Trigonometry, Prove the Identity and more

  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. 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. 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. lol omg ... i see it now, i'm so exhausted! thanks
  6. dextercioby

    dextercioby 12,328
    Science Advisor
    Homework Helper

    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. 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. dextercioby

    dextercioby 12,328
    Science Advisor
    Homework Helper

    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. 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. dextercioby

    dextercioby 12,328
    Science Advisor
    Homework Helper

    Okay. Raise to the second power and do all multiplications. What do you get ?
  11. from where you got these questions ;P?
  12. 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. 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. dextercioby

    dextercioby 12,328
    Science Advisor
    Homework Helper

    It's okay so far. Now cross multiply.
  15. 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. 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. [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. dextercioby

    dextercioby 12,328
    Science Advisor
    Homework Helper

    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. WTF!!! two steps? lol ...
  20. i love their synthesis problems, they're pretty good
  21. 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.
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