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Trig Identies

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Use properties of the trigonometric functions to find the exact value of each expression. Do not use a calculator

    sin (-pi/12) csc (25 pi)/12


    2. Relevant equations
    sin (negative angle) = - sin (angle)
    csc theta = (sin theta)^-1 = 1/(sin theta)


    3. The attempt at a solution
    Ok there is obviously some sort of property of trig identies I do not know so I'm struggling here...

    sin(-pi/12) csc( (25 pi)/12 ) = (- sin(pi/12) )/(sin( (25 pi)/12))
    know if I'm correct I just treat them as exponents correct? so I subtract them

    - sin( (pi/12) - ( (25 pi)/12 )
    - sin( (-24 pi)/12 )
    + sin( (24 pi)/12
    sin 2 pi = 0

    OK THIS IS WRONG I put this into my calculator and i get negative one. I postulated for like two hours on how to do this. I think I figuered it out but am not sure why it works can someone please explain it to me...

    ok I start out here
    sin(-pi/12) csc( (25 pi)/12 ) = (- sin(pi/12) )/(sin( (25 pi)/12))


    then
    - sin( (pi/12) - ( (25 pi)/12 )

    - sin( (24 pi)/12 )

    if i just ignore the negative sign next to the 24... something here is wrong with what I'm doing HELP

    - sin( (24 pi)/12 )

    now I take the recipical but leave the pi on top of the angle not sure why or why i leave pi on top

    - sin( (12 pi)/24

    simplify

    - sin( pi/2

    pi/2 is ninety degrees which has the coordinates (0,1) and sense sine is equal to the y cordinate I get one but sense it's the opposite i get negative 1 as my answer which is correct

    So I obviously have no idea how to really do this problem if someone could tell me how to do this that would be great. There isn't really much I can do because I'm obviously don't know some property or something here.

    Thanks
     
  2. jcsd
  3. Nov 13, 2009 #2

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    None of your algebraic manipulations make any sense. You are just making up operations that don't exist.

    Notice that Pi/12 is half of pi/6 and you know the functions for multiples of pi/6. Look at the half angle formulas instead of trying to make up your own.
     
  4. Nov 13, 2009 #3
    I have yet to have been taught these formulas yet so there must be some other way to do it
     
  5. Nov 14, 2009 #4
    Here are the two special triangles I've used in the past: http://fouss.pbworks.com/f/special triangle 3.JPG and http://fouss.pbworks.com/f/special triangle 2.JPG and recall that sin (a-b)=(sin a)(cos b)-(sin b)(cos a)

    Now sin [tex]\frac{-pi}{12}[/tex] = sin ([tex]\frac{pi}{6}[/tex] - [tex]\frac{pi}{4}[/tex]) = sin [tex]\frac{pi}{6}[/tex] * cos [tex]\frac{pi}{4}[/tex] - sin [tex]\frac{pi}{4}[/tex] * cos [tex]\frac{pi}{6}[/tex]

    Use the special triangles (unless you have them memorized, which you should have) and solve.

    Edit: Sorry I thought you were doing two different equations. I now see that sin (-pi/12) * csc (25 pi)/12 is what you want. I guess you can do the second one and after solving it, multiple both answers.
     
    Last edited by a moderator: Apr 24, 2017
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