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Homework Help: Chapter Summary: Trigonometric Functions

  1. Apr 7, 2005 #1
    The problem reads: Find [itex]\sin\theta[/itex] and [itex]\cos\theta[/itex]

    Part a gives me the coordinates [itex]\left(-1,1\right)[/itex]

    The triangle I got had the [itex]x-length[/itex] as [itex]-1[/itex], while the [itex]y-length[/itex] was [itex]1[/itex]. The hypotenuse I got was [itex]\sqrt{2}[/itex]

    Since [itex]\sin[/itex] is [tex]\frac{opposite}{hypotenuse}[/tex] I got [tex]\sin\theta=\frac{1}{\sqrt{2}}[/tex]

    The book says it is [tex]\sin\theta=\frac{\sqrt{2}}{2}[/tex]

    What did I mess up on? Data I'm waiting for you :biggrin:
     
  2. jcsd
  3. Apr 7, 2005 #2

    dextercioby

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    Heh,he's not here.:tongue2:

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

    Do u see why?

    Daniel.
     
  4. Apr 7, 2005 #3
    :rolleyes: Wow, I was freaking stupid. Thanks for pointing that out lol.

    I think I will go hide in shame now.
     
  5. Apr 7, 2005 #4
    Okay, Houston we have a real problem now.

    It reads: Give the exact value of each expression in simplest radical form.

    a. [tex]\sin\frac{5\pi}{4}[/tex] b. [itex]\cos90[/itex] c. [itex]\sin150[/itex] d. [tex]\cos\frac{11\pi}{6}[/tex]

    The only one I could figure out was b.

    How exactly do I go about finding the radical form of these? (especially the ones in increments of [itex]\pi[/itex])
     
  6. Apr 7, 2005 #5

    dextercioby

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    HINT:Use addition & subtraction formulas for sine & cosine


    Daniel.
     
  7. Apr 7, 2005 #6
    Hi!

    Let's look at a. We want

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

    so the angle we're looking for is [itex]\pi / 4[/itex] past the negative x-axis (ie. it's in quadrant 3 [using the same terminology as last time], and [itex]45^\circ[/itex] from each axis). Does that help? :smile:

    If you're able to answer that question now, try to do similar manipulations on the others to figure out where the angles are.
     
  8. Apr 7, 2005 #7
    And dextercioby's suggestion will work just as well, if you feel like taking a more algebraic approach. Some identities that might help (and you should try to figure out for yourself why they work) follow:

    [tex] \sin (-\theta) = -\sin \theta[/tex]
    [tex] \cos (-\theta) = \cos \theta[/tex]
    [tex] \tan (-\theta) = -\tan \theta[/tex]
    [tex] \sin (\pi - \theta) = \sin \theta[/tex]
    [tex] \cos (\pi - \theta) = -\cos \theta[/tex]
    [tex] \tan (\pi + \theta) = \tan \theta[/tex]
    [tex] \sin (\pi/2 - \theta) = \cos \theta[/tex]
    [tex] \cos (\pi / 2 - \theta) = \sin \theta[/tex]
     
    Last edited: Apr 7, 2005
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