1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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))

    - 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


    - 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.

  2. jcsd
  3. Nov 13, 2009 #2


    User Avatar
    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook