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Derivation of trigonometric identities form rotation on the plane

  1. Jul 9, 2013 #1
    1. The problem statement, all variables and given/known data

    I want to derive the trig identities starting with rotation on the plane.


    2. Relevant equations
    One rotation through a given angle is given by
    $$x' = xcosθ - ysinθ $$
    $$y' = xsinθ + ycosθ$$


    3. The attempt at a solution

    What if I wanted to rotated through any angle $$ψ$$.

    Then

    $$ x'' = x'cosψ - y'sinψ $$
    $$ = (xcosθ - ysinθ)cosψ - (xsinθ + ycosθ)sinψ $$
    $$ = xcosθcosψ - ysinθcosψ - xsinθsinψ +ycosθsinψ $$

    I'm concerned about having x's and y's still. Hint?
     
  2. jcsd
  3. Jul 9, 2013 #2

    Ray Vickson

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    No problem: having x and y is *exactly* what you want, because they allow you to find the new sine and cosine values. Remember: new_x = cos(angle)*old_x - sin(angle)*old_y, etc. So, with old_x = x and new_x = x'', the coefficients are just the sine and cosine.
     
  4. Jul 9, 2013 #3
    Wait wait,

    What do you mean the coefficients are just the sine and cosine? I'm just a little bit confused because what if you wanted to arrive at the sum to product identity from where I felt off. You are saying I can replace x and y by cosine and sine? Or x'' by .... Maybe that is more specific for what I struggle with.
     
  5. Jul 9, 2013 #4

    Ray Vickson

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    Well, YOU wrote ##x' = \cos(\theta) x - \sin(\theta) y.## The right-hand-side is a linear function of x and y. What is the coefficient of x? Of y?
     
  6. Jul 9, 2013 #5
    OK. I see. But I'm saying how would I get the sum to product identity or any other from the form I have. That is why I want to get rid of x's and y's because they are not in those identities. You see what I'm asking now?
     
  7. Jul 9, 2013 #6

    Ray Vickson

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    I already told you. Now I give up.
     
  8. Jul 9, 2013 #7
    Yeah I understand that you already told me. But I feel like you did not. I don't understand how to get from my last formula with x and y in it and x'' on the left side to one of the identities.
     
  9. Jul 9, 2013 #8

    LCKurtz

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    If you call the rotation through ##\alpha## by ##T_\alpha## and express your original formulas as matrices you have$$
    T_\alpha(x,y) =\begin{bmatrix}
    \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha
    \end{bmatrix}\begin{bmatrix}x\\y \end{bmatrix}$$Now, rotation through an angle ##\theta=\alpha+\beta## can be represented by one rotation followed by another:$$
    T_\theta = T_{\alpha + \beta} = T_\beta T_\alpha$$So the rotation through ##\alpha + \beta## is represented by the matrix multiplication$$
    \begin{bmatrix}
    \cos(\alpha+\beta) & -\sin(\alpha+\beta)\\ \sin(\alpha+\beta) & \cos(\alpha+\beta)
    \end{bmatrix}
    =\begin{bmatrix}
    \cos\beta & -\sin\beta\\ \sin\beta & \cos\beta
    \end{bmatrix}\begin{bmatrix}
    \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha
    \end{bmatrix}$$Multiplying those matrices out will give you the addition formulas, if that's what you are asking.
     
    Last edited: Jul 9, 2013
  10. Jul 9, 2013 #9
    $$ x'' = x'cosψ - y'sinψ $$
    $$ = (xcosθ - ysinθ)cosψ - (xsinθ + ycosθ)sinψ $$
    $$ = xcosθcosψ - ysinθcosψ - xsinθsinψ -ycosθsinψ $$
    $$ = x(cosθcosψ - sinθsinψ)-y(sinθcosψ +ycosθsinψ) $$

    On the other hand we have

    $$ x'' = x'cosψ - y'sinψ = xcos(θ+ψ) - sin(θ+ψ)$$
     
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