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Find the number of real solutions for $|y|=\sin x$ and $y=\cos^{-1}(\cos x)$, for $x\in [-2\pi,2\pi]$

Now the easiest way to solve this is by sketching the graphs for both functions enter image description here

I used a calculator for this.

Clearly there are three solutions, because the lines are tangential for the curves, but this isn’t exactly obvious when sketching them free hand.

There is no way of knowing whether graphs are tangential to each other just by looking at. You need a fair bit of algebra, which is particularly difficult to do in this question. What is the best of sketching the graphs?

Aditya
  • 6,191

1 Answers1

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You only need to know a few basic facts and the shape of the sine curve to graph this freehand.

  • The definition of $\cos^{-1}$ tells you that on $[0,\pi], \cos^{-1}\cos x = x$.
  • Cosine is even. Therefore on $[-\pi, 0], \cos^{-1}\cos x = \cos^{-1}\cos |x| = |x|$
  • The rest of its graph follows from periodicity.
  • Since $|y| = \sin x$ can have no solution when $\sin x < 0$, the graph only exists on $[0,\pi]$ and its periodic translations.
  • On $[0,\pi]$ it is obviously the upper lobe of $y = \sin x$, and its mirror image though the $x$-axis.

The solutions at $-2\pi, 0, 2\pi$ are obvious. To be sure these are the only solutions, one only needs one more well-known inequality: $$\forall x > 0, \sin x < x$$

Paul Sinclair
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