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Pre-Calc Homework Help

  1. Nov 6, 2004 #1
    Simplify the given expression:
    2) (sec^2 x)(csc x)/(csc^2 x)(sec x)
     
  2. jcsd
  3. Nov 6, 2004 #2

    arildno

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    What have you tried?
     
  4. Nov 6, 2004 #3
    well...i've done this:

    (tan^2 x + 1)(csc x)/(cot^2 x + 1)(sec x)

    but...i don't know if i'm going in the right direction, for my teacher is horrible, and i don't know where to go from here if i am going in the right direction

    any help would be greatly appreciated
     
  5. Nov 6, 2004 #4

    arildno

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    OK: Let's make the EASIEST cancellations first:
    If you look at the expression like this:
    [tex]\frac{sec^{2}(x)csc(x)}{sec(x)csc^{2}(x)}[/tex]

    isn't there a couple of cancellations which immediately spring to your mind?
     
  6. Nov 6, 2004 #5
    so is it (sec x)/(csc x) ???
     
  7. Nov 6, 2004 #6

    arildno

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    Precisely!
    Now, knowing the relation between sec and cos and csc and sin, can you simplify even further?
     
  8. Nov 6, 2004 #7
    well, since sin/cos = tan....then would sec/csc = 1/tan ?

    That's all i can think of
     
  9. Nov 6, 2004 #8

    arildno

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    No, you have:
    [tex]\frac{\frac{1}{\cos(x)}}{\frac{1}{\sin(x)}}=\frac{1}{\cos(x)}\frac{1}{\frac{1}{\sin(x)}}=\frac{\sin(x)}{\cos(x)}=tan(x)[/tex]
     
  10. Nov 6, 2004 #9
    alright...thanks, i'm starting to get it a little better....
     
  11. Nov 6, 2004 #10
    Wait...this baffles me...

    26.) Find the exact value of sin 5pi/12

    Is there any way to do this logically w/ a calculator or anything?
     
  12. Nov 6, 2004 #11

    cepheid

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    Of course there is! (hopefully you meant without a calculator) That's why the angle is given to you in radians, as a rational multiple of [itex] \pi [/itex].

    Draw the unit circle: what coordinate points do certain angles represent? [itex] \pi, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2} [/itex] etc.
     
  13. Nov 6, 2004 #12
    but 5pi/12 isn't on my unit circle...the one's you listed are though...i just don't get how exactly you can find 5pi/12 with information of pi/2, etc...
     
  14. Nov 6, 2004 #13
    would 1/4 make sense for the answer since i got the sin of 5pi/6 to equal 1/2?
     
    Last edited: Nov 6, 2004
  15. Nov 6, 2004 #14
    Try to do this using the trig identity
    [tex]\cos{2a}=1-2\sin^{2}{a}[/tex]
     
  16. Nov 6, 2004 #15
    What? So you're saying 1/4 isn't right then?
    Do I even need to use trig identities for this type of question?

    (I'm not arguing with that identity..i'm just very confused :bugeye:)
     
    Last edited: Nov 6, 2004
  17. Nov 6, 2004 #16
    1/4 is incorrect. Sometimes identities are necessary.
     
  18. Nov 6, 2004 #17

    cepheid

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    Yeah, sorry if I gave you the wrong idea. 1/4 is incorrect. You are making the assumption that if I halve the angle, I halve the sine. You can see why that would only be true for a linear relationship right? (which sine is not). I think Sirus has the right technique, since the trig identity involves a term with twice the angle and another with just the angle itself. We know how to work with [itex] \frac{\pi}{6} [/itex], and multiples of it, so find the cosine of the angle [itex] \frac{\5pi}{6} [/itex] and work from there.
     
  19. Nov 7, 2004 #18

    kreil

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    Another way is to convert the radians to degrees and work from there:

    sin5pi/12=sin5(180)/12=sin75=sin(30+45)

    now you can just use the formula for the sum of angles:
    sin (a+b) = cos(b) sin(a) + sin(b) cos(a)
     
  20. Nov 7, 2004 #19
    use the sum formulas of trig:
    sin (a+b)=sin(a) cos(b)+cos(a) sin (b)

    now, sin 5pi/12=sin (2pi/12 + 3pi/12), so the exact value of sin 5pi/12 is?
     
  21. Nov 7, 2004 #20
    alright....root 2/4 + root 6/4


    Thanks for all help
     
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