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Homework Help: Calculate the probability that a head will show

  1. Aug 10, 2008 #1
    A kind of the trigonometry question...

    Let θ be an angle in standard position such that cotθ=-4/3 and sinθ<0.
    Determine the exact value of sec θ




    Determine the number of solutions for (asinx-b)(acosx-a)(bsinx+a)=0
    where 0≦x<2[tex]\pi[/tex], if 0<a<b.




    A fair coin is tossed 60 times. Using the normal approximation to the binomial distribution, calculate the probability that a head will show between 32 and 36 times.




    Please give me as many explanations as possible~ thanks
     
  2. jcsd
  3. Aug 10, 2008 #2
    Re: Trigonometry III

    Well, where are sine and cot negative at the same time? Quadrant IV. So, draw a triangle whose adjacent side is 4, and whose opposite side is (negative) three. By doing this, you are in essence drawing cot(-4/3). Solve for the hypotenuse of this triangle, and find out what secant is.

    For problem two:
    sinx=b/a, cosx=1, sinx=-a/b. Since a<b, sinx=b/a does not exist, because |b/a|>1, which is not in the range of the arcsin function. Cosx=1 at one place (x=0), and sinx=b/a in two places: where sinx is positive. I'm not sure on this, but I think you can say that there are only two solutions, because the coefficient of the x is only 1. So, on the interval [0,2[tex]\Pi[/tex]], there are three solutions.

    For problem three:
    I'm not sure on these...haven't done probability in a very long time. Here's my guess though. Type this on your calculator: binomcdf(60,0.5,36)-binomcdf(60,0.5,32). Another guess would be normalcdf(32,36) (also on your calculator).
     
    Last edited: Aug 10, 2008
  4. Aug 10, 2008 #3

    Defennder

    User Avatar
    Homework Helper

    Re: Trigonometry III

    Interesting username.
     
  5. Aug 10, 2008 #4
    Re: Trigonometry III

    and the thread title :smile:
     
  6. Aug 10, 2008 #5
    Re: Trigonometry III

    Please show your own efforts (in future) before asking questions (see rules).
     
  7. Aug 11, 2008 #6
    Re: Trigonometry III

    It's ok ==~~ not a creative name...
     
  8. Aug 11, 2008 #7
    Re: Trigonometry III

    how's the title interesting @@

    You mean the form?
     
  9. Aug 11, 2008 #8
    Re: Trigonometry III



    I got the second one ~~

    however... I still don't know how to do the first one and the third one
     
  10. Aug 11, 2008 #9
    Re: Trigonometry III


    so many rules =o=
     
  11. Aug 11, 2008 #10
    Re: Trigonometry III

    1) draw x-y axis.

    plot cot (t) = -4/3 (it can be at two positions)
    cot(t) = x/y
    so you know what x and y would give that

    once you have two vectors(or lines) find where the second condition is met,
    you would left with one vector (or line)

    using that line find your sec t


    3. no clue (can't remember how to do those questions)

    remove [solved] if possible from the thread ...

    ------------------------

    @thread title:
    It's font. I know when you use chinese font and write in english it looks really cool to me!
     
  12. Aug 11, 2008 #11
    Re: Trigonometry III


     Oh~~ I did get the first one。。。 

     Thanks for the inspiration

     the third one is 。。。still confused。。。

     haha。。。 I guess it’s cuter。。 so that’s why i use it all the time
     
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