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Question:

If $\sin(\pi\cos\theta)=\cos(\pi\sin\theta)$, then show that $\theta=\pm\frac{1}{2}\sin^{-1}\frac{3}{4}$.

My book's solution:

$$\sin(\pi\cos\theta)=\cos(\pi\sin\theta)$$

$$\sin(\pi\cos\theta)=\sin(\frac{\pi}{2}\pm\pi\sin\theta)\ [\text{Formula:}\cos\theta=\sin(\frac{\pi}{2}\pm\theta)]$$

$$\sin^{-1}(\sin(\pi\cos\theta))=\sin^{-1}(\sin(\frac{\pi}{2}\pm\pi\sin\theta))...(i)$$

$$\pi\cos\theta=\frac{\pi}{2}\pm\pi\sin\theta...(ii)$$

$$\cos\theta=\frac{1}{2}\pm\sin\theta$$

$$\cos\theta\pm\sin\theta=\frac{1}{2}$$

$$\cos^{2}\theta\pm2\sin\theta\cos\theta+\sin^{2}\theta=\frac{1}{4}$$

$$1\pm\sin2\theta=\frac{1}{4}$$

$$\sin2\theta=\pm\frac{3}{4}$$

$$\sin^{-1}(\sin2\theta)=\sin^{-1}(\pm\frac{3}{4})$$

$$2\theta=\pm\sin^{-1}(\frac{3}{4})$$

$$\theta=\pm\frac{1}{2}\sin^{-1}(\frac{3}{4})\ \text{(showed)}$$

This solution is good and all, but if we input the value of $\theta$ in line (i) we will see something interesting:-

$$\sin^{-1}(\sin(\pi\cos\theta))=\sin^{-1}(\sin(\frac{\pi}{2}\pm\pi\sin\theta))$$

$$[\text{Let's input $\theta=\frac{1}{2}\sin^{-1}(\frac{3}{4})$}]$$

$$\sin^{-1}(\sin(164.058809^{\circ}))=\sin^{-1}(\sin(164.058809^{\circ})$$

$$164.058809^{\circ}=164.058809^{\circ}$$

This is what my book did essentially. However, isn't $164.058809^{\circ}$ outside the restricted range of $\sin^{-1}(x)$: $[\frac{\pi}{2},-\frac{\pi}{2}]$? So, is the line (ii) in the solution of the book valid?

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    $\begingroup$ +1 to your question, not merely because I agree with you, but because of your work shown. In fact, my reaction is the same as yours. The range of the arcsin function is $-\pi/2 \leq \theta \leq \pi/2.$ This is clearly inappropriate when identifying all satisfying values of $\theta$ within a modulus of $(2\pi)$. This is regardless of whether you are considering the range of angles to be $-\pi < \theta \leq \pi$, or $-180^{\circ} < \theta \leq 180^\circ$ or $0^\circ \leq \theta < 360^\circ.$ ...see next comment $\endgroup$ Sep 24 '21 at 6:09
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    $\begingroup$ It may be that I am overlooking something in the author's analysis that prevents his use of the arcsin function from being relevant. Even if that is true, I personally regard it as irrelevant. The student should not have to reverse-engineer the analysis to determine that the invalidity of a certain step is harmless. $\endgroup$ Sep 24 '21 at 6:12
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    $\begingroup$ @user2661923 "The student should not have to reverse-engineer the analysis to determine that the invalidity of a certain step is harmless." PRECISELY. $\endgroup$
    – ryang
    Sep 24 '21 at 6:34
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    $\begingroup$ To see why it's wrong: it would be perfectly valid to replace the $\pm$ with just a $+$. ($\cos(\theta) = \sin(\frac{\pi}{2}+\theta)$ is a valid identity), and therefore arrive at the conclusion (following the author's reasoning) that $\theta=\frac12\sin^{-1}\frac34$ (without the $\pm$). But this would be an incorrect conclusion, since $\theta=-\frac12\sin^{-1}\frac34$ is also a correct solution. That means that the reasoning is incorrect. $\endgroup$ Sep 24 '21 at 13:37
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  1. Even though the spirit of the exercise is really asking you to solve the given equation (in other words, for the final line to be equivalent to the first line), it technically presents an if-then statement, so let's not quibble about the author squaring both sides of the equation without justification, thereby potentially creating extraneous solutions. Let's also not quibble about them needlessly applying $\cos\theta=\sin\left(\frac{\pi}{2}\pm\theta\right)$ instead of simply $\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right).$

    However, your book is blithely discarding solutions in the the fourth and eleventh lines, so those steps are invalid; for example

    • $$\arcsin(\sin(\pi\cos\theta)) \not\equiv \pi\cos\theta$$

      (counterexample: $\frac\pi6)$

    and

    • $$\arcsin(\sin2\theta) \not\equiv 2\theta$$

      (counterexample: $\frac\pi3).$

  2. As for your main question: the first three lines of the given solution are equivalent to one another, and $\arcsin\left(\sin\left(\pi\cos\theta\right)\right)$ accepts all real values of $\theta.$


