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Not sure how to approach problem

  1. Jan 3, 2005 #1
    Hi,

    During a review this question has popped up (not sure what unit)

    "Determine the exact value*

    ````5pi``````11pi
    sin ___ - cos ___
    ````4`````` 6
    I am not sure what to do with the sin and cos?

    Thanks
     
  2. jcsd
  3. Jan 3, 2005 #2

    dextercioby

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    So you have to find this value...
    [tex] \sin\frac{5\pi}{4}-\cos\frac{11\pi}{6}[/tex]...??

    Okay.
    [tex] \sin\frac{5\pi}{4}=\sin(\pi+\frac{\pi}{4}})=-\sin\frac{\pi}{4} [/tex]
    [tex] \cos\frac{11\pi}{6}=\cos(2\pi-\frac{\pi}{6})=\cos\frac{\pi}{6} [/tex]

    Can u take from here?Can u see the properties of sine and cosine i used to get to my results??

    Daniel.
     
  4. Jan 3, 2005 #3
    Can you not factorise the [tex]\pi[/tex] out for a start and then see if there is a relationship between sin and cos. I know that one is a translation of the other and that this can be shown by [tex]sin(x+90)[/tex] = [tex]cos(x)[/tex].

    This might help you but I think some maths brainbox will come along and say it is wrong. :tongue2:

    The Bob (2004 ©)
     
  5. Jan 3, 2005 #4
    How did you get to these conclusions??? (I know they must be right but I cannot see how).

    The Bob (2004 )
     
  6. Jan 3, 2005 #5

    quasar987

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    Simpler (or at least more fundamental) is to just look at the trigonometric circle (this one's not very good cuz it doesn't show the angles) and notice that for every integer multiple of pi, the sine of that value is 0, and for every odd integer multiple of pi, the cosine of that value is -1. In mathematical language,

    [tex]sin(n\pi) = 0 \ \forall n \in \mathbb{Z}[/tex]

    and

    [tex]cos((2n+1)\pi)=-1 \ \forall n \in \mathbb{Z}[/tex]

    Now, 5 is an integer ==> sin(5pi) = 0. 11 is an odd integer ==> cos (11pi) = -1.


    (TheBob: look at the trigonometric circle, you will see that they are true.)
     
  7. Jan 3, 2005 #6
    Unfortunately time is not on my side for tonight but I will look properly tomorrow. However on my scanning of the web page I saw these functions:

    [tex]sec[/tex] and [tex]csc[/tex]. Could you explain what they are please?

    The Bob (2004 ©)
     
  8. Jan 3, 2005 #7

    dextercioby

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    They are the 2 less used circular trigonometric functions:

    [tex] \sec x=:\frac{1}{\cos x} [/tex]
    [tex] \csc x=:\frac{1}{\sin x} [/tex]

    I would rather not those two functions.Sine and cosine are much easier to work with.

    Daniel.

    PS.If u can't manage with the trigonometric circle (which would be really bad),u can prove my formulas using the addition formulas for sine and cosine.
     
  9. Jan 4, 2005 #8
    Cheers for this.

    Going for the vote of confidence I see. Thanks for this as well.

    My problem is that I need some numbers in place of the [tex]x[/tex]'s and the [tex]i[/tex]'s.

    If anyone could do that so I can see how you all got from:
    [tex] \sin\frac{5\pi}{4}=\sin(\pi+\frac{\pi}{4}})=-\sin\frac{\pi}{4} [/tex]

    And:[tex] \cos\frac{11\pi}{6}=\cos(2\pi-\frac{\pi}{6})=\cos\frac{\pi}{6} [/tex]

    then that woul be a big help. I have tried to see for myself but nothing I do makes any sense.

    Cheers.

    The Bob (2004 ©)
     
  10. Jan 4, 2005 #9

    dextercioby

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    [tex] \sin\frac{5\pi}{4}=\sin(\frac{4\pi+\pi}{4})=\sin(\pi+\frac{\pi}{4})=\sin\pi\cos\frac{\pi}{4}+\sin\frac{\pi}{4}\cos{\pi} [/tex]

    Now you know that [tex] \sin\pi=0;\cos\pi=-1[/tex]
    Then
    [tex] \sin\frac{5\pi}{4}=-\sin\frac{\pi}{4} [/tex]


    And following the same pattern,u prove the other formula as well.

    Daniel.
     
