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The problem:

$$\sin\theta+\cos\theta=\sqrt{2\sin2\theta}$$

My solution:

$$\sin\theta+\cos\theta=\sqrt{2\sin2\theta}$$

$$\sin\theta+\cos\theta=\sqrt{2\times2\sin\theta\cos\theta}$$

$$\sin\theta+\cos\theta=2\sqrt{\sin\theta\cos\theta}$$

$$\sin\theta-2\sqrt{\sin\theta\cos\theta}+\cos\theta=0$$

$$(\sqrt{\sin\theta}-\sqrt{\cos\theta})^2=0$$

$$\sqrt{\sin\theta}=\sqrt{\cos\theta}$$

Either $\cos\theta$ or $\sin\theta$ can be equal to zero. Both can't be equal to zero at the same time. As we can see that $\sqrt{\sin\theta}=\sqrt{\cos\theta}$, neither of $\cos\theta$ and $\sin\theta$ is equal to zero. So, dividing both sides by $\cos\theta$ is valid.

$$\sqrt{\tan\theta}=1$$

$$\tan\theta=1$$

$$\theta=n\pi+\frac{\pi}{4}\tag{1}$$

My question:

  1. We've stumbled upon an interesting solution in (1). Here, $\theta$ satisfies our original equation only when $n$ is even or $0$. Why is that? Isn't $n$ supposed to belong to the set of integers?

PS: This might help you in answering the question.

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    $\begingroup$ Note that the square root is only defined for positive values of $\sin\theta$ or $\cos\theta$, or whatever you put under the square root. When you devide by $\sqrt{\cos\theta}$ you may change the set of solutions, the same applies when you square. $\endgroup$
    – Thomas
    Commented Sep 27, 2021 at 8:22

5 Answers 5

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My solution: $$\sin\theta+\cos\theta=\sqrt{2\sin2\theta}$$

$$\sin\theta+\cos\theta=\sqrt{2\times2\sin\theta\cos\theta}$$

$$\sin\theta+\cos\theta=2\sqrt{\sin\theta\cos\theta}$$

$$\sin\theta-2\sqrt{\sin\theta\cos\theta}+\cos\theta=0$$

$$(\sqrt{\sin\theta}-\sqrt{\cos\theta})^2=0\tag{*}$$

$$\sqrt{\sin\theta}=\sqrt{\cos\theta}$$

$$\sqrt{\tan\theta}=1\tag#$$

$$\tan\theta=1$$

$$\theta=n\pi+\frac{\pi}{4}$$

  1. Note that $$\frac{\sqrt{\sin\theta}}{\sqrt{\cos\theta}}=1\implies\sqrt{\frac{\sin\theta}{\cos\theta}}=1$$ but $$\sqrt{\frac{\sin\theta}{\cos\theta}}=1\kern.6em\not\kern -.6em \implies\frac{\sqrt{\sin\theta}}{\sqrt{\cos\theta}}=1;$$ so, $$\dfrac{\sqrt{\sin\theta}}{\sqrt{\cos\theta}}\not\equiv\sqrt{\tan\theta}.$$ In step $(\#)$, you introduced extraneous solutions—expanding the set of candidate solutions—by allowing $\sin\theta$ and $\cos\theta$ to be both negative (as well as both positive).

    Nevertheless, $(\#)$ is a valid step, as the forward implication is correct.

  2. However, the above may be moot, since the prior step $(*)$ may have discarded solutions and thus may be invalid: without justification, it is not apparent that $$\sin\theta-2\sqrt{\sin\theta\cos\theta}+\cos\theta=0\implies(\sqrt{\sin\theta}-\sqrt{\cos\theta})^2=0,$$ since \begin{align}&{}{}\sin\theta-2\sqrt{\sin\theta\cos\theta}+\cos\theta\\\not\equiv&{}{}\sin\theta-2\sqrt{\sin\theta}\sqrt{\cos\theta}+\cos\theta\\\equiv&{}{}(\sqrt{\sin\theta}-\sqrt{\cos\theta})^2.\end{align}

    (The converse is certainly true though.)

    Here, the forward-implication does turn out to be justifiable: $\theta$ happens to reside in the first quadrant, so $\sin\theta$ and $\cos\theta$ are indeed both positive.


Addendum 1: suggested solution \begin{align}&{}\sin\theta+\cos\theta=\sqrt{2\sin2\theta}\\ \color{red}\implies &{}1+\sin2\theta=2\sin2\theta\\ \iff &{}\sin2\theta=1\\ \iff&{}\theta=(4n+1)\frac\pi4.\tag1\end{align} The first step above turns out to have introduced extraneous solutions $(\text{for example, }\frac{5\pi}4),$ which we must prune out. Alternatively, we could note the implicit restriction (changing the $\color{red}\implies$ to $\color{red}\iff)$ $$\sin\theta+\cos\theta\geq0\\ \sqrt2\sin\left(\theta+\frac\pi4\right)\geq0\\ \theta+\frac\pi4\in\bigcup[2n\pi,(2n+1)\pi]$$ $$\theta\in\bigcup[(8n-1)\frac\pi4,(8n+3)\frac\pi4].\tag2$$ Combining $(1)$ and $(2)$: $$\theta=(8k+1)\frac\pi4\\=\frac\pi4+2k\pi.$$


Addendum 2

So, steps that create extraneous roots are still valid.

We are arguing from mathematical axioms and given a context, so the step $\big(Q(x)\to R(x)\big)$ is valid precisely when it is mathematically true in that context.

