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I have been thinking about this for a day now but it seems every way is predicated on using either $\cos(0)=1$ and $\sin(0)=0$ as a giving to justify the argument. Is it possible to show this to be true without prior knowledge of sine and cosine at zero?

For example, if we use the identity $\cos^2(0) + \sin^2(0) = 1$, we have to assume knowledge of either $\cos(0)$ or $\sin(0)$, maybe I missed something. Then I thought about using the Taylor series at $x=0$ but not assuming it to be giving. By doing so, I end up with \begin{align} \cos(x) &= \cos(0) - x\sin(0) - \frac{x^2}{2!}\cos(0) + \frac{x^3}{3!}\sin(0) + \cdots\\ &= \cos(0)\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!} - \sin(0)\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}\tag{1}\\ \cos(0) &= \cos(0)\cdot 1 - \sin(0)\cdot 0\\ \cos(0) &= \cos(0) \end{align} Therefore, I haven't reached $\cos(0) = 1$ unless I use it as a giving. Then I thought about the exponential $e^{ix} = \cos(x) + i\sin(x)$. Now this would work $$ e^{i\cdot 0} = \cos(0) + i\sin(0) = 1 + 0\cdot i $$ so the imaginary part has to be zero, but if we derive the power series and not take Euler's formula for granted, we get \begin{align} e^{ix} &= \cos(0)\biggl[\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}+i\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}\biggr]\\ &+ \sin(0)\biggl[\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}-\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}\biggr]\\ e^0 &= \cos(0)[1+0\cdot i]+\sin(0)[1-0]\\ 1 &= \cos(0) + \sin(0) \end{align} Now, I need to use $\sin(0)=0$ as a giving but the point is to not take these identities for granted.

For why I accept the derivative of $\cos$ to be $\sin$, the power series would be $$ \sin(x) = \sin(0)\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}+\cos(0)\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{(2n+1)!}\tag{2} $$ Then derivative of $(2)$ is $(1)$ after a change of variables.

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closed as unclear what you're asking by anomaly, Hans Lundmark, RE60K, Jonas Meyer, Micah Mar 6 '15 at 21:08

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    $\begingroup$ It's a definition. You can define cosine as the solution to $y''=-y$ satisfying $y(0)=1$. $\endgroup$ – symplectomorphic Mar 6 '15 at 18:03
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    $\begingroup$ How do you defined "algebraically" the cosine and the sine ? $\endgroup$ – user171326 Mar 6 '15 at 18:04
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    $\begingroup$ I think it would help if you told us what axioms you would feel comfortable using. You have to start with a definition of cosine somehow, otherwise you can't prove anything about it because it doesn't mean anything. $\endgroup$ – zeldredge Mar 6 '15 at 18:05
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    $\begingroup$ @dustin Then whats your definition of cosine? $\endgroup$ – Marm Mar 6 '15 at 18:06
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    $\begingroup$ That Taylor series is not the definition. It's just the expanion of an arbitary (presumably analytic) function $f$ at $0$, with the coefficient $f^{(n)}(0)$ replaced by $\pm \cos(0)$ and $\pm \sin(0)$. In other words, you're taking $\cos x$ to be the function with $y'(x) = -\sin x$ and $y''(x) = -\cos x$...except that that definition is obviously circular. If you want to take $\cos x$ to be the function with $y''(x) = -y(x)$, then it's still not going to be uniquely determined unless you set boundary conditions, such as specifying $y(0) = \cos 0$ and $y'(0) = \sin 0$. $\endgroup$ – anomaly Mar 6 '15 at 18:21
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If you define $\cos$ by $\cos x = \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}$, then $\cos 0 = 0^0 = 1$ (all other terms are $0$). If you define $0^0$ differently, you are not actally talking about the $\cos$ function. If you define $\cos x = \Re(e^{ix})$, then you get the same.

If you want to define $\cos$ via the differential equation:

$$\cos''(x) = -\cos(x)$$ then you need an initial value. Again, if you don't take $\cos(0)=1$, then you are not talking about the $\cos$ function. The choice is $\cos(0)=1$ is canonical, the only other canonical choice is mapping $0$ to $0$, but this is what $\sin$ already does.

Also, if $e^{ix} = \cos(x) + i\sin(x)$, then $1 + i0 = 1 = e^0 = e^{i0} = \cos(0) + i\sin(0)$, so by taking the real value: $\cos(0) = 1$.

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  • $\begingroup$ Start with a function called $\cos(x)$. We don't know its definition so we say hey let's determine the taylor series. We get $f(0) = \cos(0)$, $f'(0) = -\sin(0)$ and so on. The power series then has cosine and sine involved. Unless we know something about the values at zero, the Taylor series doesn't lose the values. $\endgroup$ – dustin Mar 6 '15 at 18:13
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    $\begingroup$ @dustin How? In what way do you get the taylor series? Derivatives? Then where did the definition of the derivative of cosine come from? proofwiki.org/wiki/Derivative_of_Sine_Function Each proof somehow relies on the value of at least one of cos(0) or sin(0), knowledge of the other's derivative, or from a richer definition of cosine than you are using. $\endgroup$ – JMoravitz Mar 6 '15 at 18:17
  • $\begingroup$ @JMoravitz see edit. No knowledge is assumed to determine they are derivatives. $\endgroup$ – dustin Mar 6 '15 at 18:23
  • $\begingroup$ I also stated not to take Euler's formula for granted. $\endgroup$ – dustin Mar 6 '15 at 18:29
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    $\begingroup$ @dustin You are mistaken. If you don't "know" $\cos(x)$, then how are you going to reason about it, derive it's taylor series for instance? Let me define the term "bleeeerg". No, I won't tell you what "bleeeerg" means, but as you can "obviously" see, "bleeeerg" is green, wears pants and likes eating candy. Doesn't make sense? Of course not! Now replace "bleeeerg" by $\cos(x)$ and "is green, wears pants and likes eating candy" by "has taylor expansion ..." and we are back to our initial situation. $\endgroup$ – Stefan Perko Mar 6 '15 at 20:16
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For me, cosine was defined analytically using the infinite sum $$\cos(x):=\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\ldots$$

from which you can see that $\cos(0)=1$

Which definition do you want to use?

