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Suppose $\cos \theta = \cos \alpha$. If we give no restrictions to the angles, I understand that we must $\mathbf{not}$ conclude that $\theta = \alpha $. Can we conclude then that $\theta - \alpha = 2 \pi $ ? Is the same true for the sine function?

ILoveMath
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5 Answers5

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Consider the diagram below.trigonometric_functions_symmetry

By symmetry, $\cos\theta = \cos(-\theta)$ and $\sin\theta = \sin(\pi - \theta)$.

If $\cos\theta = \cos\alpha$, then $\alpha$ must be coterminal with $\theta$ or with $-\theta$. Thus, $$ \alpha = \begin{cases} \theta + 2n\pi, n \in \mathbb{Z}\\ -\theta + 2n\pi, n \in \mathbb{Z} \end{cases} $$

If $\sin\theta = \sin\alpha$, then $\alpha$ must be coterminal with $\theta$ or $\pi - \theta$. Hence, $$ \alpha = \begin{cases} \theta + 2n\pi, n \in \mathbb{Z}\\ \pi - \theta + 2n\pi, n \in \mathbb{Z} \end{cases} $$

N. F. Taussig
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Think about points on the unit circle. The point on the circle corresponding to the angle $\theta$ is $(\cos\theta,\sin\theta)$. If you specify both the x and the y coordinate, the point is determined and so the the angle is determined up to an integral multiple of $2\pi$.

But if you only specify one of the coordinates, then there are two points on the unit circle that have that coordinate (they happen to coincide in degenerate cases, but that doesn't affect the final result I'm about to state).

Specifying the x coordinate means you are specifying the value of $\cos\theta$. The reflection of this point across the x axis has the same x coordinate. If the first point corresponds to angle $\theta$, then the reflected point corresponds to the angle $-\theta$. As before, these are the only possible angles up to integral multiples of $2\pi$.

Likewise, specifying the y coordinate means you are specifying the value of $\sin\theta$. The reflection of this point across the y axis corresponds to the angle $-\theta +\pi$, and again these are the only possible angles up to integral multiples of $2\pi$.

The following summarizes what I've said above (here, $k$ can be any integer).

If $\cos\theta =\cos\alpha$, then either $\alpha=\theta +2k\pi$ or $\alpha = -\theta +2k\pi$.

If $\sin\theta=\sin\alpha$, then either $\alpha=\theta +2k\pi$ or $\alpha = - \theta + (2k+1)\pi$.

MPW
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  • @HenryLebesgue: Yes. I think what you meant to type is correct: $\varphi =n\theta + 2\pi k$ are the only solutions. (Oh--your comment has disappeared! I will leave this response for you, though.) – MPW Dec 07 '14 at 04:17
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Try to show the following

$$\cos\theta=\cos\alpha\iff\begin{cases}\theta=\alpha\\\text{or}\\\theta=-\alpha\end{cases}\;\;+2k\pi\;,\;\;k\in\Bbb Z$$

Similarly, but not equally:

$$\sin\theta=\sin\alpha\iff\begin{cases}\theta=\alpha\\\text{or}\\\theta=\pi-\alpha\end{cases}\;\;+2k\pi\;,\;\;k\in\Bbb Z$$

Timbuc
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    I think I got it. Do I have to use the formulas $$ \sin x - \sin y = 2 \sin ( \frac{x+y}{2} ) \cos ( \frac{ x -y }{2} ) $$ ? – ILoveMath Dec 07 '14 at 03:45
  • @MarvinGaye Why would you? The above follows at once from the very definition of the trigonometric functions on the unit (or trigonometric) circle. – Timbuc Dec 07 '14 at 03:46
  • I dont understand why they follow at once – ILoveMath Dec 07 '14 at 03:47
  • For example, the extension of the cosine function from the first to the fourth quadrant makes sure that $;\cos x=\cos(-x);$, and likewise with the other quadrants. Something similar is true for the sine functions. – Timbuc Dec 07 '14 at 03:50
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    @MarvinGaye The sum to product identity you mentioned would be a great tool to use here. – David H Dec 07 '14 at 03:59
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$\cos\alpha=\cos(-\alpha)$ so there are two possible sets of angles that $\theta$ could belong to. Either $\theta=\alpha+2n\pi$, or $\theta=-\alpha+2n\pi$. $n$ can be any whole number.

Empy2
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Hint: $$\forall x\in\mathbb{R} \sin(x\pm2\pi n)=\sin(x)$$

$$\forall x\in\mathbb{R} \cos(x\pm2\pi n)=\cos(x)$$ This is because sine and cosine have a period of $2\pi$, a full revolution. With the result, we see that $sin(x)=sin(y)\implies x=y+2\pi k,k\in\mathbb{Z}$ Your conjecture is false for, $\alpha=0,\beta=4\pi.(k=2)$

Teoc
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  • I must say that, to the contrary, you have stated the trivial part, and have omitted the part that requires more thought. I won't downvote, but this is useless. OP already stated your essential content, and you have failed to mention the other angles with the same sine or cosine. – MPW Dec 07 '14 at 04:09
  • Now this is just plain wrong. $\sin 0=\sin\pi$, yet there is no $k\in\mathbb Z$ such that $0=\pi +2\pi k$. – MPW Dec 07 '14 at 15:23
  • $k=1, sin(3\pi)=0$ – Teoc Dec 07 '14 at 19:11
  • You stated "$\sin x =\sin y \implies x=y + 2\pi k,k\in\mathbb Z$". This is wrong. Taking $k=3$ as you suggest produces the result "$0=3\pi$" which is false. – MPW Dec 07 '14 at 21:10
  • Taking $ k=3 \implies sin(0)=sin(0+23\pi) \implies sin(0)=sin(6\pi)$ which is true. – Teoc Dec 07 '14 at 21:20