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Homework Help: Trigonometry, Prove the Identity and more

  1. Aug 24, 2007 #1
    1. The problem statement, all variables and given/known data

    Prove the identity
    11. 1 - co5xcos3x - sin5xsin3x = 2sin^2x

    50. ln |secx + tanx| = -ln |secx - tanx|

    52. The following equation occurs in the study of mechanics:
    [tex]\sin\theta = \frac{I_1\cos\phi}{\sqrt{(I_1\cos\phi)^2 + (I_2\sin\phi)^2}}[/tex]. It can happen that [tex]I_1 = I_2[/tex]. Assuming that this happens, simplify the equation.

    2. Relevant equations

    3. The attempt at a solution

    11. Idk what I'm doing wrong

    50. GRRR!!!

    53. Can I make this assumption that [tex]\phi = 45^\circ[/tex] and that [tex]\theta = 45^\circ[/tex]?
    Last edited: Aug 24, 2007
  2. jcsd
  3. Aug 24, 2007 #2
    For 50

    ln(secx + tanx) = -ln(secx - tanx)
    ln(secx + tanx) = ln(1 / (secx - tanx))
    secx + tanx = 1 / (secx - tanx)

    Cross multiply, and you're good.
  4. Aug 24, 2007 #3


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    11) You should know the identity [tex] 1-cos(2 \phi) = 2 sin^2 (\phi) [/tex], that should help.
    Last edited: Aug 24, 2007
  5. Aug 24, 2007 #4
    lol omg ... i see it now, i'm so exhausted! thanks
  6. Aug 24, 2007 #5


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    So you'd know, 11 is almost done, in case you didn't bother to write the whole computation, if you have already made it. Just cut 1 from both sides of the equation, multiply the result by -1 and then use the fact that sin^2 +cos^2 =1.
  7. Aug 24, 2007 #6
    oh wow, very nice. thank you! i didn't even think about taking the e of both sides.

    anyone for 53? can i make that assumption or did i screw up? i did it about 3x.
    Last edited: Aug 24, 2007
  8. Aug 24, 2007 #7


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    You wrote 52 but i think you meant 53. You have

    [tex]I_{1}^{2}\cos^{4}\theta=I_{2}^{2}\sin^{4}\theta [/tex]

    You can't take sqrt like that.
  9. Aug 24, 2007 #8
    how did you get the sin/cos to the 4th power?

    idk if my handwriting is clear, but [tex]\sin\theta = \frac{I_1\cos\phi}{\sqrt{(I_1\cos\phi)^2 + (I_2\sin\phi)^2}}[/tex]
  10. Aug 24, 2007 #9


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    Okay. Raise to the second power and do all multiplications. What do you get ?
  11. Aug 24, 2007 #10
    from where you got these questions ;P?
  12. Aug 24, 2007 #11
    do you like them?

    it's from Algebra and Trigonometry, Beecher (Calc-prep book) ... i love it! i didn't learn Trigonometry properly b/c i took it in a 3-week mini-mester ... i love Math now so i'm studying harddd
  13. Aug 24, 2007 #12
    ok i'm getting the same answer ... i even substituted b/c the damn I's were making me dizzy

    check my work up to this point plz ... i can't think straight anymore

    [tex]\sin\theta = \frac{I_1\cos\phi}{\sqrt{(I_1\cos\phi)^2 + (I_2\sin\phi)^2}}[/tex]

    squaring both sides ...

    [tex]\sin^2\theta = \frac{I_1^2\cos^2\phi}{I_1^2\cos^2\phi + I_2^2\sin^2\phi}[/tex]

    is my denominator right, or did i screw it up?
    Last edited: Aug 24, 2007
  14. Aug 24, 2007 #13


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    It's okay so far. Now cross multiply.
  15. Aug 24, 2007 #14
    Thanks you. :smile:
    Yes I do. Just finished the first one. :biggrin:

    wow, I almost missed these ones!
    I only did examples from 6.3, and nothing exciting was there, so I moved to the next exercise also because I had less time that day, and was tired from the work. I am on the last chapter of that book.

    I am doing this because I thought I need to brush up my basic skills before I start my first year of university(in EE). In calculus, I got to integration by parts, and then got reluctant.
  16. Aug 24, 2007 #15
    for #50:
    can I just take derivatives of both sides, and then prove it?
    that way it takes only two steps to solve all the problem
  17. Aug 24, 2007 #16
    [tex]\sin^2\theta I_1^2\cos^2\phi + \sin^2\theta I_2^2\sin^2\phi = I_1^2\cos^2\phi[/tex]


    [tex]I_1^2\cos^2\phi(\sin^2\theta-1) = -I_2^2 \sin^2\theta\sin^2\phi

    (\sin^2\theta-1) = -\cos^2\theta[/tex]

    [tex]I_1^2\cos^2\phi = \frac{-I_2^2\sin^2\theta\sin^2\phi}{-\cos^2\theta}[/tex]

    right so far?
  18. Aug 24, 2007 #17


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    Unfortunately [tex]f_{1}'(x)=f_{2}'(x) \nRightarrow f_{1}(x)=f_{2}(x) [/tex].

    EDIT: At Rocophysics. Yes. Now do you see the 4-th powers coming up ? You have obtained what i had already written b4.

    EDIT at EDIT: OOOps, i didn't see there were 2 angles, theta and phi. Then disregard my comments about the 4-th power.
    Last edited: Aug 24, 2007
  19. Aug 24, 2007 #18
    WTF!!! two steps? lol ...
  20. Aug 24, 2007 #19
    i love their synthesis problems, they're pretty good
  21. Aug 24, 2007 #20
    lol!!! i told you that on post #8, is all good! this is good latex practice

    EDIT ... oops i didn't say it, i just made it neater to show there were 2 angles, argh! my fault.
  22. Aug 24, 2007 #21
    how about this one:
    [tex]\frac{1}{{sin\left( t\right) }^{2}}=\frac{{I}_{2}^{2}\,{tan\left( p\right) }^{2}}{{I}_{1}^{2}}+1[/tex]
  23. Aug 24, 2007 #22
    take 1 to the left side, and this further simplifies the relationship :D

    so cot(t) = [I2/I1]tan(p)
  24. Aug 24, 2007 #23
    Nope, otherwise you could prove that 1=2.
  25. Aug 24, 2007 #24
    is it okay to take the square root of both sides ...

    [tex]I_1^2\cos^2\phi = \frac{I_2^2\sin^2\phi\sin^2\theta}{\cos^2\theta}[/tex]

    then it would become

    [tex]\frac{I_1\cos\phi}{\sin\phi} = \frac{I_2\sin\theta}{\cos\theta}[/tex]

    i'm looking over yours ...
  26. Aug 24, 2007 #25
    it gives the same answer:
    cot(t) = [I2/I1]tan(p)
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