Question 479447
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Place one endpoint of the diagonal at the origin and then the other end point is *[tex \Large (a,\,b)].


Convert *[tex \Large (a,\,b)] to polar coordinates *[tex \Large (r,\,\varphi)] :


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ \sqrt{a^2\ +\ b^2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \varphi\ =\ \tan^{-1}\left(\frac{b}{a}\right)]


Since the diagonal of a square bisects a vertex angle of *[tex \Large \frac{\pi}{2}], each of the sides of the square which can be represented by a vector: *[tex \Large \left(r_i\,\varphi_i\right)] must be offset in angle from the diagonal by a measure of *[tex \Large \frac{\pi}{4}].


Furthermore, if the diagonal of a square is of measure *[tex \Large d], then the side *[tex \Large s\ =\ d\frac{\sqrt{2}}{2}].


Hence, the polar coordinates of the side rotated counterclockwise from the diagonal are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_1\ =\ r\frac{\sqrt{2}}{2}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \varphi_1\ =\ \varphi\ +\ \frac{\pi}{4}]


Which can then be converted back to rectangular coordinates by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(a_1,\,b_1\right)\ =\ \left(r_1\sin(\varphi_1),\,r_1\cos(\varphi_1)\right)]


First look at *[tex \Large \sin\left(\varphi_1\right)\ =\ \sin\left(\varphi\ +\ \frac{\pi}{4}\right)]


Then using the sum formula for sin:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \sin\left(\varphi\ +\ \frac{\pi}{4}\right)\ =\ \frac{\sqrt{2}}{2}\left(\sin(\varphi)\ +\ \cos(\varphi)\right)]


Next, note that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_1\ =\ r\frac{\sqrt{2}}{2}\ =\ \sqrt{\frac{a^2\ +\ b^2}{2}}]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_1\sin\left(\varphi_1\right)\ =\ \frac{\sqrt{a^2\ +\ b^2}}{2}\left(sin(\varphi)\ +\ \cos(\varphi)\right)]


The sum formula for cos is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \cos(A\ +\ B)\ =\ \cos(A)\cos(B)\ -\ \sin(A)sin(B)]


And from that I'll let you derive that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_1\ =\ r\frac{\sqrt{2}}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_1\cos\left(\varphi_1\right)\ =\ \frac{\sqrt{a^2\ +\ b^2}}{2}\left(cos(\varphi)\ -\ \sin(\varphi)\right)]


And then using the fact that the other side of the square is rotated clockwise from your given diagonal and that all sides of a square are identical in measure, you should be able to see that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_2\ =\ r\frac{\sqrt{2}}{2}]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \varphi_2\ =\ \varphi\ -\ \frac{\pi}{4}]


I'll let you look up the difference formulas and verify that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_2\sin\left(\varphi_2\right)\ =\ \frac{\sqrt{a^2\ +\ b^2}}{2}\left(sin(\varphi)\ -\ \cos(\varphi)\right)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_2\cos\left(\varphi_2\right)\ =\ \frac{\sqrt{a^2\ +\ b^2}}{2}\left(cos(\varphi)\ +\ \sin(\varphi)\right)]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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