How to prove that cosine function is decreasing on the interval $(0,\pi)$ if we are allowed to use only the definition of cosine through exponential function (or Taylor series)?
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3Well, one definition of $\pi$ is the lowest real number greater than 0, for which $\sin{x} = 0$, and with that it should be pretty easy to prove. – Lieven Nov 22 '12 at 22:11
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What are you allowed to use? How do you define $\pi$? – levap Nov 22 '12 at 22:18
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You can use the definition of decreasing function directly and try to bound $\frac{cos(y)}{cos(x)}$ by $1$ for $x \leq y$. Express $y$ as $x + \Delta$ and use the definition of cosine through exponential function.
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You need to show that the terms of the series are decreasing, but to do that you need to take the derivative of Taylor series. This might also help.
glebovg
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If so, you could approximate $\sin(x)$ by the first few terms of Taylor series and plug in some values, but that would be an informal proof. – glebovg Nov 22 '12 at 22:17
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It is enough to show that for coplex numbers $z<t\in(0,\pi) cost<cosz$ am I right? – Nasibabuba Nov 22 '12 at 22:21
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The individual terms of the Taylor series are not all decreasing on $[0,\pi]$. – Robert Israel Nov 22 '12 at 22:22
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The teacher might be trying to get across the idea that "decreasing" means f(y)<=f(x) for x <= y. That's a hard one for students to grasp once they get the notion that it means the derivative is negative. – Adam Cross Nov 22 '12 at 22:31
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This may not be the answer you are looking for but I thought it might be worthy enough to be posted.
As $\theta$ increases from $0$ to $\pi$, from the figure, $x$ decreases from $r$ to $-r$. Since $\cos(\theta) = \dfrac{x}r$ and $r$ is fixed $\cos(\theta)$ decreases as $x$ decreases. Hence, $\cos(\theta)$ decreases from $1$ to $-1$ as $\theta$ increases from $0$ to $\pi$.