For example, given $(x-3)(x-4)=0,$ the step $\big(x\in\{3,4\}{\implies} x\in\{3,4,7\}\big)$ is valid, even though it makes the candidate solution set less precise; however, it isn't valid to then deduce that $\big((x-3)(x-4)=0{\iff} x\in\{3,4,7\}\big),$ that is, that the actual solution set is $\{3,4,7\}.$

However, steps that discard solutions are invalid, right?

Yes, $\big(x\in\{3,4\}{\implies}x\in\{4\}\big)$ is an invalid step.

Also, I think that (*) might be valid because $\theta$ is in the first quadrant: see @user's answer. If $\theta$ is in the first quadrant, then $\sin\theta=|\sin\theta|$ & $\cos\theta=|\cos\theta|.$

Yes, $(*)$ turns out to be a valid step. But since you had neither shown (using the given equation's implicit conditions) nor even asserted that $θ$ in fact lies in quadrant $1$ (rather than quadrant $3),$ your presentation there may be too sketchy/hand-wavy.

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  • $\begingroup$ You squared both sides in the 2nd line of your solution. Didn't you create extraneous roots? $\endgroup$ Commented Sep 28, 2021 at 4:23
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    $\begingroup$ @tryingtobeastoic Squaring does potentially (though not necessarily) create extraneous solutions, which were exactly what we were accounting for in $(2).$ (With this restriction, the second line effectively becomes an $\iff$.) $\endgroup$
    – ryang
    Commented Sep 28, 2021 at 4:36
  • $\begingroup$ So, steps that create extraneous roots are still valid. However, steps that discard solutions are invalid, right? $\endgroup$ Commented Sep 28, 2021 at 5:24
  • $\begingroup$ Also, I think that (*) might be valid because $\theta$ is in the first quadrant: see @user's answer. If $\theta$ is in the first quadrant, then $\sin\theta=|\sin\theta|$ & $\cos\theta=|\cos\theta|$. What do you think? $\endgroup$ Commented Sep 28, 2021 at 6:25
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    $\begingroup$ @tryingtobeastoic : Steps that create extraneous roots are not automatically valid. If you use a step that potentially creates extraneous roots, all solutions that you find must be checked for correctness. $\endgroup$ Commented Sep 28, 2021 at 6:28
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$\sin \theta+\cos \theta$ is given to be positive square root of $2 \sin (2 \theta)$. Odd values of $n$ make $\sin \theta+\cos \theta$ negative, so only even values are permitted.

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We need that $\sin2\theta\ge 0$ that is

$$0+2n\pi\le2\theta \le \pi+2n\pi \quad \iff \quad n\pi\le \theta \le \frac \pi 2+n\pi$$

therefore only solutions in the first or third quadrant are allowed but since in the third quadrant $\sin \theta + \cos \theta <0$, only solutions in the $\color{red}{\text{first quadrant}}$ are allowed that is by $n=2k$

$$\color{red}{2k\pi\le \theta \le \frac \pi 2+2k\pi}$$

Then, since all terms involved in the expression are positive, and we have that for $A,B\ge 0$

$$A=\sqrt B \iff A^2=B$$

we can then proceed by squaring both side to obtain an $\color{magenta}{\text{equivalent equation}}$

$$\sin\theta+\cos\theta=\sqrt{2\sin2\theta} \iff 1+\sin 2\theta =2 \sin 2\theta \iff \color{magenta}{\sin 2\theta =1} $$

$$\iff2\theta=\frac \pi 2+2n\pi \iff \theta=\frac \pi 4+n\pi\iff \color{magenta}{\theta=\frac \pi 4+2k\pi}$$


Edit

As an alternative approach, since all terms involved in the expression are positive, by AM-GM we obtain

$$\sin\theta+\cos\theta \ge 2\sqrt{\sin \theta \cos \theta}=\sqrt{2\sin 2\theta}$$

with equality for $\sin\theta=\cos\theta \implies \theta=\frac \pi 4+2k\pi$.

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  • $\begingroup$ Pardon me, but I don't completely understand the conditions you set up to make $\sin2\theta$ positive. I'm talking about $0+2kπ≤2θ≤π+2kπ⟺kπ≤θ≤\frac{π}{2}+kπ$. If possible, could you please try to present a more dumbed-down picture with more words? $\endgroup$ Commented Sep 28, 2021 at 8:19
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    $\begingroup$ $\sin 2\theta$ is not negative in $[0,\pi]$, $[2\pi,3\pi]$,$\ldots$ therefore for $\theta\in[0,\pi/2],[\pi,3\pi/2],\ldots$. $\endgroup$
    – user
    Commented Sep 28, 2021 at 10:56
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Since $\theta$ must lie in the 1st quadrant in order to make both $\sqrt{\cos \theta} $and $\sqrt{\sin \theta}$ real, therefore the general solution of the equation should be $\theta= 2 n\pi+\frac{\pi}{4} \text {, where } n \in \mathbb{Z}.$

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By squaring, you have established that any solution of the equation must be of the form $\theta = \frac{\pi}{4}+k \pi$. However, not all numbers with this form need to be solutions of the equation. Imagine a simpler situation, where you are solving $x = -\sqrt{x}$, and thus assuming that $x\ge 0$. When you square both sides of the equality, you obtain $x^2=x \Leftrightarrow x = 0 \vee x = 1$, but $x=1$ is clearly not a solution to the initial equation. This boils down to realising that $ a = b \Rightarrow a^2=b^2$ but $a^2=b^2 \not \Rightarrow a = b$.

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