  11. Jan 4, 2005 #10
    I wish either I had your understanding or this countries educations system was better.

    This now makes sense. but the next stage I do not understand. I cannot see how you get from [tex]\sin(\pi+\frac{\pi}{4})[/tex] to [tex]\sin\pi\cos\frac{\pi}{4}+\sin\frac{\pi}{4}\cos{\pi} [/tex].

    [tex]\cos\frac{11\pi}{6}=\cos(\frac{12\pi-\pi}{6})=\cos(2\pi-\frac{\pi}{6})[/tex] but the next bit [tex]\cos\frac{\pi}{6}[/tex] does not follow like the [tex]\sin[/tex] bit does. So cosine is different to sine in some respect and so I cannot even try that final stage like the sine section.

    As you can see, without this understanding here it is impossible for me to understand how to answer the question. I will be honest and say that none of this has been covered yet at my college but it interests me and so if you have the time and the patience I would like to learn this please.

    Cheers.

    The Bob (2004 ©)
     
  12. Jan 4, 2005 #11

    dextercioby

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    The fundamental formula is
    [tex] \sin(x+y)=\sin x\cos y+\sin y\cos x [/tex](1)
    I know a beautiful geometrical proof to it,but unfortunately i can't share it with you.
    It's the formula i applied for those 2 angles ([itex] \pi \text{and} \frac{\pi}{4} [/itex]).

    Use formula (1) to prove that:
    [tex] \sin(x+\frac{\pi}{2})=\cos x [/tex] (2)
    Use formula (1) to prove that:
    [tex] \sin(x+\pi)=-\sin x [/tex] (3)

    Use formulas (1),(2) and (3)to prove that:
    [tex] \cos(x+y)=\cos x\cos y-\sin x\sin y [/tex](4)

    'Sine'is an odd function:
    [tex] \sin(-x)=-\sin x[/tex](5)
    'Cosine'is an even function:
    [tex]\cos(-x)=\cos(x) [/tex] (6)

    Use formulas (1),(4),(5) and (6) to prove:
    [tex] \sin(x-y)=\sin x\cos y-\sin y\cos x [/tex] (7)
    [tex] \cos(x-y)=\cos x\cos y+\sin x\cos y [/tex] (8)

    Daniel.
     
  13. Jan 4, 2005 #12
    Right so:
    [tex]\sin(x+\frac{\pi}{2}) = \sin x \cos\frac{\pi}{2}+\sin\frac{\pi}{2} \cos x[/tex] but the stage that allows you to get to [tex]\cos x[/tex] is missing in my mind. Is it:

    [tex]\cos \pi = -1 = \cos 180[/tex] therefore [tex]\cos \frac{\pi}{2} = \cos 90 = 0[/tex]

    [tex]\sin \pi = 0 = \sin 180[/tex] therefore [tex]\sin \frac{\pi}{2} = \sin 90 = 1[/tex]

    Therefore [tex]\sin x \cos\frac{\pi}{2}+\sin\frac{\pi}{2} \cos x = (\sin x \times 0)+(1 \times \cos x) = 0+\cos x = \cos x[/tex]

    And: [tex]\sin(x+\pi) = \sin x \cos\pi+\sin\pi \cos x = (\sin x \times -1)+(0 \times \cos x) = -\sin x + 0 = -\sin x[/tex]

    The next bit is going to take a little more thinking.

    The Bob (2004 ©)
     
  14. Jan 4, 2005 #13
    [tex] \cos(x+y)=\cos x\cos y-\sin x\sin y [/tex]

    Therefore: [tex] \cos(x+y)=\sin (x + \frac{\pi}{2})\cos y+\sin (x+\pi) \sin y [/tex] but then what???

    The Bob (2004 ©)

    EDIT: This is the wrong way round.
     
    Last edited: Jan 4, 2005
  15. Jan 4, 2005 #14

    dextercioby

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    Well,did u prove the 'cos' addition formula??
    HINT:
    [tex] \cos(x+y)=\sin(x+y+\frac{\pi}{2})=\sin[x+(y+\frac{\pi}{2})]=... [/tex]
    ,and make use of the 'sin' addition formula for
    [tex] x=x;y=y+\frac{\pi}{2} [/tex]

    Daniel.
     
  16. Jan 4, 2005 #15
    But if [tex]\cos x=\sin(x+\frac{\pi}{2})[/tex] then [tex]\cos y=\sin(y+\frac{\pi}{2})[/tex]

    which means that [tex]\cos x \cos y = \sin(x+\frac{\pi}{2}) \sin(y+\frac{\pi}{2})[/tex]


    [tex]\cos (x+y) = \cos x \cos y - \sin x \sin y[/tex] therefore:

    [tex]\cos (x+y) = \sin(x+\frac{\pi}{2}) \sin(y+\frac{\pi}{2}) - \sin x \sin y[/tex]

    But this must be wrong because it does not leave what I need.

