# Derivation of trigonometric identities form rotation on the plane

1. Jul 9, 2013

### Jbreezy

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. Jul 9, 2013

### Ray Vickson

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.

3. Jul 9, 2013

### Jbreezy

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.

4. Jul 9, 2013

### Ray Vickson

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?

5. Jul 9, 2013

### Jbreezy

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?

6. Jul 9, 2013

### Ray Vickson

I already told you. Now I give up.

7. Jul 9, 2013

### Jbreezy

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.

8. Jul 9, 2013

### LCKurtz

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
9. Jul 9, 2013

### szynkasz

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

On the other hand we have

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