4
$\begingroup$

I'm trying to solve the following problem:

The graphs for $y = \sin x$ and $y = \cos x$ has two points of intersection in the interval $[-\pi, \pi]$. Determine the equation for the line that passes through these two points.

What I've done (and failed):

A point of intersection is when $\sin x = \cos x$. By drawing the unit circle, I can easily see that the two points when the x-value (cos) is the same as the y-value (sin) is at $x = \frac{\pi}{4}$ and $x = \frac{-3\pi}{4}$. Those points are $\left(\frac{\sqrt 2}{2}, \frac{\sqrt 2}{2}\right)$ and $\left(-\frac{\sqrt 2}{2}, -\frac{\sqrt 2}{2}\right)$.

After that, I tried calculating the slope of the line (y1-y2 / x1-x2) and then plug in the x/y value of a point and solve for b. In the end, I ended up with $y = 1 \cdot x + 0$ or $y = x$ which is... wrong.

I don't know what to do.

$\endgroup$
2
  • 1
    $\begingroup$ $5\pi/4$ is not in the given interval. You want your second point to lie in the interval $[-\pi,0]$. $\endgroup$ Sep 17, 2012 at 16:37
  • 1
    $\begingroup$ Note that the points you should have used to calculate the slope of the (incorrect) line are $(\pi/4,\sqrt2/2)$ and $(5\pi/4,-\sqrt2/2)$. This gives a slope of $-\sqrt2/\pi$, not $1$, as you have. $\endgroup$ Sep 17, 2012 at 16:45

2 Answers 2

2
$\begingroup$

The points $\left(\frac{\sqrt 2}{2}, \frac{\sqrt 2}{2}\right)$ and $\left(-\frac{\sqrt 2}{2}, -\frac{\sqrt 2}{2}\right)$ are not on the graphs of those two functions. To avoid confusion, let's say the equation of the circle is $x^2+y^2=1$ and then use a different letter, $\theta$, for the angle, and yet another letter, $w$, for the values of the trigonometric function, so that instead of writing $y=\cos x$ and $y=\sin x$, we'll write $w=\cos\theta$ and $w=\sin\theta$. The graphs of the sine and cosine functions are in the $(\theta,w)$ plane, not in the $(x,y)$ plane. So figure out what $\theta$ is, and what $w$ is, at the points of intersection. Then you'll have two points, and you can find the equation of the line that passes through them.

$\endgroup$
1
$\begingroup$

The question specified that the values of $x$ are in the interval $[-\pi,\pi]$.

The two values of $x$ in this interval are $\pi/4$ and $-3\pi/4$.

Now the rest should go well.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .