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Between $0$ to $2π$, I have noticed that $\sin x$, $\cos x$ and $\tan x$ values repeat for different values of $x$.

For example, $\sin 30 = \sin 150$

What exactly is the interval between two successive values of $x$ such that the value of $\sin x$, $\cos x$ or $\tan x$ are equal?

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There are several ways of doing this. Have you tried drawing out their graphs? Do you know the geometric meaning of $\sin, \cos, \tan$? – Henry Swanson Aug 30 '13 at 16:19
As indicated in the answer below, the sine and cosine repeat every $360^{\circ}$, and the tangent repeats every $180^{\circ}$. These are called the periods of these functions. – user84413 Aug 30 '13 at 17:30
Please, please, please, use the degree symbol "^\circ" if you want degrees. If you don't use it, you mean radians, whether that's what you want or not. – Stefan Smith Aug 30 '13 at 23:10
up vote 4 down vote accepted

$\sin(x) = \sin(\pi - x)$, or, equivalently
$\sin(x) = \sin(180^o - x)$.

$\sin(x) = \sin(2\pi n + x) \forall n \in \mathbb{Z}$, or, equivalently,
$\sin(x) = \sin(360^o n + x) \forall n \in \mathbb{Z}$.

$\cos(x) = \cos(-x)$. Alternately,
$\cos(x) = \cos(360^o - x)$, or (equivalently to the latter)
$\cos(x) = \cos(2\pi - x)$.

$\cos(x) = \cos(2\pi n + x) \forall n \in \mathbb{Z}$, or, equivalently
$\cos(x) = \cos(360^o n + x) \forall n \in \mathbb{Z}$.

$\tan(x) = \tan(\pi + x)$, or, equivalently,
$\tan(x) = \tan(180^o + x)$.

$\tan(x) = \tan(\pi n + x) \forall n \in \mathbb{Z}$, or, equivalently,
$\tan(x) = \tan(180^o n + x) \forall n \in \mathbb{Z}$.

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I had to downvote this because of the formatting. It's unreadable to me in its current form. – AnonSubmitter85 Aug 30 '13 at 18:25
What makes it unreadable? – qaphla Aug 30 '13 at 18:39
It's a giant jumbled mess with too many equations crammed too close together. Moreover, half of the equations say the same thing (e.g., the very first line). – AnonSubmitter85 Aug 30 '13 at 18:47
Tried to make it better. Half the equations say the same thing because OP wasn't very clear on whether to use degrees or radians, so I just gave both. I'm not really sure what you want from the formatting, but I'd hate for it to be so bad that you can't read it at all. – qaphla Aug 30 '13 at 18:53
Much better. I think tables work best in these cases, but I know that there is only so much time one is willing to spend on these answers (and rightfully so, too). – AnonSubmitter85 Aug 30 '13 at 18:56


Let $\sin x=\sin y$

Applying $\sin C-\sin D=2\sin\frac{C-D}2\cos\frac{C+D}2,$

we have $2\sin\frac{x-y}2\cos\frac{x+y}2=0$

If $\sin\frac{x-y}2=0, \frac{x-y}2=n180^\circ\implies x-y=n180^\circ$ where $n$ is any integer

If $\cos\frac{x+y}2=0, \frac{x+y}2=(2m+1)90^\circ\implies x+y=(2m+1)180^\circ$ where $m$ is any integer


Apply $\cos C-\cos D=-2\sin\frac{C-D}2\sin\frac{C+D}2,$


$\displaystyle\tan x=\tan y\iff \frac{\sin x}{\cos x}=\frac{\sin y}{\cos y}$

$$\implies \sin x\cos y-\cos x\sin y=0\implies \sin(x-y)=0$$

$\implies x-y=r180^\circ$ where $r$ is any integer

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