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Trigonometric Relation Formulas

  1. Dec 1, 2003 #1
    I know some of them, such as :

    cos (2x) = (cos x)^2 - (sin x)^2

    sin (2x) = 2(cos x)(sin x)

    tan (2x) = (2tan (x))/(1-(tan x)^2)

    sin (a+b) = sin a cos b + cos a sin b

    cos (a+b) = cos a cos b - sin a sin b

    I need the other formulas such as

    sin x/2
    cos x/2
    sin x^2
    cos x^2
    sin x^3
    cos x^3
    sin (x^(1/2))
    cos (x^(1/2))

    And any others, everyyything about them,

    Anyone can help? or guide me to a link?
     
  2. jcsd
  3. Dec 1, 2003 #2
    [tex]Tan[x +- y] = \frac{Tan(x) +- Tan (y)}{1 -+ Tan(x)Tan(y)}[/tex]

    [tex]Sin^{2}(x) + Cos^{2}(x) = 1[/tex]

    [tex]\frac{Sin A}{A} = \frac{Sin B}{B}[/tex]

    [tex]A^{2} = B^{2} + C^{2} - 2BC Cos A[/tex]

    [tex]Cos A = \frac{A^{2} - B^{2} - C^{2}}{-2BC}[/tex]

    [tex]Sin(\frac{x}{2}) = +-\sqrt{\frac{1 - Cos(x)}{2}}[/tex]

    [tex]Cos(\frac{x}{2}) = +-\sqrt{\frac{1 + Cos(x)}{2}}[/tex]

    [tex]Sin^{2}(\frac{x}{2}) = \frac{1 - Cos(x)}{2}[/tex]

    [tex]Cos^{2}(\frac{x}{2}) = \frac{1 + Cos(x)}{2}[/tex]

    [tex]Sin^{2}(x) = \frac{1 - Cos(2x)}{2}[/tex]

    [tex]Cos^{2}(x) = \frac{1 + Cos(2x)}{2}[/tex]
     
    Last edited: Dec 1, 2003
  4. Dec 2, 2003 #3
    Thanks But ...

    Thank you, but I still need sin (x^1/2), cos (x^1/2), sin (x^2), cos (x^2), sin (x^3), cos (x^3), sin (x^1/3), cos (x^1/3)

    The roots and poweres are for the angles not for the function

    :)
     
  5. Dec 2, 2003 #4
    Sin[sqrt(x)] ? ... i'm gonna leave that for the expert of this forum to answer lol. But I will cogitate on it.


    Edit: After playing around with a right triangle a bit, I got this from a right triangle with sides a,b,c, with c being the hyp, and a being the opp side with angle x.

    [tex]
    sin(x) = \frac{a}{c}[/tex]

    [tex]x = sin^{-1}\frac{a}{c}[/tex]

    [tex]\sqrt(x) = \sqrt{sin^{-1}\frac{a}{c}} = \phi[/tex]

    [tex]
    \begin{equation*}
    \begin{split}
    sin \phi = +- \sqrt{\frac{1 - cos(2\phi)}{2}}
    &= +- \sqrt{\frac{1 - cos(2\sqrt{sin^{-1}(\frac{a}{c})})}{2}}
    \end{split}
    \end{equation*}
    [/tex]
     
    Last edited: Dec 2, 2003
  6. Dec 3, 2003 #5
    :)

    Thank you for making a try, anyone else can help also?
     
  7. Dec 3, 2003 #6
    google search "trigonometric identities" will give you thousands of sites
     
  8. Dec 3, 2003 #7

    but not the ones he want.
     
  9. Dec 3, 2003 #8

    chroot

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    How about

    http://functions.wolfram.com

    It's the base of relationships used by the Mathematica software and has every identity, I believe, known to man.

    - Warren
     
  10. Dec 3, 2003 #9
    I doubt you'll find any identities for those functions. At least nothing that isn't even messier than the original function. I can think of some ugly expansions, like:

    [tex]
    \sin\left(x^2\right)=\sum_{n=0}^\infty(-1)^n\frac{x^{2+4n}}{(2n+1)!}
    [/tex]

    Of course, I could be wrong. I don't know how to prove that there isn't a simple identity.
     
  11. Dec 4, 2003 #10
    how about sin(arctanx)? can this be simplified?
    how about sin(arccosx)?
    I can get this:
    sinx=cos(pi/2-x)
    y=sinx=cos(pi/2-x)
    x=arcsiny=pi/2-arccosy
    so arccosy=pi/2-arcsiny
    sin(arccosy)=sin(pi/2-arcsiny)=cos(arcsiny)
     
  12. Dec 4, 2003 #11
    [tex]
    \begin{align*}
    \sin(\arctan(x))&=\frac{x}{\sqrt{1+x^2}} \\
    \sin(\arccos(x))&=\sqrt{1-x^2}
    \end{align*}
    [/tex]
     
  13. Dec 4, 2003 #12
    Now that I think about it we also have:

    [tex]
    \begin{align*}
    \cos(\arcsin(x))&=\sqrt{1-x^2} \\
    \cos(\arctan(x))&=\frac{1}{\sqrt{1+x^2}} \\
    \tan(\arcsin(x))&=\frac{x}{\sqrt{1-x^2}} \\
    \tan(\arccos(x))&=\frac{\sqrt{1-x^2}}{x}
    \end{align*}
    [/tex]

    Using those identities, I can spot a few more identities, like

    [tex]
    \sin(\arccos(x))=\cos(\arcsin(x))
    [/tex]

    which was already mentioned, as well as

    [tex]
    \begin{align*}
    \sin(\arctan(x))&=x\cos(\arctan(x)) \\
    \tan(\arcsin(x))&=\frac{1}{\tan(\arccos(x))}
    \end{align*}
    [/tex]
     
  14. Dec 4, 2003 #13
    [tex]= dSin^{-1}(x)/dx[/tex]
     
  15. Dec 4, 2003 #14
    I thought

    [tex]
    \frac{d}{dx}(\arcsin(x))=\frac{1}{\sqrt{1-x^2}}
    [/tex]
     
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