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HW question in Trig relations

  1. Apr 11, 2013 #1
    If cos(alpha)=-sqrt3/5 and alpha is in the third quadrant, find exact values for sin(alpha) and tan(alpha).

    A. Well what I did was use the pythagorean theorem so (-sqrt3)^2+(opposite)^2=5^2. then in the end I got sqrt22 for opposite so then sin(alpha)=sqrt22/5 and tan alpha =sqrt22/-sqrt3,
    but here is the problem then the alpha is outside the third quadrant. then no matter the switchs made to my answs for sin tan +- they angles never fit properly. am I just making a dumb mistake of is the question off?
  2. jcsd
  3. Apr 12, 2013 #2

    Simon Bridge

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    You should sketch the situation.
    The opposite said should be -√22 (spot the minus sign). Pythagoras will only tell you the lengths, not the directions.

    Try it this way:
    Construct the right-angled triangle in the third quadrant. The angle to the hypotenuse, measured anticlockwise from the x axis is ##\alpha## but you want to use trigonometry on interior angles.
    The interior angle is ##\theta = \alpha-180## ##H=5## and ##A=\sqrt{3}##... all positive values.
    So ##\alpha = 180+\arccos(\sqrt{3}/5)## will be in the third quadrant.
  4. Apr 12, 2013 #3
    Ok I see so sin(alpha)=-sqrt22/5 and tan alpha =-sqrt22/-sqrt3,
  5. Apr 12, 2013 #4

    Simon Bridge

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    Care is needed:
    tan(θ)=1 could come from O/A = 1/1 or O/A -1/-1 ... are they the same angle?
    ... a calculator evaluating tan-1(1) will say 45deg for both.
  6. Apr 12, 2013 #5
    Ok I need to come back to this its almost 2 am where I live and I cannot think right now. Thanks A ton for the help, my brain is just not functioning
  7. Apr 12, 2013 #6

    Simon Bridge

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    2am effect - know it well - sweet dreams.
  8. Apr 12, 2013 #7


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    It would help a lot if you used parentheses. Do you mean sin(alpha)= -sqrt(3)/5 or -sqrt(3/5)?
  9. Apr 12, 2013 #8
    Its sqrt(3)/5
    Ok so could I just ust sin(alpha)=sin(pi/2+arccos(sqrt(3)/5)=-sqrt(22)/5
    And tan (alpha)=the answer above ^/cos (pi/2+arccos(sqrt(3)/5)=-sqrt(22)/-sqrt3.
  10. Apr 12, 2013 #9
    Because by doing this I have given an exact angle in the third quadrent correct.
  11. Apr 12, 2013 #10
    Also not pi/2 just pi
  12. Apr 12, 2013 #11


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    (You can use the Edit feature to edit a previous post, rather than tacking on sentence fragments in additional posts.)

    Sure. If [itex]\ \alpha\ [/itex] is in the third quadrant and [itex] \displaystyle \ \cos(\alpha)=-\frac{\sqrt{5}}{3}\,,\ [/itex] then [itex] \displaystyle \ \alpha\ =\pi+\arccos\left(\frac{\sqrt{5}}{3}\right)\ . [/itex]

    Then [itex] \displaystyle \ \sin(\alpha)=\sin\left(\pi+\arccos\left(\frac{\sqrt{5}}{3}\right)\right)[/itex]

    [itex]\displaystyle \quad\quad\ \quad\quad\ \quad =\sin(\pi)\cdot\cos\left(\arccos\left(\frac{\sqrt{5}}{3}\right)\right)
  13. Apr 13, 2013 #12


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    Are you sure you read and wrote the question right?

    cos α = -√3/5 or cos α = -√(3/5) ? :smile:
  14. Apr 13, 2013 #13
    I'm positive. cos(a)=-sqrt(3)/5
  15. Apr 13, 2013 #14


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    You know, there are ways to solve this that are a lot easier. For instance, I found the following trig identities to be very useful:

    sin2(α) + cos2(α) = 1

    tan(α) = sin(α)/cos(α)
  16. Apr 14, 2013 #15
    Ok so now I see the easier solution. sin a = -sqrt(1-(-sqrt(3)/5)) and tan a =-(sin a)/-(cos a)
    Thanks cepheid and everyone else
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