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Yesterday we were having a lecture on point coordinates after rotation. The prof. explained that the position of a point after a counterclockwise rotation is obtained from the following formula,

$x=x_0 \operatorname{Cos}(\theta)-y_0 \operatorname{Sin}(\theta), \qquad y=y_0 \operatorname{Cos}(\theta)+x_0\operatorname{Sin}(\theta)$,

where $\theta$ is the angle of rotation.

And then, out of nowhere, for an exercise he used the following formula for the displacements of the point after rotation,

$u_x=(-x_0 \operatorname{Cos}(\frac{\pi-\theta}{2})-y_0\operatorname{Sin}(\frac{\pi-\theta}{2}))\times 2\operatorname{Sin}(\frac{\theta}{2}), \qquad \\ u_y=(x_0 \operatorname{Sin}(\frac{\pi-\theta}{2})-y_0\operatorname{Cos}(\frac{\pi-\theta}{2}))\times 2\operatorname{Sin}(\frac{\theta}{2}).$

Why the signs are different? why instead of just $\theta$ he has used $\frac{\pi-\theta}{2}$? and most importantly what is the role of $2\operatorname{Sin}(\frac{\theta}{2})$?

Thanks in advance.

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    $\begingroup$ Next time, ask the professor, don't be ashamed. You can even send him a e-mail with a link to this post. $\endgroup$ May 18, 2018 at 16:28
  • $\begingroup$ @GiuseppeNegro for the next time definitely. $\endgroup$
    – KratosMath
    May 18, 2018 at 16:29

1 Answer 1

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You must use some trig equalities: $$ \cos\frac{\pi-\theta}{2}=\sin\frac{\theta}{2},\quad \sin\frac{\pi-\theta}{2}=\cos\frac{\theta}{2},\\ \cos\theta=1-2\sin^2\frac{\theta}{2}=2\cos^2\frac{\theta}{2}-1,\quad \sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2} $$ to obtain: $$ \begin{align} &\left(-x_0 \cos\frac{\pi-\theta}{2}-y_0\sin\frac{\pi-\theta}{2}\right)\cdot 2\sin\frac{\theta}{2}\\ =&\left(-x_0 \sin\frac{\theta}{2}-y_0\cos\frac{\theta}{2}\right)\cdot 2\sin\frac{\theta}{2}\\ =&-2x_0 \sin^2\frac{\theta}{2}-2y_0\sin\frac{\theta}{2}\cos\frac{\theta}{2}\\ =&x_0 (1-\cos\theta)-y_0\sin\theta\\ =&x_0-x_0\cos\theta-y_0\sin\theta\\ =&x_0-x, \end{align} $$ and the analogous for the second equation.

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  • $\begingroup$ Excellent!! just one typo at the beginning. $\operatorname{cos}(\frac{\pi-\theta}{2})= \operatorname{sin}(\frac{\theta}{2})$ and also for the other one. Thanks a lot. $\endgroup$
    – KratosMath
    May 18, 2018 at 20:04
  • $\begingroup$ I corrected the typo, thank you. $\endgroup$ May 18, 2018 at 20:33

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