I'm sorry to ask this question , but should ask it may help me to know more

about series theory , It is well known that $\cos x $ and $\sin x$ are

represented by alternative series which hard for me to show wether sinx and

cosx are in this range $[-1,1]$ .

My question here :Is there analytical proof show that $\sin x$ and $\cos x$ both are really in $[-1,1]$ using series theory if it is possible

Note: I do not want to use geometric interpretation because it is standard at all

Thank you for any help

  • 2
    $\begingroup$ Quick note, you mean range not domain. $\endgroup$ – user223391 Aug 9 '15 at 23:03
  • $\begingroup$ yes , i meant range $\endgroup$ – zeraoulia rafik Aug 9 '15 at 23:05
  • 1
    $\begingroup$ You can prove that $\sin^2x+\cos^2x=1$ just from the series — multiply the sine and cosine series by themselves and add. This implies that $-1\le\sin x\le 1$ (why?). $\endgroup$ – Akiva Weinberger Aug 9 '15 at 23:37
  • $\begingroup$ I've not scrolled down but immediately I'd look at calculus. Finding a minimum and maximum will work. $\endgroup$ – Alec Teal Aug 9 '15 at 23:59
  • $\begingroup$ it is showed in the answer of themaker, f(0)=1 ?,?,,,, $\endgroup$ – zeraoulia rafik Aug 10 '15 at 0:00

From the series is easy to prove that (using the fact that power series can be differentiated term by term inside the disk of convergence) $$\frac{d}{dx} \cos(x) = -\sin(x)$$ $$\frac{d}{dx} \sin(x) = \cos(x)$$ If we define $$F(x) := \cos^2(x) + \sin^2(x)$$ then $F(0) = 1$ and $$\frac{d}{dx}F(x) = 2\sin(x)\cos(x) - 2\cos(x)\sin(x)= 0\ \ \forall x\in \mathbb R$$ hence $F$ is constant $$F(x) = \cos^2(x) + \sin^2(x) = 1 \ \ \forall x\in \mathbb R$$ From this is clear that $$|\cos(x)|\leq 1 \ \ \forall x\in \mathbb R$$ $$|\sin(x)|\leq 1 \ \ \forall x\in \mathbb R$$

  • $\begingroup$ I love this answer. somehow never realized the derivative of sin^2+cos^2 is 0. $\endgroup$ – MichaelChirico Aug 10 '15 at 1:47

Using series, you can prove (by defining $\exp$, $\sin$ and $\cos$ as series) that

$$\forall x\in\Bbb C,e^{ix} = \cos x + i \sin x$$

$$\forall x \in \Bbb R, |e^{ix}|=1$$

And from there, it's easy to see that

$$|\sin x|\le |e^{ix}|=1$$

$$|\cos x|\le |e^{ix}|=1$$

  • $\begingroup$ I think sinx and cosx are both represented by alternative series and only you are used Euler formula . $\endgroup$ – zeraoulia rafik Aug 9 '15 at 23:29
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    $\begingroup$ @zeraouliarafik Euler's formula is PROVEN using series. $\endgroup$ – user223391 Aug 9 '15 at 23:30
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    $\begingroup$ Given that Euler's formula is proven by series, it's not immediately clear why $|e^{ix}|=1$, as this fact requires $\cos(x)^2+\sin(x)^2=1$ as in themaker's answer. It's not hard to show this with some more work but, without it this does feel like cheating. $\endgroup$ – Alex R. Aug 9 '15 at 23:38
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    $\begingroup$ How did you conclude that $|e^{ix}|=1$? That's trivial if you know already that $\cos x$ and $\sin x$ are real and $(\cos x)^2+(\sin x)^2=1$, but how would you know that if you just start with the two series? Whatever your answer to that question, it should be explained above. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 10 '15 at 0:01
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    $\begingroup$ @xavierm02 : I think you should say explicitly in your answer you you conclude that. What happens when you take Cauchy products doesn't seem self-evident. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 10 '15 at 16:49

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