    It isn't making much sense to me.

    The Bob (2004 ©)
     
  17. Jan 4, 2005 #16
    I'm tired and need sleep so I am off. I will think about it but I am really sorry: I do not get it. I can get parts but not the rest.

    The Bob (2004 ©)
     
  18. Jan 4, 2005 #17

    BobG

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    The unit circle (or trigonometric circle) is the easiest method. And you really have very few numbers to memorize.

    [tex]sin 0 = \frac {\sqrt{0}}{2}[/tex]

    [tex]sin \frac{\pi}{6} = \frac {\sqrt{1}}{2}[/tex]

    [tex]sin \frac{\pi}{4} = \frac {\sqrt{2}}{2}[/tex]

    [tex]sin \frac{\pi}{3} = \frac {\sqrt{3}}{2}[/tex]

    [tex]sin \frac{\pi}{2} = \frac {\sqrt{4}}{2}[/tex]

    Obviously, it's a little over the top to use the square root of zero or the square root of one, but it shows the progression very clearly. In other words, in practice:

    [tex]sin 0 = 0[/tex]

    [tex]sin \frac{\pi}{6} = \frac {1}{2}[/tex]

    [tex]sin \frac{\pi}{4} = \frac {\sqrt{2}}{2}[/tex]

    [tex]sin \frac{\pi}{3} = \frac {\sqrt{3}}{2}[/tex]

    [tex]sin \frac{\pi}{2} = 1[/tex]

    Cosine regresses in the same way, starting from 1 for cosine 0 going to 0 for cosine pi/2.

    If you've got one quarter of the circle down, the only thing to visualize is the sign of each as you move to different quadrants of the trig circle and how far the radius is from the x-axis for the sine or from the y-axis for the cosine.

    If you have a good grasp of the basics, you don't actually have to get into the sum/difference laws until you start getting the 'tougher' angles, like [tex]\frac{\pi}{12}[/tex] and so on.

    A good look at the unit circle and you'll see that the end point of [tex]\frac{\pi}{6}[/tex] is just as far away from the y-axis as the end point of [tex]-\frac{\pi}{6}[/tex].

    The end point of [tex]\frac{\pi}{6}[/tex] is just as far away from the x-axis as [tex]\frac{5 \pi}{6}[/tex].

    Being able to visualize that makes the odd/even identities for sine and cosine intuitively obvious even before you prove them (and, yes, in trig class, you get to step along proving one identity after another, but it's a lot easier if you can already visualize these things in your head before you start).

    Edit: And in light of dextercioby's brilliant observation about the difference in time zones, I've deleted that insult about being such a slacker you have to take a nap in the middle of the afternoon. :wink:
     
    Last edited: Jan 4, 2005
  19. Jan 4, 2005 #18

    dextercioby

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    Trigonometry,whether circular,elliptic or hyperbolic,is wonderful and not really difficult.Since it's almost midnight over the Channel,i'll let u sleep on them... :tongue2:

    Daniel.
     
    Last edited: Jan 4, 2005
  20. Jan 4, 2005 #19
    OK I have received the answer but I am still dont understand the problem
    ``````````__ ___
    Answer: -(\/2 + \/3 )
    ````````` __________
    `````````````2

    Please Help!

    by the way the teacher said you use the trignometric circle
     
  21. Jan 5, 2005 #20

    BobG

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    [tex]\pi[/tex] radians is halfway around the circle, or along the negative x-axis.

    [tex]\frac {5 \pi}{4}[/tex] radians is [tex]\frac {\pi}{4}[/tex] radians past halfway. If you know your sines for one quarter of the circle, and can visualize where [tex]\frac {5 \pi}{4}[/tex] radians is on the unit circle, you know that the sine of [tex]\frac {5 \pi}{4}[/tex] has to be [tex]- \frac{\sqrt {2}}{2}[/tex] .

    You do the same for the cosine of the second angle - you learn the unit circle well enough that you know what the cosine of [tex]\frac{11 \pi}{6}[/tex] has to be.

    Substitute the values for the sine and the cosine into the original problem. Since they both have a denominator of 2, you can combine them into the answer you received:

    [tex]\frac{-\sqrt{2} - \sqrt{3}}{2} = \frac{-(\sqrt{2} + \sqrt{3})}{2}[/tex]
     
    Last edited: Jan 5, 2005
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