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How does one prove that if $\cos(t) = \cos(t')$ and $\sin(t) = \sin(t')$ then $t = t' + 2k\pi$ ?

I've tried proving the above statement, which I think is valid.

I know $\sin(t)$ is injective on $[-\pi/2; \pi/2]$ and $\cos(t)$ is injective on $[0; \pi]$, but until now I've not been able to use this to prove the statement rigorously.

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  • $\begingroup$ Can't you just square both equations and add them together? $\endgroup$ – Michael Angelo Jan 26 '15 at 12:03
  • $\begingroup$ The fundamental question is: how do you define sine and cosine? There are several definitions and it is important to know what you use. $\endgroup$ – Josué Tonelli-Cueto Jan 26 '15 at 14:04
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also can $$\cos{x}-\cos{y}=0\Longrightarrow 2\sin{\dfrac{x-y}{2}}\sin{\dfrac{x+y}{2}}=0\tag{1}$$ $$\sin{x}-\sin{y}=0\Longrightarrow 2\sin{\dfrac{x-y}{2}}\cos{\dfrac{x+y}{2}}=0\tag{2}$$ with $(1),(2)$ must $$\sin{\dfrac{x-y}{2}}=0$$ since we kown $$\sin{(k\pi)}=0,k\in Z$$ so $$\dfrac{x-y}{2}=k\pi,\Longrightarrow x=y+2k\pi,k\in Z$$

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\begin{align*} \cos(t-t') &= \cos(t)\cos(t') + \sin(t)\sin(t') \\ &= \cos^2(t) + \sin^2(t) = 1 \\ \sin(t-t') &= \sin(t)\cos(t') - \cos(t)\sin(t') \\ &= \sin(t)\cos(t) - \cos(t)\sin(t) = 0. \end{align*} As $\sin(t-t')=0$ we have $t-t'=k\pi$ for some $k\in\mathbb{Z}$, and so $\cos(t-t')=\cos(k\pi)=(-1)^k$. However, we have also seen that $\cos(k\pi)=1$, so $k$ is even, so $t-t'=2m\pi$ for some $m\in\mathbb{Z}$.

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  • $\begingroup$ I wonder if you thought of this by first noting that $e^{it} = e^{i t'} \implies e^{i(t - t')} = 1 \implies \cos(t - t') =1$ and $\sin(t - t') = 0$. $\endgroup$ – littleO Jan 26 '15 at 12:29
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if $\cos(t) = \cos(t^\prime), $ then $t=\pm t^\prime + 2k\pi $ and if $\sin t = \sin t^\prime,$ then $t = t^\prime + 2k\pi, \pi - t^\prime + 2k\pi$ these follow form $\cos$ being an even function while $\sin$ is an odd function and both are $2\pi$-periodic. that is all you need.

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If you want to do it rigorously, you could use the fact that $\sin$ and $\cos$ are orthonormal in an inner product space. See: http://www.jimworthey.com/orthoquestions.html

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The point $P_t=(\cos t,\sin t)$ is the point where the ray at angle $t$ meets the unit circle. Your statements are saying that $P_t=P_{t'}$. This means the two rays coincide, so their angles must differ by a multiple of a full rotation. That is, $t-t'=2k\pi$ for some integral $k$.

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i dont know if this help but.

we have that for $k\in\mathbb{Z}$ $$\begin{align} \sin t=\sin t'&\iff t'=2k\pi+\pi-t\vee t'=2k\pi+t\\ \cos t=\cos t'&\iff t'=2k\pi-t\vee t'=2k\pi+t \end{align}$$ then to have $\sin t=\sin t'$ and $\cos t=\cos t'$ at same time you get $t'=2k\pi+t$ because

if $t'=2k\pi+\pi-t$ you have $\sin t'=\sin(2k\pi+\pi-t)=\sin(\pi-t)=\sin t$ but $\cos t'=\cos(2k\pi+\pi-t)=\cos(\pi-t)=-\cos t$

if $t'=2k\pi-t$ you have $\cos t'=\cos(2k\pi-t)=\cos(-t)=\cos t$ but $\sin t'=\sin(2k\pi-t)=\sin(-t)=-\sin t$

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Cos(x) and sin(x) are periodic functions of period 2k(pi).hence sin(x)=sin(x') and cos(x)=cos(x').

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  • $\begingroup$ Well you have that $\sin(\pi)=\sin(0)$ but it's not a multiple of $2k\pi$,clearly cosine doesn't follow but your argument isn't valid. $\endgroup$ – kingW3 Jan 26 '15 at 12:07
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    $\begingroup$ This addresses the converse of the original question. :) $\endgroup$ – Andrew D. Hwang Jan 26 '15 at 12:41

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