If we look at a unit circle we can see that the values of cosine are between -1 and 1 however is there a particular proof for this fact? I have tried using the Eulers identity to arrive at a proof with no luck. I have also tried using the expanded form of cosine : $cos(x)=\sum_{n=0}^{\infty } \frac{(-1)^n x^{2n}}{2n!}$. Using the expansion I tried to manipulate as much as I could however it seems that I am not finding a proof in this fashion either. I also tried going from the fact that it is a cauchy sequence.Any ideas?
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2If you define $f(x)=\sum (-1)^nx^{2n}/(2n)!$ you can prove $f(x)^2+f'(x)^2=1$. – Angina Seng Jul 12 '20 at 10:24
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2How are you defining $\cos(x)$ – MATHS MOD Jul 12 '20 at 11:16
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@MATHSMOD I am defining it as the sum of the terms posted above. In general the cosine function. – lambdaepsilon Jul 12 '20 at 13:23
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@AnginaSeng I realize that you can use the derivatives but I did not want to approach it in that fashion. This relies on the sine function and I did not wish to use it. – lambdaepsilon Jul 12 '20 at 13:36
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We can very simply prove this using the identity $\sin^2(x)+\cos^2(x)=1$ derived from Pythagoras and the unit circle. This is equivalent to $\cos^2(x)=1-\sin^2(x)$ and since $\sin(x)\in\mathbb{R},\forall{x}\in\mathbb{R}$, it follows that $\sin^2(x)\in\mathbb{R}^+,\forall{x}\in\mathbb{R}$. Our inequality $\cos^2(x)=1-\sin^2(x)$ then turns into $\cos(x)\leq1$, with equality being achieved when $\sin^2(x)$ is minimum (i.e. when $\sin^2(x)=0$ or $x=0$). Therefore this inequality sets the maximum bound of $\cos(x)$ to be 1.
JC12
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Yes but I wanted to avoid using the sine function as well . I wanted to prove the same for the sine function and should not like to begin from things on the unit circle. I came up with this proof of course but wanted to fins some other form – lambdaepsilon Jul 12 '20 at 13:30
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Further I want to find a proof that shows that $\forall x\in\mathbb{R}$ it holds that $cos(x)\in[-1,1]$ – lambdaepsilon Jul 12 '20 at 13:32
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1@lambdaepsilon, the proof has not relied on any characteristic of $\sin(x)$ other than the fact that it can obviously attain a value of $0$ when $x=0$ (which is trivial), creating the upper bound for $\cos(x)$. Furthermore, the only thing inferred is that since $\sin(x)\in\mathbb{R},\forall{x}\in\mathbb{R}$, it follows that $\sin^2(x)\in\mathbb{R}^+,\forall{x}\in\mathbb{R}$, which is true for any real number even if not mentioned in the sine function. Since we determined the maximum to be 1, the amplitude is 1 and thus the minimum of the function $\cos(x)$ is one. This proves your question. – JC12 Jul 12 '20 at 22:39
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1You may realise that we need to rely on the fact that $\cos(x)$ is periodic as a pre-assumed piece of information to prove that $\cos(x)\in[-1,1],\forall{x}\in\mathbb{R}$. If this assumed, then the function obviously has an amplitude and it is easy to prove. Proving the function's periodicity is not easy and requires some quite a lot of series: see https://math.stackexchange.com/questions/63102/how-to-prove-periodicity-of-sinx-or-cosx-starting-from-the-taylor-seri. – JC12 Jul 12 '20 at 22:41