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Trigonmetry Functions (radian circles)

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data
    If sin (x) = -6/7

    what is cos(x) = ?
    what is tan(x) = ?

    the answers have to be in integer form, as in no decimals.
    the "x" is supposed to be in quadrant 3

    2. The attempt at a solution

    This is in radians

    so i did sin^-1 (-6/7) = -1.02969

    the answer im assuming is (x)

    than cos (x) --> 0.5150 and since its in quadrant 3 this will -.5150
    than tan(x) --> -1.664

    im not sure if i am right soo far, but assuming i am. i tried using pythagoras to figure our what -0.5150 will be in decimal --> -(sqrt(13))/7

    and tan --> and im lost cant figure this out

    any guide or assistance would be great thanks
     
  2. jcsd
  3. Aug 25, 2009 #2

    rock.freak667

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    sin(x)=opp/hyp.

    so if sin(x)=-6/7, what is opp and hyp?

    When you get that, draw the angle x in the 3rd quadrant and construct the right angled triangle.
     
  4. Aug 25, 2009 #3

    Mark44

    Staff: Mentor

    Draw a right triangle inside a circle of radius 7. The sine function is positive in quadrants I and II and is negative in quadrants III and IV. Since you are given that sin(x) = -6/7, which quadrants can the angle x be in? For each of these, how long does the side adjacent have to be?

    Your answers won't be in integers: they will be either rational numbers (the quotient of two integers) or the quotient of a radical and an integer.
     
  5. Aug 25, 2009 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You could also use the fact that [itex]sin^2(x)+ cos^2(x)= 1[/itex] to find cos(x) (remembering that the angle is in the third quadrant so cos(x) is negative.)

    Then use the fact that tan(x)= sin(x)/cos(x).

    In essence, both rock.freak667 and I are telling you, "yes, use the Pythagorean theorem".
     
  6. Aug 25, 2009 #5
    The teacher asked for fractions, so my mistake as putting it as an integer sorry..

    I get how to get what tan(x) is but i dont understand how to find the fraction or what the fraction looks like

    the cos one worked, ( i can check online if i got the right answer and i did) but tan is confusing me,
     
  7. Aug 25, 2009 #6

    rock.freak667

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    tan(x)=sin(x)/cos(x)

    or if you drew the triangle properly, you could find tan(x) by it's definition.
     
  8. Aug 26, 2009 #7
    Use the fact that:

    [tex]cos(x)=\sqrt{1-sin^2x}[/tex]

    If you know the value of sin(x) you can very easy find the value for cos(x) using the formula above ^^.

    Then, when you know the values for sin(x) and cos(x), just substitute here tan(x)=sin(x)/cos(x) and find tan(x).

    Regards.
     
  9. Aug 26, 2009 #8

    HallsofIvy

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    Rock.freak667's point is that, since sine= opp/hyp, you can represent sin(x)= -6/7 as a right triangle with "opp"= -6 (on a graph, down 6 on the y-axis) and hypotenuse 7. You can use the Pythagorean theorem to determine that "adj" (measured on the x-axis) is [itex]-\sqrt{7^2- 6^2}[/itex] (negative square root because you are told that the angle is in the third quadrant and both x and y are negative there). Now that you know the lengths of all three sides of the right triangle, it is easy to find any of the trig functions. In particular, cos(x)= adj/hyp and tan(x)= opp/adj.
     
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