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Solve $\displaystyle \cos(x-\alpha)\cos(x-\beta) = \cos{\alpha}\cos{\beta}+\sin^2{x}$.

My attempt:

$\displaystyle \cos(x-\alpha)\cos(x-\beta) = \cos{\alpha}\cos{\beta}+\sin^2{x} \Rightarrow \cos(x-\alpha)\cos(x-\beta)-\cos{\alpha}\cos{\beta} = \sin^2{x},$ and

$$\begin{aligned}LHS & = \frac{1}{2}\cos\left(2x-\alpha-\beta\right)+\frac{1}{2}\cos\left(\alpha-\beta\right)-\frac{1}{2}\cos\left(\alpha-\beta\right)-\frac{1}{2}\cos\left(\alpha+\beta\right)\\& = \frac{1}{2}\cos\left(2x-\alpha-\beta\right)-\frac{1}{2}\cos\left(\alpha+\beta\right) \\& = -\sin\left(\frac{2x-\alpha-\beta+\alpha+\beta}{2}\right)\sin\left(\frac{2x-\alpha-\beta-\alpha-\beta}{2}\right) \\& = -\sin{x}\sin\left(x-\alpha-\beta\right)\end{aligned}$$

Thus, we have:

$$\begin{aligned} & -\sin{x}\sin\left(x-\alpha -\beta\right) = \sin^2{x} \\& \Rightarrow \sin^2{x}+\sin{x}\sin\left(x-\alpha-\beta\right) = 0 \\& \Rightarrow \sin{x}\left(\sin{x}+\sin\left(x-\alpha-\beta\right)\right) = 0\end{aligned} $$

So either $\sin{x} = 0$ or $\sin{x} = \sin\left(\alpha+\beta-x\right)$. If $\sin{x} = 0$, then we have $\sin{x} = 0 \Rightarrow \sin{x}$ $= \sin\left(0\right)$ $\Rightarrow x = n\pi$ or $(2n+1)\pi$ for $n\in\mathbb{Z}$ -- we can write this as $k\pi$, where $k\in\mathbb{Z}$. If, on the other hand, $\sin{x} = \sin\left(\alpha+\beta-x\right)$ then $x = 2n\pi+\alpha+\beta-x \Rightarrow x = n\pi+\frac{1}{2}\left(\alpha+\beta\right)$, or $ x = 2n\pi+\alpha+\beta-x $ (EDIT: this was meant to be $x = (2n+1)\pi-\alpha-\beta+x$), which contains no solutions (is that the right way to put it?). Thus the solutions for the equation are $n\pi+\frac{1}{2}\left(\alpha+\beta\right)$ and $k\pi$, for any integers $k$ and $n$.

Question: The answer in the book has the condition $\alpha+\beta \ne (2m+1)\pi$ -- why is that?

Request: If you have the time, please critique the way my solution is presented (reasoning, use of notation, flow etc - I've an admission test in which this plays big part soon). Is my use of $\Rightarrow$ ok?

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The second term in the second line of your display is missing a $\cos$ before $\alpha-\beta$. –  Arturo Magidin Mar 12 '11 at 21:34
    
Fixed that, thanks. –  Lyrebird Mar 12 '11 at 21:39
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1 Answer

up vote 5 down vote accepted

One point: why are you going back from $x = n\pi+\frac{1}{2}(\alpha+\beta)$ to $x=2n\pi + \alpha+\beta-x$? And why do you say there are no solutions?

That said: you are incorrect in claiming that $\sin(x) = \sin(\alpha+\beta-x)$ implies that $x = 2n\pi + \alpha+\beta-x$. That is not the only ways in which two values of sine may be equal. After all, sine takes the same value at $0$ and at $\pi$, even though $0$ and $\pi$ don't differ by a multiple of $2\pi$. And it takes the value $\frac{1}{2}$ at both $\frac{\pi}{6}$ and at $\frac{5\pi}{6}$, even though the difference is not even an integer multiple of $\pi$. So that part of the derivation is incorrect.

Added/Edit. From the comments, it seems you meant to write the correct conditions: either $x$ and $\alpha+\beta-x$ differ by a multiple of $2n\pi$ (this is the condition you checked), or else they are "symmetric around $\pi/2$ or $3\pi/2$ up to a multiple of $2\pi$", which gives the condition $x = (2m+1)\pi - (\alpha+\beta-x)$.

Your error was discarding/completely ignoring the second condition. The second condition does not give an equation for $x$, but it does give an equation for $\alpha+\beta$! It says that you must have $0 = (2m+1)\pi - (\alpha+\beta)$, or that $\alpha+\beta=(2m+1)\pi$ must hold. In other word: if $\alpha+\beta = (2m+1)\pi$, then you will always have $\sin(x) = \sin(\alpha+\beta-x)$, regardless of what the value of $x$ is.

So, in summary: if $\alpha+\beta = (2m+1)\pi$ for some integer $m$, then every $x$ is a solution. And if $\alpha+\beta\neq (2m+1)\pi$, then your derivation is correct (once you fix the analysis of this second case).

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Oh, I didn't go back. I meant $(2n+1)-\alpha-\beta+x$ in the second case. We have $\sin{x} = \sin\left(\alpha+\beta-x\right)$ implies $x = 2n\pi+\alpha+\beta-x$ or $x = (2n+1)\pi-\alpha-\beta+x$. In the first case, we get $x = n\pi+\frac{1}{2}\left(\alpha+\beta\right)$ but in the second case the x on the RHS and the x on the LHS cancel, so it no solutions there. That was what I meant, but I didn't write. –  Lyrebird Mar 12 '11 at 22:11
    
@StudentOfMaths: Ehr... okay... still incorrect, though, because those are not the only cases in which two values of sine may agree. –  Arturo Magidin Mar 12 '11 at 22:14
    
@Arturo Magidin - Just one more question (then I'll be off to doing it in your way). I was taught (in the book) that if $\sin{\theta} = \sin{\alpha}$ then $x = 2n\pi+\alpha$ or $(2n+1)\pi-\alpha$. So apparently, that's not true? –  Lyrebird Mar 12 '11 at 22:19
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@Student of Math: Ehr... Sorry; yes, that works. When sine takes the same value in two different points on $[0,2\pi]$, the two points must be symmetrically placed around $\pi/2$ or around $3\pi/2$. The conditions reduce to $\alpha = \pi-\theta$. By the periodicity, you get $\alpha = (2n+1)\pi-\theta$ as one case; and of course $\alpha = 2n\pi + \theta$ is the other, which are your two conditions. Let me edit this a bit then. –  Arturo Magidin Mar 12 '11 at 22:25
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@StudentOfMath: I've fixed the answer; you are indeed fine with your derivation, except you should not have simply discarded that second case, you should have taken a closer look at what it implied. Sorry for the confusion I may have caused. –  Arturo Magidin Mar 12 '11 at 22:32
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