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  • $\begingroup$ Derive that definition without assuming $\cos(0)=1$ and you have a bunch of $\cos(0)$ hanging around. $\endgroup$ – dustin Mar 6 '15 at 18:08
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    $\begingroup$ It's a definition, and therefore cannot be derived. We can define $\cos$ in this way, in which case it is apparent that $\cos(0) =1$, or we can define $\cos$ in terms of triangles, in which case it is also apparent that $\cos(0) = 1$. $\endgroup$ – gj255 Mar 6 '15 at 18:08
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    $\begingroup$ What do you mean 'derive that definition'? This is my definition. $\endgroup$ – Tyr Curtis Mar 6 '15 at 18:08
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    $\begingroup$ This agrees with the analysis definition of the cosine used here: proofwiki.org/wiki/Definition:Cosine $\endgroup$ – JMoravitz Mar 6 '15 at 18:19
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you know that $\lim_{x\rightarrow \infty}\frac{sinx}{x}=1$ geometrically. from here you know that $\lim_{x\rightarrow \infty}\sin x=0$. if you assume that the sine function is continuous, then you have that $sin(0)=0$, and from here it follows from $\sin^2(0)+\cos^2(0)=1$ that $cos(0)=\pm 1$. if you also assume that the cosine function is continuous, then $cos(0)$ couldn't be negative, because $\cos x\geq 0$ for $0\leq x\leq \frac{\pi}{2}$, which is defined geometrically.

so if $\cos(0)$ can't be negative, we've established that $\cos(0)=1$

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  • $\begingroup$ cos0=0????????? $\endgroup$ – RE60K Mar 6 '15 at 19:39
  • $\begingroup$ here, happy? ;) $\endgroup$ – tzoorp Mar 6 '15 at 20:15
  • $\begingroup$ You have write @name otherwise adg won't know you said anything. $\endgroup$ – dustin Mar 6 '15 at 21:14
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If we define cos and sin on the unit circle as x and y coordinates on the periphery.. then at length 0 along the circle ("angle" 0) we will be at coordinate (1,0) so cos will be 1 and sin will be 0.

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  • $\begingroup$ I want to show this algebraically not relying on geometry. $\endgroup$ – dustin Mar 6 '15 at 18:02
  • $\begingroup$ Ok so you are fine with trigonometrical one i.e. sum of squares is 1. Then the taylor expansion of sin(0) will be 0 so cos(0) must be either -1 or 1 according to that sum of squares = 1. $\endgroup$ – mathreadler Mar 6 '15 at 18:05
  • $\begingroup$ this is in no way wrong! $\endgroup$ – RE60K Mar 7 '15 at 6:46
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Anything you do would be useless, since all this is defined once and everything is based upon it. Even Euler's Formula and Taylor series use the definition that $\cos0=1,\sin0=0$. The following could be taken as a starting point while neglecting any geometrical consequence: $$y''=-y$$ And the solution will be of the form of: $$a\sin x+b\cos x$$ (where a,b are variable constants)

Suppose someone unknown about trigonometric functions wanted to define them, he would take two starting functions: $$\sin_2x=c\sin x+d\cos x\\\cos_2x=e\sin x+f\cos x$$ (where c,d,e and f are constants)

And then to him the solution of above equation would be: $$a_2\sin_2x+b_2\cos_2x$$

Now if did took $(c,d,e,f)\equiv(1,2,3,4)$, then $\sin_20$ would be $2$ and $\cos_20=4$.

Now the special solution we use is $(1,0,0,1)$ , because it follows a nice property: $$\sin^2x+\cos^2x=1$$ but his does: $$(\cos_2x-2\sin_2x)^2+\left(\frac{3\sin_2x-\cos_2x}2\right)^2=1$$ There are also other unmentioned advanted, one involving range too, since $|\sin x|,|\cos x|<1$ but $|\sin_2x|<\sqrt{5},|\cos_2x|<5$. There is surely an advantage with that solution if considering without any geometrical basis. When geometry comes into play, there's no choice left...

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  • $\begingroup$ Prior to having the knowledge we have today, mathematicians had to find a way to prove it to be true without saying I know this to be true. How was it done in this case, because unless I know something useful about $\cos(0)$ or $\sin(0)$, making the conclusion $\cos(0)=1$ doesn't seem to come so naturally. Once $\cos(0)=1$ is known the other will follow. $\endgroup$ – dustin Mar 6 '15 at 18:52
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    $\begingroup$ @dustin that follows from geometrical ease. Also you could have $\cos_2x=\sin x$ an $\sin_2x=\cos x$then $\cos_2(0)=0$ and all formulas are interchanged, it's just an easy definition which when defined like this become the functions related to triangles. $\endgroup$ – RE60K Mar 6 '15 at 19:05
  • $\begingroup$ I found a solution from ancient Greece. $\endgroup$ – dustin Mar 6 '15 at 19:06

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