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Homework Help: Special triangles and Unit Circles HELP

  1. Jan 9, 2005 #1
    If [tex] sin theta = \frac {-4} {7} and \frac {3 pi} {2} < theta < 2 pi [/tex]then determine the exact value of [tex] \frac {1} {cot (theta)} [/tex]

    I dont know where to start I know to set up a circle with a cartesian plane but what am I supposed to do? :uhh:
    Last edited: Jan 9, 2005
  2. jcsd
  3. Jan 9, 2005 #2


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    Okay,lemme get it straight...
    You're given that
    [tex] \sin\theta=-\frac{4}{7} [/tex] (1)
    [tex] \frac{3\pi}{2}<\theta <2\pi [/tex] (2)
    and u're asked
    [tex] \frac{1}{\cot\theta} [/tex] (3)

    Then the angle is in the IV-th quadrant where the 'sine' is negative and the cosine is positive.

    Use the formula
    [tex] \sin^{2}\theta+\cos^{2}\theta =1 [/tex](4)
    and the fact that 'cosine' is positive to find the 'cosine'.
    [tex] \frac{1}{\cot\theta}=\tan\theta=\frac{\sin\theta}{\cos\theta} [/tex](5)


    Last edited: Jan 9, 2005
  4. Jan 9, 2005 #3
    Ok how do u know that the angle is in the 4th quadrant? How do u use that formula?? Im way too confused on this one y does it say <theta , 2 pi?

  5. Jan 9, 2005 #4


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    I guess,though u've been working with the trig.circle,u've never made a sign table for 'sine' and 'cosine'.
    Besides,the 4-th quadrant is defined by
    [tex] \frac{3\pi}{2}<\theta < 2\pi [/tex]

    Which formula to use???

  6. Jan 9, 2005 #5
    ok yes I understand how the angle is in the forth quadrant, how do u use the eqn [tex] \sin^{2}\theta+\cos^{2}\theta =1 [/tex]
    ?? What do I have to do to solve this question are there any diagrams involved?
  7. Jan 9, 2005 #6


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    That is a second order algebraic eq.in 'cosine' of theta,since u know that the sine is -4/7.Solve it,knowing that u must accept the positive value.


    PS.No diagrams,just algebra.
  8. Jan 9, 2005 #7
    when u say solve what do u mean?? Does -4/7 mean that the opposite is -4 and 7 is the hypotenuse?

    Im not sure how to solve for cos?
  9. Jan 9, 2005 #8
    sin(theta) = -4/7, cos(theta) = unknown.

    but you know:

    sin^2(theta) + cos^2(theta) = 1
    (-4/7)^2 + cos^2(theta) = 1

    solve for Cos, and since it's a squareroot you're going to get a positive and negative answer, take the positive answer (since cos's positive in the restrictions given of 3pi/2 < theta < 2pi)
  10. Jan 10, 2005 #9


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    [tex] \sin^{2}\theta=(\sin\theta)^{2} =\frac{(-4)^{2}}{7^{2}}=\frac{16}{49}[/tex]
    [tex] \cos^{2}\theta=1-\sin^{2}\theta=1-\frac{16}{49} =\frac{33}{49} [/tex]
    [tex] \cos\theta=\pm\sqrt{\cos^{2}\theta}=\pm \frac{\sqrt{33}}{7} [/tex]
    U need to chose the plus sign (remember that the angle is in the IV-th quadrant,where the 'cos' is positive)
    [tex] \cos\theta=\frac{\sqrt{33}}{7} [/tex]

    So now find the tangent.

  11. Jan 10, 2005 #10
    OR you could draw a right triangle, make one angle theta, and put in 4 as the side opposite theta, and 7 as the hypotonose (since Sin = opposite/hypotonose = 4/7 [change -4 to 4 since you can't have a negative length of a side])

    Use pythagorem theorm to find that the 3rd side = radical 33, and see that Cos(theta) = adjacent / hypotonose = radical 33 / 7.

