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More trig homework

  1. Aug 18, 2005 #1
    given cos theta=4/9, where 3pi/2 is less than/equal theta greater than/equal 2pi.
    Find exact value of sin1/2theta+cos1/2theta

    I have come up with sq rt10 + sq rt26 / 6

    Just don't have a warm fuzzy
     
  2. jcsd
  3. Aug 18, 2005 #2

    VietDao29

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    Homework Helper

    Nope, I don't think it's right.
    You have:
    [tex]\frac{3\pi}{2} \leq \theta \leq 2\pi[/tex]
    [tex]\Leftrightarrow \frac{3\pi}{4} \leq \frac{\theta}{2} \leq \pi[/tex]
    So [tex]\sin \frac{\theta}{2} > 0[/tex], and [tex]\cos \frac{\theta}{2} < 0[/tex]
    You also have:
    [tex]\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta[/tex]
    Therefore:
    [tex]\cos 2 \theta = \cos ^ 2 \theta - \sin ^ 2 \theta = 2\cos ^ 2 \theta - 1[/tex]
    In other word:
    [tex]\cos \theta = \cos ^ 2 \frac{\theta}{2} - \sin ^ 2 \frac{\theta}{2} = 2\cos ^ 2 \frac{\theta}{2} - 1[/tex]
    From the equation, you will work out cos(theta / 2). Remember that cos(theta / 2) < 0.
    You can then use
    [tex]\cos ^ 2 \alpha + \sin ^ 2 \alpha = 1[/tex]
    to find out sin(theta / 2). Remember sin(theta / 2) > 0.
    Viet Dao,
     
  4. Aug 18, 2005 #3
    How would I express that in radical form?
     
  5. Aug 18, 2005 #4

    VietDao29

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    What do you mean?
    Have you covered:
    [tex]\cos 2 \theta = \cos ^ 2 \theta - \sin ^ 2 \theta = 2\cos ^ 2 \theta - 1 = 1 - 2\sin ^ 2 \theta[/tex] yet?
    Viet Dao,
     
  6. Aug 18, 2005 #5
    I came up with:
    sq rt 10 - sq rt 26 all over 6
     
  7. Aug 18, 2005 #6

    VietDao29

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    Homework Helper

    Yup, that's correct.
    Viet Dao,
     
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