This is the general solution: $$\sin(\pi\cos\theta)=\cos(\pi\sin\theta)\\ \cos\left(\frac\pi2-\pi\cos\theta\right)=\cos(\pi\sin\theta)\\ \frac\pi2-\pi\cos\theta=2n\pi\pm\pi\sin\theta\\ \cos\theta\pm\sin\theta=\frac12-2n\\ \sqrt2\cos\left(\theta\mp\frac\pi4\right)=\frac12-2n\\ \cos\left(\theta\mp\frac\pi4\right)=\frac1{2\sqrt2}-\sqrt2n\\ =\frac1{2\sqrt2}\\ \theta\mp\frac\pi4=2k\pi\pm\arccos\left(\frac1{2\sqrt2}\right)\\ \theta=\left(8k\pm1\right)\frac\pi4\pm\arccos\left(\frac1{2\sqrt2}\right).$$ N.B. The two $\pm$ signs above are independent and not to be combined; in other words, there are four—not two—independent general solutions.

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    $\begingroup$ As an engineer, I approve of "$\pi=\frac{\pi}{6}$" and "$\pi=\frac{\pi}{3}$" (the counterexamples). $\endgroup$
    – JamesA
    Sep 24 '21 at 7:49
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    $\begingroup$ @tryingtobeastoic You're asking why line (i) is true for the particular solution $\frac{1}{2}\arcsin\left(\frac{3}{4}\right).$ And my answer explains why line (i) is (and is expected to be) true for all solutions of the given equation, even solutions that exceed $100000^\circ.$ $\endgroup$
    – ryang
    Sep 24 '21 at 12:40
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    $\begingroup$ @tryingtobeastoic When you say, "Look at what my book did in (i)," I compare line (i) to the line before and see that it is a correct inference. But what they did on the line after (i) is completely unjustified, as shown in the middle part of this answer. In other words, we all seem to agree that line (ii) is wrong. $\endgroup$
    – David K
    Sep 24 '21 at 13:18
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    $\begingroup$ @tryingtobeastoic Taking the $\arcsin$ on $\sin Q$ was a valid move since $\arcsin$ is a bijection (one-to-one function), and since $\sin Q$'s codomain $[-1,1]$ falls within $\arcsin$'s domain. $\endgroup$
    – ryang
    Sep 24 '21 at 13:30
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    $\begingroup$ @DavidK $\sin(\frac{\pi}{2}+\theta)=\sin(\frac{\pi}{2}-\theta)$ for all $\theta;$ so, writing $$\cos\theta=\sin(\frac{\pi}{2}\pm\theta),$$ while needless, is actually equivalent to simply writing either $\sin(\frac{\pi}{2}-\theta)$ or $\sin(\frac{\pi}{2}+\theta).$ $\endgroup$
    – ryang
    Sep 24 '21 at 14:07
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The reasoning shown in the book is incorrect. Indeed, were it correct, then we could make the following deduction.

$$\sin(\pi\cos\theta)=\cos(\pi\sin\theta)$$

$$\sin(\pi\cos\theta)=\sin(\frac{\pi}{2}+\pi\sin\theta)$$ This is exactly the same as your step (i), but with a $+$, rather than a $\pm$. That's perfectly fine, since $\cos\theta=\sin(\frac{\pi}{2}+\theta)$ is a correct identity for any $\theta$.

$$\sin^{-1}(\sin(\pi\cos\theta))=\sin^{-1}(\sin(\frac{\pi}{2}+\pi\sin\theta))$$

$$\pi\cos\theta=\frac{\pi}{2}+\pi\sin\theta$$

This is the dodgy step. In your example, it works out for the author, but now the statement is genuinely incorrect.

$$\cos\theta=\frac{1}{2}+\sin\theta$$

$$\cos\theta-\sin\theta=\frac{1}{2}$$

$$\cos^{2}\theta-2\sin\theta\cos\theta+\sin^{2}\theta=\frac{1}{4}$$

$$1-\sin2\theta=\frac{1}{4}$$

$$\sin2\theta=\frac{3}{4}$$

$$\sin^{-1}(\sin2\theta)=\sin^{-1}(\frac{3}{4})$$

(assuming that $\theta$ lies in $[-\pi,\pi]$)

$$2\theta=\sin^{-1}(\frac{3}{4})$$

$$\theta=\frac{1}{2}\sin^{-1}(\frac{3}{4})\ \Box$$

But this is now incorrect, since $\theta = -\frac12\sin^{-1}\frac34$ is also a solution.

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  • $\begingroup$ There are also two other solutions within the interval $[-\pi,\pi]$. Even in the book (with the $\pm$ symbol) the dodgy step is incorrect. $\endgroup$
    – David K
    Sep 24 '21 at 17:36
  • $\begingroup$ @JohnGowers Sir, why do you assume that $\theta$ lies in $[-\pi,\pi]$? $\endgroup$ yesterday

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