1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    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 Proof

  1. Oct 11, 2007 #1

    danago

    User Avatar
    Gold Member

    If [tex]z=cis\theta[/tex], verify that [tex]
    \tan \theta = \frac{{z - z^{ - 1} }}{{i(z + z^{ - 1} )}}
    [/tex]. Use this result to prove that [tex]
    \cos (2\theta ) = \frac{{1 - \tan ^2 \theta }}{{1 + \tan ^2 \theta }}
    [/tex]


    Ok, ive managed to verify the first equation given, but im not really sure how to use it to prove the second identity. Im really not sure where to start. If somebody could give me a hint about where to start id be very appreciative.

    Thanks,
    Dan.
     
  2. jcsd
  3. Oct 11, 2007 #2

    mjsd

    User Avatar
    Homework Helper

    Does your [tex]z^{-1}[/tex] actually mean conjugate of z?
     
  4. Oct 11, 2007 #3

    danago

    User Avatar
    Gold Member

    Nah its the reciprocal of z.
     
  5. Oct 11, 2007 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Haven't seen "cis" in a long time! It's engineering shorthand for [itex]cos(\theta)+ i sin(\theta)= e^{i\theta}[/itex]. From the last form, or comparing [itex]\theta[/itex] for z and z-1, it should be clear that if [itex]z= cis(\theta)[/itex] then [itex]z^{-1}= cis(-\theta)= cos(\theta)- i sin(\theta)[/itex]. Putting those in for z and z-1 in
    [tex]\frac{z- z^{-1}}{i(z+ z^{-1}}[/itex]

    By the way, when |z|= 1, as is the case here, z-1 is the complex conjugate:
    [tex]\frac{1}{x+ iy}= \frac{1}{x+iy}\frac{x-iy}{x-iy}= \frac{x-iy}{x^2+ y^2}= x- iy[/tex]
     
  6. Oct 11, 2007 #5

    mjsd

    User Avatar
    Homework Helper

    basically, you have [tex]\tan \theta[/tex] equal something, so all you need to do to prove your 2nd identity is just sub in this where [tex]\tan \theta[/tex]'s appear and simplify then you shall see that it actually equals to [tex]\cos (2\theta)[/tex]
    hint:
    [tex]\cos (x)=\frac{e^{ix}+e^{-ix}}{2} [/tex]
     
  7. Oct 11, 2007 #6

    mjsd

    User Avatar
    Homework Helper

    Nah, my question was a rhetorical one :smile:

    also note that [tex]z+z^*=2\text{Re}(z)[/tex] , [tex]z-z^* = 2i\text{Im}(z)[/tex] and [tex]z z^* = |z|[/tex]
     
  8. Oct 11, 2007 #7

    danago

    User Avatar
    Gold Member

    Got it :smile: thanks for the help guys :D
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Trig Proof
  1. Trig Proof (Replies: 1)

  2. Trig Proof (Replies: 5)

  3. Trig proof (Replies: 8)

  4. Trig proof (Replies: 5)

  5. Trig Proof (Replies: 4)

Loading...