan A is negative and \sin A is positive,therefore A is in 2nd quadrant
     <br />
\sin A=\frac{\tan A}{\sqrt{1+\tan^2 A}}<br />
;  <br />
\cos A=-\frac{1}{\sqrt{1+\tan^2 A}}<br />

    Consequently,you can find \sec A and \csc A by taking reciprocals
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