May Kristy
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Hi. I'm currently studying for A levels FP1 exam. I just came across this theory so can anyone please explain to me how reflection in the line y= x tan thetha works? Like I understand other matrix transformations like stretch of scale factor k from the x-axis and such. Perhaps a good web link will do.
It would be great if anyone could explain.
Thanks.....
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RDKGames
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(Original post by May Kristy)
Hi. I'm currently studying for A levels FP1 exam. I just came across this theory so can anyone please explain to me how reflection in the line y= x tan thetha works? Like I understand other matrix transformations like stretch of scale factor k from the x-axis and such. Perhaps a good web link will do.
It would be great if anyone could explain.
Thanks.....
If you consider 0 \leq \theta <180 where it is measured from the positive horizontal axis anticlockwise, then any line inclined at that angle will have the form y=x\tan(\theta). Any object will simply be reflected in that line.

I can't find any videos on the topic but I will attempt to explain it here. Best to follow along and sketch yourself a diagram:

Sketch an xy-plane. Draw some arbitrary line that goes through the origin, preferably one with +ve gradient. Label the angle that this line is inclined at as \theta (ie the angle from the positive x-axis anticlockwise). The equation of this line is y=x\tan(\theta). Now label an arbitrary point between this line and the x-axis, call it A(x,y). Now draw a dotted line from the origin to this point. The length of this line is r=\sqrt{x^2+y^2} from Pythagoras, and the angle at which the dotted line sits call it \alpha. Then we know that \tan(\alpha)=\frac{y}{x}. This means we can express our coordinates of A as A(r\cos(\alpha),r\sin(\alpha)) (these are what we call the polar coordinates that you will touch upon in FP2). Next, consider what happens when you reflect this point in your line of reflection. Do so (doesn't have to be terribly accurate) and label it A'(X,Y). Now draw a dotted line between OA' and one between AA'. The triangle you create from the dotted lines is an iscosceles triangle which has y=x\tan(\theta) as the line of symmetry. Since it's an iscoscles triangle, you know that distances | OA' |=|OA| =r. Next we want to find out at what angle the line OA' is inclined at. Some simple angle labelling and addition should reveal that the angle is 2\theta-\alpha. Now we can return to our polar coordinates and say that this point has coordinates A'(r\cos(2\theta-\alpha),r\sin(2\theta-\alpha)).

So we have
X=r\cos(2\theta-\alpha)
and
Y= r\sin(2\theta-\alpha)

Next we proceed to simplify these by using a C4 compound angle formulae:
\cos(A-B)= \cos(A)\cos(B)+\sin(A)\sin(B) and \sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B).

So we get:
X=r\cos(2\theta)\cos(\alpha)+r \sin(2\theta)\sin(\alpha)
Y=r\sin(2\theta)\cos(\alpha)-r\cos(2\theta)\sin(\alpha)

But we know that x=r\cos(\alpha) and y=r\sin(\alpha) so we get:

X=x\cos(2\theta)+y\sin(2\theta)
Y=x\sin(2\theta)-y\cos(2\theta)

Therefore, those are the transformed coordinates which are only dependent on the angle \theta at which the reflection line is inclined.

For completeness, we have a transformation R which gave us:

\displaystyle R \begin{bmatrix} x \\y \end{bmatrix} = \begin{bmatrix} x\cos(2\theta)+y\sin(2\theta) \\x\sin(2\theta)-y\cos(2\theta) \end{bmatrix}

Meaning that R=\begin{bmatrix} \cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta) \end{bmatrix} and this is the reflection matrix that you use.
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May Kristy
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(Original post by RDKGames)
If you consider 0 \leq \theta <180 where it is measured from the positive horizontal axis anticlockwise, then any line inclined at that angle will have the form y=x\tan(\theta). Any object will simply be reflected in that line.

I can't find any videos on the topic but I will attempt to explain it here. Best to follow along and sketch yourself a diagram:

Sketch an xy-plane. Draw some arbitrary line that goes through the origin, preferably one with +ve gradient. Label the angle that this line is inclined at as \theta (ie the angle from the positive x-axis anticlockwise). The equation of this line is y=x\tan(\theta). Now label an arbitrary point between this line and the x-axis, call it A(x,y). Now draw a dotted line from the origin to this point. The length of this line is r=\sqrt{x^2+y^2} from Pythagoras, and the angle at which the dotted line sits call it \alpha. Then we know that \tan(\alpha)=\frac{y}{x}. This means we can express our coordinates of A as A(r\cos(\alpha),r\sin(\alpha)) (these are what we call the polar coordinates that you will touch upon in FP2). Next, consider what happens when you reflect this point in your line of reflection. Do so (doesn't have to be terribly accurate) and label it A'(X,Y). Now draw a dotted line between OA' and one between AA'. The triangle you create from the dotted lines is an iscosceles triangle which has y=x\tan(\theta) as the line of symmetry. Since it's an iscoscles triangle, you know that distances | OA' |=|OA| =r. Next we want to find out at what angle the line OA' is inclined at. Some simple angle labelling and addition should reveal that the angle is 2\theta-\alpha. Now we can return to our polar coordinates and say that this point has coordinates A'(r\cos(2\theta-\alpha),r\sin(2\theta-\alpha)).

So we have
X=r\cos(2\theta-\alpha)
and
Y= r\sin(2\theta-\alpha)

Next we proceed to simplify these by using a C4 compound angle formulae:
\cos(A-B)= \cos(A)\cos(B)+\sin(A)\sin(B) and \sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B).

So we get:
X=r\cos(2\theta)\cos(\alpha)+r \sin(2\theta)\sin(\alpha)
Y=r\sin(2\theta)\cos(\alpha)-r\cos(2\theta)\sin(\alpha)

But we know that x=r\cos(\alpha) and y=r\sin(\alpha) so we get:

X=x\cos(2\theta)+y\sin(2\theta)
Y=x\sin(2\theta)-y\cos(2\theta)

Therefore, those are the transformed coordinates which are only dependent on the angle \theta at which the reflection line is inclined.

For completeness, we have a transformation R which gave us:

\displaystyle R \begin{bmatrix} x \\y \end{bmatrix} = \begin{bmatrix} x\cos(2\theta)+y\sin(2\theta) \\x\sin(2\theta)-y\cos(2\theta) \end{bmatrix}

Meaning that R=\begin{bmatrix} \cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & -\cos(2\theta) \end{bmatrix} and this is the reflection matrix that you use.
Hi! I did not see your reply earlier. Thank you so much! You're a genius.
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