How do you find the coordinates flipped for Trig question 11 part c?

[Coco12]

For question 11 , how do you do part c?
I know that (cos theta corresponds to x value and sin theta corresponds to y value. Using that I found the angle to be 318 degrees for part a. For part c, how would you start that? The answer is the coordinates flipped , x and y values with a positive sign. How do you get there?

 

[Student100]

It's generally not recommended to use phone speak here, also you show what ever work you've attempted even if wrong.

If the points are given such that p(theta) is known, then what would you do to find p(theta + pi)?

 

[Integrand]

View attachment 65315
For question 11 , how do u do part c?
I know that (cos theta corresponds to x value and sin theta corresponds to y value. Using that I found the angle to be 318 degrees for part a. For part c, how would you start that? The answer is the coordinates flipped , x and y values with a positive sign. How do you get there?

Are you sure the x-value in the answer doesn't have a negative sign?

Are you familiar with radian measure? If not, it might be a good idea to look it up. If so, think about the amount of turning measured by π.

 

[Student100]

Are you sure the x-value in the answer doesn't have a negative sign?

Are you familiar with radian measure? If not, it might be a good idea to look it up. If so, think about the amount of turning measured by π.

Degrees are fine for this.

 

[Simon Bridge]

You should use a diagram.
Draw the x and y-axis ... put point P on it, and draw a line from the origin to P.
Now you know which quadrant it is in without having to do sines and cosines.

The measure of the angle is probably taken anticlockwise from the +x axis.
Looking at the context, I think they want angles in radians, not degrees.

Do you know how to rotate a line in a diagram?
Part c is telling you to rotate the line you just drew by pi radians.
Thats 180 degrees - so sketch it in and you can see where it ends up.

[edit] the x value would have a negative sign if the point is in the second quadrant.
Sketch it out and see.

 

[Coco12]

You should use a diagram.
Draw the x and y-axis ... put point P on it, and draw a line from the origin to P.
Now you know which quadrant it is in without having to do sines and cosines.

The measure of the angle is probably taken anticlockwise from the +x axis.
Looking at the context, I think they want angles in radians, not degrees.

Do you know how to rotate a line in a diagram?
Part c is telling you to rotate the line you just drew by pi radians.
Thats 180 degrees - so sketch it in and you can see where it ends up.

[edit] the x value would have a negative sign if the point is in the second quadrant.
Sketch it out and see.

That is what I thought too but the answer in the book is different. Ok I understand that however when are the coordinates flipped?

 

[Student100]

The coordinates don't flip, they're an arbitrary creation made by you. The point moves 180 degrees, the sin and cosine oscillate between one and negative one. So if you move the point 180 degrees what happens? Originally the sin was negative and cosine was positive, so if you shift them 180 degrees what will be the sign?

 

[Coco12]

The signs would change but why is that in the book the coordinates that was given at the top flip the x value becomes the y value??

 

[Simon Bridge]

The signs would change but why is that in the book the coordinates that was given at the top flip the x value becomes the y value??

That would happen if the rotation was pi/2 ... sketch it out.

There are two possibilities:
1. you are looking at the answer to (d) rather than (c)
2. there is a mistake in the book

Always sketch it out:
The geometry does not lie.
Geometry is your friend.
Trust the geometry.

 

[Coco12]

I understand part c but part d I still don't really understand

 

[Coco12]

That would happen if the rotation was pi/2 ... sketch it out.

There are two possibilities:
1. you are looking at the answer to (d) rather than (c)
2. there is a mistake in the book

Always sketch it out:
The geometry does not lie.
Geometry is your friend.
Trust the geometry.

The one highlighted in yellow would be the quadrant it is in if you add 90 degrees on to it.? The answer in the back of the book for d is (2/3, sqrt (5)/3)

I thought the signs will just change to positive as it is in the first quadrant. Why would the coordinates switch?

 

[vanhees71]

That would happen if the rotation was pi/2 ... sketch it out.

The geometry does not lie.
Geometry is your friend.
Trust the geometry.

The ideal geometry book needs no drawings (of course, I'm kidding).

Some hints for how to deal with polar coordinates and transforming between them and Cartesian coordinates:

(1) polar coordinates -> cartesian coordinates
$$x=r \cos \theta, \quad y=r \sin \theta, \quad r>0, \quad \theta \in ]-\pi,\pi].$$

Polar coordinates are singular at the origin, so the null-vector cannot be mapped uniquely to polar coordinates. For the angle, you can also use any other semi-open interval of length $2 \pi$.

