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I have no idea how to do this one, they intercept at some random point that I can not calculate.

$y= \cos x$, $y=\sin 2x$, $x= 0$ and $x= \pi/2$

I know the graph will go from $0$ to $0$ for $\sin 2x$ and then $1$ to $0$ for $\cos x$ so there are two areas I have to compute but I have no idea how to figure out the bounds on them.

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Which area are you after? Can you clarify your question? Check the graph –  user13655 Apr 28 '12 at 22:35
2  
use $\sin(2x) = 2\cos(x)\sin(x)$ to help you find that point. –  user29743 Apr 28 '12 at 22:35
    
@user13655: It's clear to me the question wants the absolute area between the two curves between $x=0$ and $=\pi/2$. –  anon Apr 28 '12 at 22:37
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1 Answer 1

up vote 2 down vote accepted

Rewriting $$\sin 2x = \sin x \cos x + \cos x \sin x = 2\sin x\cos x$$ we can compute the intersection: $\cos x = \sin(2x)$ is the same as $$\begin{align*} \cos x&= 2\sin x\cos x\\ \cos x - 2\sin x\cos x &= 0\\ \cos x(1 - 2\sin x) &= 0. \end{align*}$$ The product is zero if and only if $\cos x = 0$ (which on $[0,\pi/2]$ occurs only at $x=\pi/2$), or if $1-2\sin x = 0$, which is the same as $2\sin x = 1$, which is the same as $\sin x = \frac{1}{2}$; on $[0,\pi/2]$, this happens once and only once: at $x=\pi/6$.

So the point of intersection is at $x=\pi/6$.

On $[0,\pi/6]$, we have that $\cos(x)$ is greater than $\sin(2x)$. On $[\pi/6,\pi/2]$, we have that $\sin(2x)$ is greater than $\cos x$. So the area is given by $$\begin{align*} \text{Area} &= \int_0^{\pi/2}|\cos x-\sin(2x)|\,dx\\ &= \int_0^{\pi/6}|\cos x - \sin(2x)|\,dx + \int_{\pi/6}^{\pi/2} |\cos x - \sin(2x)|\,dx\\ &= \int_0^{\pi/6}(\cos x - \sin (2x))\,dx + \int_{\pi/6}^{\pi/2}(\sin(2x) - \cos x)\,dx. \end{align*}$$ Now you can simply compute the integrals and add up the appropriate quantities.

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I am getting 2 as the answer but it is not the correct answer and I am not sure why. –  user138246 Apr 28 '12 at 22:56
    
Neither am I, since you don't say how you got the answer. As I tell my students, the Government doesn't let me read minds without a warrant. –  Arturo Magidin Apr 28 '12 at 23:02
    
cosx - sin2x for pi/6 - 0 = -1 and then sin2x - cosx for pi/2 - pi/6 = 0 –  user138246 Apr 28 '12 at 23:09
    
Wrong, and wrong. You can't get a negative number for the first integral: you are integrating a positive function. You can't get $0$ for the second integral: You are integrating a positive function. And $-1$ and $0$ don't add up to $2$, so how could you possibly be getting 2 as the answer? Of course, since you continue to only give answers and not explain your process, I cannot tell you why you are wrong, just that you are definitely wrong. –  Arturo Magidin Apr 28 '12 at 23:23
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And since you don't say what you did, nobody will be able to tell you what you did wrong. I'm sure it's Stewart's fault, too. If I had to bet, I probably would put my money on you getting the integral of $\sin(2x)$ wrong. –  Arturo Magidin Apr 28 '12 at 23:45
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