KAMP
Kategori Umum => Matematika => Topik dimulai oleh: trfrm pada Juli 19, 2018, 07:33:24 PM
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Queer Case of Circle
Let in $\mathbb{R}^2$ there is a circle $\{(x,y)\in\mathbb{R}^2~|~x^2+y^2=r^2\}$.
Then, let the position of the point $A$ in this space is $\mathbf{a}=r(\mathbf{i}\cos\alpha+\mathbf{j}\sin\alpha)$
and the position of the point $B$ is $\mathbf{b}=r(\mathbf{i}\cos\beta-\mathbf{j}\sin\beta)$.
Of course,
$\mathbf{a}\cdot\mathbf{i}=r\cos\alpha$ and $\mathbf{b}\cdot\mathbf{i}=r\cos\beta$
so $\mathbf{a}\cdot\mathbf{b}=r^2\cos(\alpha+\beta)$.
If an angle which be formed by the vector $\mathbf{a}-\mathbf{b}$ and $r\mathbf{i}-\mathbf{b}$ is $\gamma$, then here is the expression
$(\mathbf{a}-\mathbf{b})\cdot(r\mathbf{i}-\mathbf{b})=|\mathbf{a}-\mathbf{b}|\,|r\mathbf{i}-\mathbf{b}|\cos\gamma$
alias
\[ \cos\gamma=\frac{1+\cos\alpha-\cos\beta-\cos(\alpha+\beta)}{2\sqrt{1-\cos(\alpha+\beta)}\sqrt{1-\cos\beta}}. \]
In sight, the last equation is stranged enough.
But, if we plot the graph $\gamma$ vs $\beta$, we get that
$\gamma$ has value $\alpha/2$ for $0<\beta<2\pi-\alpha$,
and has value $\pi-\alpha/2$ for $2\pi-\alpha<\beta<2\pi$.
How can it? Until now, I is still vexed.