(2) cartesian coordinates -> polar coordinates

The length, $r$ is simply
$$r=\sqrt{x^2+y^2}.$$
To get the angle, $\theta$ it's dangerous to use the equation with the arctan cited in many textbooks. You rather need what's called atan2(y,x) in FORTRAN. For that function you can use
$$\theta=\begin{cases} \mathrm{sign} \; y \arccos \left (\frac{x}{\sqrt{x^2+y^2}} \right ) & \text{for} \quad y \neq 0,\\ \arccos \left (\frac{x}{|x|} \right ) & \text{for} \quad y=0. \end{cases}$$

This gives angles in the range $]-\pi,\pi]$.

 

[Simon Bridge]

The one highlighted in yellow would be the quadrant it is in if you add 90 degrees on to it.?

That's the one - the sketch is a bit sloppy though.

The answer in the back of the book for d is (2/3, sqrt (5)/3)

I thought the signs will just change to positive as it is in the first quadrant. Why would the coordinates switch?

It they didn't it wouldn't be a rotation - it would be a reflection.

Try this (easier) example:

Use a ruler to draw the axis out. Mark them off from -6 to 6 in equal intervals like you are drawing a graph.

Mark in point P:

$P=(4,-3)$​

Draw a line from the origin to point P. - that is the line OP.
Use a compass - put the needle on the origin and the pencil on P - draw a circle on the diagram.
That circle is all the places point P could end up when rotated by some angle.

Draw point Q:

$Q=(3,4)$​

That point is on the circle - so it is possible to rotate P until it is at Q.

Now use a protractor to measure than angle ∠POQ ... see?

----------------------

[aside] I'd rather do $R(\theta) \vec p = \vec q$ where $$R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}$$... which is great if you know how to handle matrixes:

$$R(\textstyle \frac{\pi}{2})p = \begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\begin{pmatrix}4\\ -3\end{pmatrix} = \begin{pmatrix}3\\4\end{pmatrix}=\vec{q}$$
... but the other way is fine too.

 

[Coco12]

That's the one - the sketch is a bit sloppy though.

It they didn't it wouldn't be a rotation - it would be a reflection.

Try this (easier) example:

Use a ruler to draw the axis out. Mark them off from -6 to 6 in equal intervals like you are drawing a graph.

Mark in point P:

$P=(4,-3)$​

Draw a line from the origin to point P. - that is the line OP.
Use a compass - put the needle on the origin and the pencil on P - draw a circle on the diagram.
That circle is all the places point P could end up when rotated by some angle.

Draw point Q:

$Q=(3,4)$​

That point is on the circle - so it is possible to rotate P until it is at Q.

Now use a protractor to measure than angle ∠POQ ... see?

----------------------

[aside] I'd rather do $R(\theta) \vec p = \vec q$ where $$R(\theta) = \begin{pmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}$$... which is great if you know how to handle matrixes:

$$R(\textstyle \frac{\pi}{2})p = \begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\begin{pmatrix}4\\ -3\end{pmatrix} = \begin{pmatrix}3\\4\end{pmatrix}=\vec{q}$$
... but the other way is fine too.

So in short, adding or subtracting 90 degrees will cause the coordinates to switch?

 

[Simon Bridge]

So in short, adding or subtracting 90 degrees will cause the coordinates to switch?

Yep - and the sign of one will also change.

Basically if P=(x,y) ... then rotating by +90deg makes P'=(x',y') so that x'=-y, and y'=x.

But if I'd just said: "the coordinates switch when you added 90deg because you added 90deg", that's not really answering the question is it?

To understand how it happen, you should draw the picture like I said.
There is no point memorizing all these little rules.

What happens if you rotate by +45deg?

 

[Coco12]

Yep - and the sign of one will also change.

Basically if P=(x,y) ... then rotating by +90deg makes P'=(x',y') so that x'=-y, and y'=x.

But if I'd just said: "the coordinates switch when you added 90deg because you added 90deg", that's not really answering the question is it?

To understand how it happen, you should draw the picture like I said.
There is no point memorizing all these little rules.

What happens if you rotate by +45deg?

If you rotated by 45 degrees, the coordinates will remain the same but the sign will change, right?

 

[Simon Bridge]

If you rotated by 45 degrees, the coordinates will remain the same but the sign will change, right?

In general, the sign may change - depending on the
Nope ... draw the picture.
If you started with P=(4,-3), then you end up with both coordinates totally different numbers, but stil on the circle.

 

[Coco12]

In general, the sign may change - depending on the
Nope ... draw the picture.
If you started with P=(4,-3), then you end up with both coordinates totally different numbers, but stil on the circle.

Yes, I see it now. U can find the exact coordinates by plugging in the angle into ( cos theta, sin theta) right?

 

[Simon Bridge]

Did you draw the picture?

The instruction I am giving you is for understanding the geometry not memorizing some formula.
When you understand, you don't need formulas.

Usually people resist understanding .. if you get used to this you will quickly get ahead of your class.