Worksheet 5. 3 Less trigonometry and more dishonesty/linear algebra knowledge Less trigonometry, less dishonesty... 4

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1 Worksheet 5 Contents 1 Just look at the book, on page Trigonometry and honesty 1 3 Less trigonometry and more dishonesty/linear algebra knowledge Less trigonometry, less dishonesty Polar coordinates and the algebra-y part of trig 4 We are trying to figure out how to rotate a system of coordinates by an angle θ. We can do this in at least four different ways. Take your pick. 1 Just look at the book, on page 74 Believe the book, it gives us the formulas. 2 Trigonometry and honesty The question is: given the point P = (x, y), what are the coordinates of P in the coordinates (X, Y ) which are rotated by an angle θ? Here: In red are the new axes, and we are trying to find the lengths X = OB and Y = BP in terms of x = OC and y = CP (that negative sign is there because we are under the X-axis). To do that, we will draw the perpendicular from C to the X-axis, and while we are at it we will draw the parallel to the X-axis through P (call it CE), that intersects CE at D. These new lines are green here: 1

2 We ve made two right triangles, OEC and P DC. If you look closely at OEC, you ll notice that: The angle at E is a right angle. The angle at O is θ. The hypothenuse is x. Therefore, OE = x cos(θ). The remaining angle at C is therefore 90 θ, which means that the angle DCP = θ. If we now look at the triangle DP C, we have that The angle at D is a right angle. The angle at C is θ. The hypothenuse is y. Therefore, P D = y sin(θ). Notice now that BP DE is a rectangle, so P D = BE. So finally! X = OE + BE = OE + BE = x cos θ + y sin θ To find Y = BP, we can do a similar thing. We are looking at the same two triangles, but now we have Y = BP = ED = EC DC = x sin θ y cos θ Y = x sin θ + y cos θ And there it is. To go the other way, we could either solve for x and y (which I don t particularly recommend), or notice that what going from (X, Y ) to (x, y) is the same process that we just did, except turning the other way, i.e. for an angle of θ. So the answers will be x = X cos( θ) + Y sin( θ) y = X sin( θ) + Y cos( θ) = X cos θ Y sin θ = X sin θ Y cos θ 2

3 3 Less trigonometry and more dishonesty/linear algebra knowledge We could do the trigonometry as before, but instead of starting with the point (x, y), starting with the points (1, 0) and (0, 1) to make life easier. For (1, 0): Looking at the triangle, we see that X = cos θ and Y = sin θ (negative because we are in the fourth quadrant). Likewise, for (0, 1): Looking at the triangle, we see that X = sin θ and Y = cos θ. Now, what happens for a point P = (x, y)? We can see that OP = x 1, 0 + y 0, 1 Now, could it be that adding and scaling vectors and then rotating is the same thing as rotating, and then adding and scaling? Well, it s true. We just have to say the magic spell: Rotation is a linear map. (Feel free to come ask me what this means, but I won t write more right now). This means that (X, Y ) = x(cos θ, sin θ) + y(sin θ, cos θ) 3

4 Which is the answer. 3.1 Less trigonometry, less dishonesty On the original picture, draw the vector u, the unit vector in the direction of the positive X-axis and v, the unit vector in the direction of the positive Y -axis. On this picture, notice that. Check it out: OP = X u + Y v We ve seen that 1, 0 = cos(θ) u sin(θ) v 0, 1 = sin(θ) u + cos(θ) v So we must have that x, y = x 1, 0 +y 0, 1 = x(cos(θ) u sin(θ) v)+y(sin(θ) u+cos(θ) v) = (x cos θ+y sin θ) u+( x sin θ+y cos θ) v Which, since we ve just said that OP = X u + Y v, must mean that X = x cos θ + y sin θ; Y = x sin θ + y cos θ 4 Polar coordinates Here s another way to do it, using polar coordinates and no drawings. Start with your point P = (x, y). In polar coordinates it s given by r = x 2 + y 2 ; α = arctan(y/x) 4

5 (Let s say x > 0...). Also I need to call the angle something which is not θ, since I ve used that letter already. Now, in the new polar coordinates, which we should call (R, A), it s not hard to see that So in the new polar coordinates, P is given by r = R; α = A + θ R = r; A = α θ Now we need to write (X, Y ) in terms of (R, A). This is the usual change from (x, y) to (r, α), but with different letters. I.e. what we need is Plug it back in! X = R cos A Y = R sin A X = r cos(α θ) Y = r sin(α θ) Now we have to remember the insanely useful formulas for the sine and cosine of a summ: sin(a + b) = sin(a) cos(b) + cos(a) sin(b) cos(a + b) = cos(a) cos(b) sin(a) sin(b) And we have the following: (remember that sin( x) = sin x and cos( x) = cos(x)): X = r cos(α θ) = r cos(α) cos(θ) + r sin(α) sin(θ) Y = r sin(α θ) = r sin(α) cos(θ) + r cos(α) sin(θ) And finally, since r cos(α) = x and r sin α = y, we have that X = x cos(θ) + y sin(θ) Y = y cos(θ) + x sin(θ) (Remember how one proved the trig identitites we ve just used? It was essentially with the drawing from our first attempt at this problem). 5