# Extra solutions when solving $\sin\theta+\cos\theta=\sqrt{2\sin2\theta}$

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?

• 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. Sep 27, 2021 at 8:22

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 $$(2)$$: $$\theta=(8k+1)\frac\pi4\\=\frac\pi4+2k\pi.$$

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.

• You squared both sides in the 2nd line of your solution. Didn't you create extraneous roots? Sep 28, 2021 at 4:23
• @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 $\iff$.) Sep 28, 2021 at 4:36
• So, steps that create extraneous roots are still valid. However, steps that discard solutions are invalid, right? Sep 28, 2021 at 5:24
• 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? Sep 28, 2021 at 6:25
• @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. Sep 28, 2021 at 6:28

$$\sin \theta+\cos \theta$$ 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.

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

• 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? Sep 28, 2021 at 8:19
• $\sin 2\theta$ is not negative in $[0,\pi]$, $[2\pi,3\pi]$,$\ldots$ therefore for $\theta\in[0,\pi/2],[\pi,3\pi/2],\ldots$.
– user
Sep 28, 2021 at 10:56

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 $$\theta= 2 n\pi+\frac{\pi}{4} \text {, where } n \in \mathbb{Z}.$$

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 $$a^2=b^2 \not \Rightarrow a = b$$.