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For example, $\sin(x)\cos(x)$ can be written as $\sin(2x)/2$, the limit as $x$ approaches $0$ of $\sin(x)\cos(x)$ is $0$, and the limit as x approaches $\pi/2$ is $0$. I don't see a reason why sine always increases/decreases faster than cosine. Why does sine overpower cosine near x=0, and why does cosine overpower sine near x=pi/2, I feel that if they did increase the same the limit should be somewhere in the middle, such as 1/2 for both of these values. If you try to take the derivative of each with respect to some other to see why one increases faster, you end up with the same problem because the derivatives just keep giving you negative versions of the original problem.

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$\lim_{x\to\pi/2}\sin(x)\cos(x)=0$ –  Damian Sobota May 17 '13 at 11:18
    
Oh I forgot it's sin(2x)/2, I was thinking sin(x)/2 –  Ovi May 17 '13 at 11:21
    
@DamianSobota but why does sine overpower cosine near x=0, and why does cosine overpower sine near x=oi/2? I feel like the limit should be somewhere in the middle, such as 1/2 –  Ovi May 17 '13 at 11:26
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It's a matter of the calculus of limits. If two functions have finite limits at some point, then their multiplication has a limit which is the multiplication of the limits. –  Damian Sobota May 17 '13 at 11:36

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the limit of $\sin(x)\cdot \cos(x)$ as $x\rightarrow \pi /2$ is $0$. So in both cases you have $\sin(x)\cdot \cos(x)=0$. Infact they do increase/decrease with the same rate. Infact $\sin(\pi/2-x)=\cos(x)$ if you vary $x$ they have the same values so they increase/decrease with the same rate.

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Well ok I got the limit wrong but why does sine overpower cosine near x=0, and why does cosine overpower sine near x=oi/2? I feel like the limit should be somewhere in the middle haha. –  Ovi May 17 '13 at 11:23
    
@Ovi what do you mean with "overpower"? please write an understandable question in correct mathematical symbols and we can try to explain. –  Amire May 17 '13 at 11:26
    
By overpower I mean the limit. sin(0) is 0, and the limit as x approaches 0 of sin(x)cos(x) is 0, so sin overpowers cosine here. Cos(pi/2)=0, and the limit as x approaches pi/2 of sin(x)cos(x)=0. So if sine and cosine increase/decrease at the same rate, why does one overpower the other in certain cases? If the increase/decrease at the same rate, why isn't the limit somewhere in the middle, such as 1/2? –  Ovi May 17 '13 at 11:31
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You are making a serious mistake here: one way to understand the increasing/decreasing rate of two functions $f(x),g(x)$ is to look at their ration $\frac { f(x)}{ g(x)}$ analyzing their product , in this case will give you no relevant informations. for example as i stated before $\sin(\pi/2-x)=\cos(x) \Leftrightarrow \frac { \pi/2-x}{ cos(x)}=1$ so infact they increase/decrease with the same rate because their ration is constant. –  Amire May 17 '13 at 11:34

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