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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 Euler's 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. I know that identities can be brought in to prove it, but I should like to find a more fundamental way of proving it. Any ideas?

J. W. Tanner
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lambdaepsilon
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    $e^{ix}e^{-ix}=\cos^2x+\sin^2x=1$ implies $\cos^2x\le1$, which implies $-1\le \cos x\le1$. – J. W. Tanner Jul 12 '20 at 13:56
  • this was what I was looking for. I was trying to prove as rigidly as possible from the exponential and sums, I proved eulers identity and wanted to expand from there into bounds however I did not think to apply this. Thanks again. – lambdaepsilon Jul 12 '20 at 14:04
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    The title says without using identities but the body says tried using the Euler's identity !? – J. W. Tanner Jul 12 '20 at 14:34
  • As seen by comments there, OP was not satisfied with the answer there – J. W. Tanner Jul 12 '20 at 14:39
  • I meant that I did not wish to start with the more elemental identies given that alot of them require assumptions relying on this fact. – lambdaepsilon Jul 12 '20 at 18:18
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    @lambdaepsilon: Why you qualify as best answer one basic identities used if you are asking unused? Do not be inconsequential. In the cosine series, for values close to zero, the term $\dfrac{-x^2}{2}$ dominates and that determines that $cos (x)$ is less than 1. I trust that it was not you who put downvote to my answer. Downvotes should be prohibited without an explanation. – Ataulfo Jul 13 '20 at 12:51
  • @Piquito I found your answer to be satisfactory but not exactly what I was looking for. On an objective measure your proof was good however in the fashion that I was trying to prove I found that the other proof was a bit more suited to my subjective taste. Further It was not I that downvoted I actually upvoted your answer given that it was a sufficient proof for the problem. – lambdaepsilon Jul 13 '20 at 22:35
  • @lambdaepsilon:- No problem, Thanks for your reply. – Ataulfo Jul 14 '20 at 23:06

2 Answers2

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Let us write the power series for the cosine as $c(x)$. Since the radius of convergence is infinite we can differentiate term by term. The derivative is $-s(x)$, where $s(x)$ denotes the series for the sine. And $s'(x)=c(x)$.

From $s(x)c(x)-c(x)s(x)=0$ we get $$(c^2(x)+s^2(x))'=0,$$ so it's constant. Evaluation at zero gives $$c^2(x)+s^2(x)=1.$$ In particular $c^2(x)\leq1$, so $|c(x)|\leq1$.

Martin Argerami
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$\cos^2x+\sin^2x=(\cos x+i\sin x)(\cos x -i\sin x)=(\cos x+i\sin x)(\cos (-x )+ i \sin (-x))$

$=e^{ix}e^{-ix}=e^0=1.$

Since $\cos x$ and $\sin x$ are real numbers, their squares are non-negative.

Therefore, $\cos^2x\le1$, which implies $-1\le \cos x\le1$.

J. W. Tanner
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