    *also keep in mind that cos must be positive since the restrictions given puts it in Q4 where cos is positive. so make cos positive and solve your equation knowing what sin= and what cos=.

    *radical means squareroot.
    Last edited: Jan 10, 2005
  12. Jan 10, 2005 #11
    does [tex] \cos = sqrt (\frac {65} {49} )[/tex]?
  13. Jan 10, 2005 #12
    cos = (sqrt(33)) / 7

    your denominator's right, your numerator's not.
  14. Jan 10, 2005 #13
    I dont know what Im doing wrong? The solution I have is [tex] \frac {-4} {sqrt 33} [/tex]
  15. Jan 10, 2005 #14


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    That's the answer to the problem and it is correct.

  16. Jan 10, 2005 #15
    lol but im still stuck :tongue2:
  17. Jan 10, 2005 #16
    Yeah 1/cot = tan = -4 / sqrt33.

    That's correct.
  18. Jan 10, 2005 #17
    but how do u get that?
  19. Jan 10, 2005 #18
    Tan = Sin / Cos.

    You're given Sin = -4/7

    Find Cos.

    Sin^2 + Cos^2 = 1
    (-4/7)^2 + Cos^2 = 1
    Cos^2 = 1 - (-4/7)^2
    Cos = SQUAREROOT ( 1 - (16/49))
    Cos = Squareroot (33 / 49)
    Cos = Squareroot(33) / 7
    [You know squareroot of 49 doesn't = -7, and it = 7, because it's in Quadrent 4, where Cosine MUST be positive.]

    So now you have: Sin = -4/7, Cos = Squareroot(33) / 7

    1/Cot = Tan = Sin/Cos = (-4/7) / (squareroot(33)/7) = -4/squareroot(33)

    Get it?
  20. Jan 10, 2005 #19
    aisha, it's best to draw a right angle triangle, with the hypotenuse being 7 and one of the sides being 4. Using Pythagoras theorem, the other side will be [tex] /sqrt 33 [/tex]

    You know that cot theta = 1/(tan theta), right?

    So, tan theta = (length_of_opposite) / (length_of_adjacent)

    Hence, cot theta = (length_of_adjacent) / (length_of_opposite)

    So, cot theta will be [tex] - \frac {/sqrt 33}{4}[/tex]
  21. Jan 10, 2005 #20


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    With what? You were just told that the solution you give is correct.

    The way I would approach this problem is:

    First draw a picture: a unit circle on a pair of axes.

    You are told that theta lies between 3pi/2 and 2 pi and you should know (MEMORIZE) that the axes correspond to 0 (at (1,0), pi/2 (at (0,1), pi (at (-1,0), 3pi/2 (at (0,-1) (and 2pi again at (1,0)). This is a circle with radius 1, diameter 2, and so circumference 2pi- those numbers just measure the distance around the circle.

    Theta is between 3pi/2 and 2pi so you should mark a point on the circle between (0,-1) and (1,0) (i.e. in the fourth quadrant as dextercioby originally said). Since points on the unit circle have coordinates (cos(theta), sin(theta)), knowing that sin(theta)= -4/7 tells you that y= -4/7. The equation of the circle is x2+ y2= 1 so you must have x2+ 16/49= 1= 49/49 . That is:
    x2= 33/49 and so x= cos(theta)= +&radic;(33)/7 (positive root because x is positive in the fourth quadrant).

    Now that you know both sin(theta) and cos(theta) you know that 1/cos(theta)= tan(theta)= sin(theta)/cos(theta)= (-4/7)(7/sqrt(33))= -4/sqrt(33).

    As for "where do I use [itex]sin^2(x)+ cos^2(x)= 1[/itex]?", you don't HAVE to use it directly.. I basically used it when I wrote the equation of the circle, x2+ y2= 1, which, since x= cos(theta) and y= sin(theta), is the same thing.
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