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I was asked to find the points of intersection between $y=(\sin x)^2$ and $y=\frac{2x}{\pi} $ but I don't know how to go about it. Obviously they intersect at the origin. I also found out the other intersection point is $x=\frac{\pi}4$ (thanks, calculator) but calculators are not allowed during my exams so I'm afraid I won't be able to solve problems like this if I'm unlucky enough to have one in the paper. Any help would be greatly appreciated. Thanks in advance!

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    $\begingroup$ There must be a typo because $\sin(x)^2 \le 1.$ $\endgroup$ Commented Oct 18, 2018 at 13:32
  • $\begingroup$ $\sin^2 x \leq 1$, but $2\frac{\pi}{2} > 1$... The second function must’ve been something else. $\endgroup$
    – KM101
    Commented Oct 18, 2018 at 13:35
  • $\begingroup$ It's $sin^2 x$ and not $sin (x)^2$. Sorry for the confusion. $\endgroup$
    – user208872
    Commented Oct 18, 2018 at 13:36
  • $\begingroup$ What do you mean with $\sin^2 x?$ This usually is $\sin(x)^2$ $\endgroup$ Commented Oct 18, 2018 at 13:37
  • $\begingroup$ There were a few typos but I edited my question so I hope it's clear now. $\endgroup$
    – user208872
    Commented Oct 18, 2018 at 13:43

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$$ y_1(x) = \sin^2 x \\ y_2(x) = \frac{2}{\pi}x $$ The first of these looks sinusoidal (but above the $x$-axis); the second has a graph that's a straight line. By drawing a graph, you can work out that they intersect at just a few points, for if $x > \pi/2$, then $y_2(x) > 1$, so it cannot be the square of the sine of anything, for sines lie between $-1$ and $1$.

As you noticed, the graphs meet at $x = 0$. For $x < 0$, we have $y_1(x) \ge 0$ but $y_2(x) < 0$, so there can be no negative solution (thanks, @Vasya!).

Now how do you find the positive solutions? The answer is disappointing, I'm afraid. You mostly use guesswork unless you know about considerably more sophisticated techniques. So (if you're me) you try out some values of $x$ for which you know the sine in your head. For instance, you try out $x = \pi/6$, and get a sine of $1/2$, so $y_1(\pi/6) = 1/4$. On the other hand, $y_2(\pi/6) = 1/3$. We're still in the region where the straight-line graph is above the sort of parabola-shaped central "bowl" of the $\sin^2$ graph.

Then you try out $\pi/4$, and you get lucky.

Then there's still something subtle: the $y_2$ graph crosses the $y_1$ graph at $x = \pi/4$, but it has to cross it again a little later. Let's look at $\pi/3$:

$$ y_1(\pi/3) = 3/4 \\ y_2(\pi/3) = 2/3 $$ Hmmm. So $y_2$ is still less than $y_1$ here. Let's go further to the right, and try $\pi/2$: $$ y_1(\pi/2) = 1 \\ y_2(\pi/2) = 1 $$

Aha! That's the second point where they cross. So the solutions are $$ x = 0, \frac{\pi}{4}, \frac{\pi}{2}. $$

But if the problem were just ever so slightly different --- if $y_2$ had been $y_2(x) = 3x/\pi$, for instance --- there'd have been no easy way to solve it, and "calculator methods" would really have been your only choice to get an approximate answer, at least with what you probably know at this point in your mathematics education.

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  • $\begingroup$ surely $x$ can't be negative $\endgroup$
    – Vasili
    Commented Oct 18, 2018 at 14:03
  • $\begingroup$ Completely right...I'll edit. I was thinking of "sine" when I wrote that stuff about oddness. D'oh! $\endgroup$ Commented Oct 18, 2018 at 14:05
  • $\begingroup$ Thanks a bunch @John Hughes!!! $\endgroup$
    – user208872
    Commented Oct 18, 2018 at 16:58

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