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(I already know the unit circle)

Why is it that the sine of an obtuse angle is the sine of its supplementary angle but the cosine of an obtuse angle is the negative of the cosine of its supplementary angle?

I can see of course on the unit circle that it is this way, but my question is:

Is it purely convention? And if it is, what was the logic behind coming up with that specific convention as opposed to another one? Was it just practical to define the unit circle this way?

Why not have the cosine of obtuse angles be defined the same as the sine of obtuse angles? Why would that not work?

Sufyan Naeem
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jeremy radcliff
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  • Don't you see that these are the most natural extensions of $\cos$ and $\sin$ to angles outside of $[0,{\pi\over2}]$? Why are you so stubborn? – Christian Blatter Apr 30 '15 at 18:52
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    This answer may be helpful. – Blue Apr 30 '15 at 19:22
  • @Blue now that's you have made me understand the OP – Sufyan Naeem Apr 30 '15 at 19:24
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    @ChristianBlatter, I'm not being stubborn, I'm sorry it comes off this way. I'm in general confused about the line in mathematics between definitions by convenience and definition that follow an actual phenomenon (maybe there is no such line). I think that exploring the limits of definitions and the reasons behind them is important for understanding anything – jeremy radcliff Apr 30 '15 at 19:25
  • @Blue, Yes! Thank you...that's exactly the kind of answer I was looking for; it's very helpful for me to see the evolution in the development of definitions. – jeremy radcliff Apr 30 '15 at 19:51

2 Answers2

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Hint

If I didn't get you wrong

We have a Fundamental Law:

$$\cos({\alpha}-{\beta})=\cos{\alpha}\cos{\beta}+\sin{\alpha}\sin{\beta}$$

We make deductions and get new identities:

Putting $\alpha=0$ we have,

$$\cos{-{\beta}}=cos{\beta}$$

Community
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Sufyan Naeem
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Those aren't definitions; they're just facts about how $\cos$ and $\sin$ behave in the second quadrant. Those features follow from how the trig functions are defined using the unit circle: If $(x,y)$ is the point on the unit circle obtained by rotating counterclockwise by $\theta$ radians from the point $(1, 0)$, then $\cos(\theta) := x$ and $\sin(\theta) := y$.

These definitions are built specifically to agree with the "SOH CAH TOA" definitions for $\theta$ between $0$ and $\pi/2$; because the hypotenuse is 1, by definition, the sine of an angle is simply equal to the opposite side, which is $y$, and the cosine of an angle is equal to the adjacent side, which is $x$.

For $\theta$ in the interval from $\pi/2$ to $\pi$ (i.e., $\theta$ an obtuse angle), $(x, y)$ is in the second quadrant, where $x<0$ and $y>0$. Reflecting across the $y$-axis corresponds to taking the supplementary angle, and this reflection negates $x$ and leaves $y$ unchanged. Hence the cosine of $\theta$ is given by the negative of the cosine of the supplementary angle to $\theta$, and the sine is equal to the sine of the supplementary angle.

Since the supplementary angle is given by $\pi-\theta$, these two facts can be summarized by the equations $$\cos(\pi-\theta) = -\cos(\theta)$$ and $$\sin(\pi-\theta)=\sin(\theta),$$ both of which are special cases of the more general angle addition and subtraction formulas.

Dustan Levenstein
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  • But when you say "they're just facts about how cos and sin behave in the second quadrant", where does that second quadrant come from? There was no analytical geometry when Trig was invented. Where does the unit circle itself come from in the first place? Why is it used to define Trig functions? And why the use of negative numbers in the coordinate system? Why not move the unit circle to the right on the x axis so that obtuse angles are still in the first quadrant? – jeremy radcliff Apr 30 '15 at 18:58
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    As I mentioned in my latest edit, the definition is built to agree with the SOH CAH TOA definition that I assume you're more familiar with. If you move the unit circle to the right, you'll affect the values of cosine and sine for acute angles. – Dustan Levenstein Apr 30 '15 at 19:02
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    @jeremyradcliff if you think to move unit circle towards right on the x-axis then do you have another plan to define sine and cosine for acute angles? – Sufyan Naeem Apr 30 '15 at 19:12
  • I guess when I used the word definition, I meant whether the coordinate system is used for convenience. Because for ex, we could imagine using a coordinate system without negative numbers (same symmetry around (0,0)); then the unit circle would obey Soh Cah Toa for angles between pi and pi/2, but we wouldn't be able to tell the difference between cos(120) and cos(240) (in degrees). – jeremy radcliff Apr 30 '15 at 19:15
  • @SufyanNaeem, yes that's a good point. But do you see the point in the latest comment I left, is there some aspect on convenience in using the coordinate system and the unit circle? – jeremy radcliff Apr 30 '15 at 19:16
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    Would you like to chat? That might be easy... – Sufyan Naeem Apr 30 '15 at 19:19
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    It sounds like you want to define your own version of sine and cosine, say let's call them $\sin_j$ and $\cos_j$, which would be related to the standard ones via $\sin_j(\theta) = |\sin(\theta)|$ and $\cos_j(\theta)=|\cos(\theta)|$. These are certainly well-defined functions that fit what you seem to desire, but if you graph $\cos_j(\theta)$ as a function of $\theta$, you'll notice a sharp bounce (a corner) in the graph at the point $\pi/2$, whereas with the standard definition it just continues smoothly through. – Dustan Levenstein Apr 30 '15 at 19:20
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    The standard definitions ensure that $\sin$ and $\cos$ are analytic functions. – Dustan Levenstein Apr 30 '15 at 19:21
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    @DustanLevenstein Ok, this makes sense. I'm just getting used to the idea that things in math are defined to be coherent with the rest of the language, and that's what I was asking thank you. Unfortunately I don't know about analytic functions, but I will read the article you linked. – jeremy radcliff Apr 30 '15 at 19:31
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    Analytic functions are essentially among the nicest possible kinds of functions we could ask for. In particular, they're "smooth" (differentiable, moreover infinitely many times). – Dustan Levenstein Apr 30 '15 at 19:36
  • @DustanLevenstein Sorry but why would there be a corner at the point $\pi/2$ in my function $\cos_j$? $\cos_j(\theta)$ would then be $|0|$, which is 0, just like $\cos(\theta)$ would be. Ah ok!! Nevermind, I got it. There would be a corner when it would go back up right after. Awesome, thank you. – jeremy radcliff Apr 30 '15 at 20:04
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    The corner I refer to is not a feature of the single value of $\cos_j$ at $\pi/2$; it's a feature of the local behavior of $\cos_j$ around the point $\pi/2$. Look at this graph around the point $\pi/2 \cong 1.57$. – Dustan Levenstein Apr 30 '15 at 20:07
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    @DustanLevenstein Yes, that totally makes sense. This is great because now I actually have some understanding of why Trig definitions were extended that way, which is really what I wanted. Initially, it seemed pretty arbitrary. – jeremy radcliff Apr 30 '15 at 20:10
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    @jeremyradcliff glad to have helped! – Dustan Levenstein Apr 30 '15 at 20:10
  • @DustanLevenstein, yes, thank you really, it makes things so much more interesting. – jeremy radcliff Apr 30 '15 at 20:12
  • Here is a simpler explanation. 1) peel glow-in-the-dark sticker 2)place sticker on hoola hoop 3) turn off lights 4) roll hoola hoop across floor – John Joy Apr 30 '15 at 20:47