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I calculated $\cos(30^{\circ})$ and it is $\frac{\sqrt(3)}{2}$.

I calculated $\cos(-30^{\circ})$ and it is $\frac{\sqrt(3)}{2}$.

And, I don't know why? Can you explain me simply? I think that figures will be very much helpful.

Nay Sie
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  • That is because the cosine is an even function (as, say, defined with the unit circle). – Bernard Aug 14 '20 at 14:30
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    Yes, diagrams are indeed helpful. What is the cosine on the unit circle? Why does your assertion follow? – Qi Zhu Aug 14 '20 at 14:32
  • See, for instance, this answer. – Blue Aug 14 '20 at 14:35
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    To understand the insight, you need to understand the concept of "continuation of a function" in a larger domain. The $\cos$ function is normally defined using right triangles, and only accepts numbers between $0$ and $\pi\over 4$, however it has some interesting properties, such as $\cos(2x)=2cos^2(x)-1$. When expanding the function $\cos$ to a larger domain, such as large number that does not fit in a triangle, negative numbers and even complex numbers, these property must be kept. And a proven fact, there is only one such continuation for each function. – cr001 Aug 14 '20 at 14:42
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    Based on your understanding, what is the cosine ? –  Aug 14 '20 at 15:01
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    @Bernard: $f(x)=f(-x)$ because $f$ is even is a tautology. –  Aug 14 '20 at 15:01
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    Why on earth does everyone think saying "because $\cos$ is an even function answers the question in any way. That's like answering "why do dogs eat meat" with "because they are carnivores". The real issue is why is $\cos$ an even function? (Which if you look at a picture of a circle and imagine to ants crawling around it, one going clockwise and the other goes counterclockwise the answer should be obvious). – fleablood Aug 14 '20 at 15:35

3 Answers3

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This is because cosine is an even function, that is $\cos(-\theta)=\cos(\theta)$ (and also sine is an odd function so $\sin(-\theta)=-\sin(\theta))$.

Here is a sketch of the cosine function:

enter image description here

Andrew Chin
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Alessio K
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    In case you are unfamiliar with the term 'even function', the term originates from looking at graphs of the form $y=x^n$, where $n$ is a positive integer. If $n$ is even, then the graph has a line of symmetry in the $y$-axis. Nowadays the term 'even function' has a much more general meaning: it refers to any graph that this property. $\cos x$ is a good example. – Joe Aug 14 '20 at 14:53
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As the others said, the reason is because cos is an even function, but I just wanted to give you a nice insight into why cosine is even. So if you take the trigonometric definition of cosine $$\cos(\theta) = \frac{x}{r}$$, where $r$ is the distance from the origin to the point ($\sqrt{x^2+y^2}$), which is always positive despite the values of $x$ and $y$.

Now, in the first case, assume that $\theta$ is in the first quadrant ($\theta\in(0,\pi/2)$), then $-\theta$ will be in the fourth quadrant($\theta\in(\pi,\frac{3\cdot\pi}{2})$), but both in the fourth and first quadrant $x$ is positive, so $$\cos(\theta)=\frac{x}{r}=\cos(-\theta)$$ For more elaboration see the picture below.

In the above picture as you can see $\alpha = -\theta$ but both of the angle corresponds to the same $x$ and opposite $y$. If you take one point as an example, you can see that both points correspond to $x = 0.707$ while for opposite $y's$, $0.707$ and $-0.707$. But as you know cos doesn't depend on the $y$ value. $$\implies \cos(\theta)=\cos(\alpha)=\cos(-\theta)= \frac{0.707}{\sqrt{0.707^2+0.707^2}}$$ This also works for the general case as long as $\theta$ is in the first quadrant.

Now, in the second case, assume that $\theta$ is in the second quadrant($\theta\in(\frac{\pi}{2},\pi)$), then $-\theta$ will be in the third quadrant($\theta\in(\pi,\frac{3\dot\pi}{2})$), but both in the second and third quadrant $x$ is negative so again $$\cos(\theta)=\frac{x}{r}=\cos(-\theta)$$ you can think of the above image reflected to get visualization.

You then don't have to think about the third and fourth quadrant because it is similar to proving it for the first and second quadrant. Therefore cosine is even which means $\cos(\theta) = \cos(-\theta)$ for any $\theta\in(0,2\cdot\pi)$

halrankard
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EHM
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Draw a picture of a circle.

$\cos \theta, \sin \theta$ represent the $(x,y)$ values of a point on a unit circle at an angle $\theta$.

Isn't it clear and obvious that if you compare going clockwise around a circle versus going counter clockwise that the $x$ values will all be the same while the $y$ values will be opposite in sign? Just look at the picture:

enter image description here

So $\cos (-\theta) = \cos (\theta)=a$ and $\sin (-\theta) = -\sin \theta=-b$

fleablood
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