# $x \sin x=2$ why is my proof that there no solutions wrong?

$\frac 12 x \sin x=1$ . Let's look at a right triangle with base $x$ and altitude $\sin x$ . Then our equation is for the area of this triangle. Let the sides of the triangle be $a=x$ , $b=\sqrt {x^2+sin^2 x}$ , and $c= \sin x$ . According to wikipedia, Heron's formula can be written as $$A=\large \frac { \sqrt {4a^2c^2-(a^2+b^2-c^2)^2}}{4}$$

Plugging in:

$4=\large \sqrt{4x^2 \sin^2 x-(x^2+x^2+\sin^2 x-\sin ^2 x)^2}$

$4=x^2 \sin^2 x -x^4$

$x^2(x^2- \sin^2 x)=-4$

$x^2$ will always be positive, and $\sin^2 x$ is never greater than $x^2$ , so this equation can have no real solutions. The original has solutions, so why is this wrong?

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an application of the intermediate value theorem to say $x=0$ and $x=12\pi +\pi/2$ shows there must be a solution. – James Jul 12 '13 at 23:33
Yea I know wolframalpha says there are infinitely many – Ovi Jul 12 '13 at 23:34
draw a graph of $y = x \sin x,$ say for $0 \leq x \leq 4 \pi.$ – Will Jagy Jul 12 '13 at 23:36
I think this particular wikipedia formula is wrong, probably it is $a^2-b^2+c^2$ correctly if starts with $4a^2c^2$. Try it out with arbitrary right triangle. There's negative under the square root in your case!! – Berci Jul 12 '13 at 23:47
I've updated the formula on Wikipedia, after verifying in Mathematica that Berci's comment is correct. – Rahul Jul 12 '13 at 23:55

## 2 Answers

This particular wikipedia formula is wrong.

It should be correctly either $$A=\large \frac { \sqrt {4a^2b^2-(a^2+b^2-c^2)^2}}{4}$$ or $$A=\large \frac { \sqrt {4a^2c^2-(a^2-b^2+c^2)^2}}{4}\,.$$ Mind the symmetry..

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The formula on Wikipedia is correct, but the OP copied it down incorrectly. – wj32 Jul 12 '13 at 23:56
@wj32 I think I copied it down correctly but Rahul Narain said he just updated it on wikipedia. – Ovi Jul 12 '13 at 23:58
Well, my browser (still) shows the cited version of the formula. However, Rahul Narain commented that he already corrected it in wiki.. – Berci Jul 12 '13 at 23:58
Yeah, Wikipedia was updated today, no doubt as a result of this problem. – Thomas Andrews Jul 12 '13 at 23:58
@Ovi: With $a=x$, $b=\sin x$, $c=\sqrt{x^2+\sin^2x}$, the formula just gives you back $\frac12x\sin x$. Not surprising, since we already know that's the area of the triangle... – Rahul Jul 13 '13 at 0:00

There are solutions on each interval $\left[2k\pi,2k\pi+\frac\pi2\right]$ for positive integer $k$ by the intermediate value theorem because $x\sin(x)$ is $0$ on the left end and $2k\pi+\frac\pi2$ on the right.

Heron's Formula should be $$A=\frac{\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}}{4}$$ Does that cause the same problem?

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Probably, but I will have time later to check – Ovi Jul 13 '13 at 0:01