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Determine domain for a trig ratio and the general solution of angle me

  1. Jan 11, 2014 #1
    Problem statement
    Given sin theta (sqrt 3)/2
    Determine the domain for the following solution
    View attachment 65505


    2) determine each general solution using the angle measure specified.
    ImageUploadedByPhysics Forums1389443143.728586.jpg

    Revelant equations

    Attempt at solution


    1What is the reasoning behind that? For example if you look at b , the answers a re all in the first and second quadrant. Would you just look at the highest and lowest value?
    2) I found the angle measure for each solution. I thought u just find the angle measure and then write the equation for the co terminal angle ( ImageUploadedByPhysics Forums1389443276.807035.jpg )

    That worked for a but in b) one of the answer was 180n.... Where did that come from? I got 180 + 360n but 180n?
    This is the same situation for d , they did not put( 2pi), instead they had the angle then plus or minus pi.
     
  2. jcsd
  3. Jan 11, 2014 #2

    haruspex

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    Not quite sure what Q1 is supposed to be. Is it this: "Given that sin(θ) = (√3)/2 and x < θ < y, find x and y (maximal interval?) such that the solutions are as given in a, b, c etc."? Asking for a domain is a bit strange since it could be [itex]\Re[/itex] minus the unwanted solutions.

    2b is a quadratic, so you would expect two sets of solutions. Please post your working.
     
  4. Jan 11, 2014 #3
    For 1, it just asked for the domain given that sine theta is (sqrt 3)/2 and the following solutions.

    For 2b, I brought over the sin x and got sin x=0 and sin x=1 when I factored it out. That would be equivalent to 90 degrees, 0 and 180 degrees. How do u write the general solution tho?
     
  5. Jan 12, 2014 #4

    haruspex

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    Depends what form you require it in. You know what the repeat interval of the sine function is, so you could just say the solutions are α+2nπ where α is any of ... etc. If you want it as a single function of some integer parameter that's a little more convoluted. You could include terms like (-1)n or even in.
     
  6. Jan 12, 2014 #5

    That is what I did: the 2 pi n . However the back of the book has 90+ (2pi)n and 180 n. where did the 180 n come from?
     
  7. Jan 12, 2014 #6

    haruspex

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    You had 0 + 360n, 90+360n, 180+360n. The first and last can be combined as 0+180n.
     
  8. Jan 13, 2014 #7

    What do you mean by the first and last can be combined ? Do you mean 0+360n and 180+360n? Since 360n is common among both, u just leave it as n?
     
    Last edited by a moderator: Jan 13, 2014
  9. Jan 13, 2014 #8

    haruspex

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    I mean that the union of the sets 0+360n, 180+360n is the set 0+180n.
    The first is ... -720, -360, 0, +360, +720, ..., the second is ... -540, -180, +180, +540 ..., and combining them gives ... -720, -540, -360, -180, 0, +180, +360, ...
     
  10. Jan 13, 2014 #9

    berkeman

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    Coco -- check your PMs. I have fixed up this post of yours...
     
  11. Jan 13, 2014 #10

    Thank you. Really sorry, sometimes when I type really fast, I forget not to use text speak. I will keep it in mind next time.
     
  12. Jan 13, 2014 #11

    Oh. So you figure out both of the plus and minus for each to find the results then combine them to give a simplified equation?
     
  13. Jan 13, 2014 #12

    haruspex

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    Yes, except that it shouldn't involve such a detailed development. I only wrote it out like that to make it absolutely clear. You should spot that 180 is a common factor through 0+360n and 180+360n, so they reduce to 180{0 + 2n, 1+2n}, which means the even and odd multiples of 180, therefore all multiples of 180.
     
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