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Homework Help: Csc theta=-17/15, cot beta=-3/4 find exact value HELP HELP HELP

  1. Jan 10, 2005 #1
    Given [tex]\csc\theta = \frac {-17} {15} where \frac {3\pi} {2} < \theta <= 2\pi and \cot \beta = \frac {-3} {4} where \frac {\pi} {2} <= \beta <= \pi [/tex] Find the exact value of [tex] \cos (\theta-\beta)[/tex]

    I think the theta angle is in quadrant 4, and the beta angle is in quadrant 2. I drew a circle with a cartesian plane and made the two right angle triangles beta and theta. The beta angle's opposite=4, adjacent=-3 and hypotenuse=5 (using pythagorean theorem) The theta angle's opposite= 15, adjacent = -17 and hypotenuse = square root of 514.

    I'm not sure If I have done this correctly so far but this is all that I did and now im stuck ..... Please help! :uhh:
  2. jcsd
  3. Jan 10, 2005 #2
    cos(x-y)=cosx cosy + sinx siny
    you made a mistake on theta..
    find cosx and sinx for each angle first...
    check the trig you did .....
  4. Jan 10, 2005 #3
    I know the formula for sum and difference but I dont know if im on the right track and what to do next??
  5. Jan 10, 2005 #4
    find sin(theta), cos(theta), sin(beta) and cos(beta) first.
    If you have adj, hyp and opp, this should be easy....
    just wanna remind you that you make a mistake finding hyp and adj of theta
  6. Jan 10, 2005 #5


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    She needn't know which is exactly 'adjacent' and which is "oppsite".For the first angle,that is.She gotta know where those angles are (in what quadrant) and the fact that,by definition
    [tex] \csc\theta=:\frac{1}{\sin\theta} [/tex](1)
    Therefore,she has the sine for the first angle & she must find the cosine,knowing the angle is in the 4-th quadrant.
    As for the second angle,she's already got both the sine and the cosine.

  7. Jan 10, 2005 #6
    how is the hyp and adj of theta wrong? I just used the fraction given in the question just like I did with beta.

    isnt csc = 1/sin?
  8. Jan 10, 2005 #7


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    For the first angle,it doesn't matter "which is which",u can solve the problem algebraically.
    U know that
    [tex] \sin^{2}\theta+\cos^{2}\theta =1[/tex]
    and u know the 'sine'.Find the cosine

  9. Jan 10, 2005 #8
    ok so [tex] \cos= \frac {8} {17}[/tex] Then what?
  10. Jan 10, 2005 #9
    dex, you are so great...you are my idol....
    back to the question, don't you think drawing the picture is more or less helping him/her "see" what actually sine,cos,sec,csc..etc is? expecially help him getting the sign right... ie. cos(theta) is positive when 2pi/3<theta<2pi...etc. if he does that algebraically, he must memorize the sign of sine and cos in each quadrant....
    ain't we here to him them understand physic/maths?
  11. Jan 10, 2005 #10
    after you get all sine and cos for each angle, apply
    cos(x-y)=cosx cosy + sinx siny

    pretty easy, right
  12. Jan 10, 2005 #11


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    You sound like memorizing "Ramayana"&"Mahabharata"... :tongue2: It's just a table of signs,for Christ's sake.
    You got the sines and the cosines,now compute what the problem has asked u.

  13. Jan 10, 2005 #12
    ok the formula is [tex] \cos (\theta-\beta)=\cos\theta \cos\beta + \sin\theta \sin \beta [/tex]

    I know that [tex] \sin=\frac {15} {-17} and \cos= \frac {8} {17} [/tex]

    but what do these two fractions mean? Where do i get beta and theta to input in the formula?

    The teacher used more diagrams of circles and cartesian planes that's why im a little lost bear with me please. :redface:
    Last edited: Jan 10, 2005
  14. Jan 11, 2005 #13


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    Your problem: Find the cosine of the difference between theta and beta. In other words, if you start at the x-axis, draw a vector at an angle beta, then draw a vector at an angle of theta. What is the angle between the two vectors? (actually, in this case, you only need to find the cosine of that angle)

    Note: I only mention the vectors because drawing them in sometimes helps to visualize just what it is that you're doing. (Also, depending how much more math you take, that cosine difference law is eventually going to turn into a dot product, a very important tool for vectors).

    You use the cosine difference identity:

    [tex]cos (\theta - \beta) = cos \theta cos \beta + sin \theta sin \beta [/tex]

    To solve it, you need the cosine and sine of theta, and the cosine and sine of beta. One step at a time:


    You know the csc of theta. You also know theta is in the fourth quadrant where sine is negative and cosine is positive.
    It's [tex]csc \theta = \frac{-17} {15}[/tex].

    The cosecant is just the reciprical of the sine. So the sine of theta is:

    [tex]sin \theta = - \frac{15}{17} [/tex]

    You then used the pythagorean theorem:

    [tex]sin^2 \theta + cos^2 \theta = 1[/tex] and found that the cosine of theta was 8/17. That gives you two of the values you need to plug into the cosine difference equation.

    You were given the cotangent of beta and the fact that it was in the second quadrant. From that you found the adjacent side (-3), the opposite side (4), and the hypotenuse (5).

    The cosine is the adjacent over hypotenuse; the sine is the opposite over hypotenuse.
    [tex]cos \beta = - \frac{3}{5}[/tex]
    [tex]sin \beta = \frac{4}{5}[/tex]

    That gives you the other two values you need to plug into your equation.

    The cosine of the difference
    You have a fraction for the cosine of theta and a fraction for the cosine of beta. Multiply the two fractions.

    You also have a fraction for the sine of theta and a fraction for the sine of beta. Multiply those two fractions.

    Add the two products together. Both products are guaranteed to have the same denominator, so the addition part is pretty easy. Both numbers you're adding are negatives, so your final answer is negative, as well.
    Last edited: Jan 11, 2005
  15. Jan 11, 2005 #14


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    So you get sin(beta)=4/5. cos(beta)=-3/5 here.

    This part isn't right. sintheta=-15/17. Opposite is -15. Hypotenuse=17. Note that sin of an angle is opposite over hypotenuse. then, using pythagorean theorem, adjacent=8.

    So you have sin(theta)=-15/17 and you get cos(theta)=8/17.

    Plug them all into:
    cos(theta-beta) = cos(theta)cos(beta) + sin(theta)sin(beta).
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