Solve the equation $\min \{ \sin x, \cos x \} = \frac{\pi}{4}$ in $[0, 2\pi]$.

Question

Consider the following equation:

$$\min \{ \sin x, \cos x \} = \dfrac{\pi}{4}$$

I have to solve this equation for $x$ in $[0, 2\pi]$. What I want to know is how can I solve this without going to a site like Desmos and plotting $y = \min \{ \sin x, \cos x \}$ and $y = \dfrac{\pi}{4}$ and see that these two functions have no intersection, therefore there are no solutions. How would I approach this on paper?

Well, $\min {\sin x,\cos x}\leq \frac{1}{\sqrt{2}}< \frac{\pi}{4}$, so the equation doesn't seem to have a solution. — WoolierThanThou, Jan 20 '20 at 23:20
The minimum of $\sin x$ and $\cos x$ for $x\in[0,2\pi)$ is $-1<\frac{\pi}{4}$, so there is no solution. — pshmath0, Jan 20 '20 at 23:25
@KaviRamaMurthy $\frac{1}{\sqrt{2}}=\frac{\sqrt{8}} {4}<\frac{3}{4}<\frac{\pi}{4}$ — WoolierThanThou, Jan 20 '20 at 23:29
@Pixel: I thought OP was asking if there is any $x$ for which $\min{\sin x,\cos x}=\dfrac{\pi}4$, not if $\min{\sin x, \cos x\mid x\in [0,2\pi]}=\dfrac{\pi}4$ — J. W. Tanner, Jan 21 '20 at 00:00
I would honestly recommend sketching the functions $\sin$ and $\cos$ on paper, then sketching the minimum of these two. It's fairly apparent where this minimum takes on the highest values and you can compute it at those points. — Milo Brandt, Jan 21 '20 at 02:20
The problem of $\min { \sin x, \cos x } + \dfrac{\pi}{4}=0$ would be very interesting. — Claude Leibovici, Jan 21 '20 at 05:40
You may be able to use $$min(x,y)=\frac{1}{2}(x+y-|x-y|)$$ somehow. — pshmath0, Jan 23 '20 at 05:08

score 1 · Answer 1 · answered Jan 20 '20 at 23:28

Since $\sin^2x+\cos^2x=1$,

either $\sin^2x\ge\frac12$, in which case $\cos^2x\le\frac12$, so $\cos x\le\frac1{\sqrt2}$,

or $\sin^2x\lt\frac12$, in which case $\sin x<\frac1{\sqrt2}$.

In either case, $\min\{\sin x,\cos x\}\le\frac1{\sqrt2}<\frac{\pi}4.$

score 0 · Accepted Answer · edited Jan 21 '20 at 02:23

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Suppose that for some $x$ we have $\min\{\sin(x),\cos(x)\}=\pi/4$, then we have $$1=\sin^2(x)+\cos^2(x)\geq(\pi/4)^2+(\pi/4)^2=\pi^2/8\approx1.2337,$$ a contradiction.

edited Jan 21 '20 at 02:23

J. W. Tanner

60,406

answered Jan 21 '20 at 01:50

YiFan Tey

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Quanto · Answer 3 · 2020-01-21T04:52:07.627

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Observe that over the given domain $[0,2\pi]$,

$$\min(\sin x, \cos x) = \bigg\{ \begin{array}{ll} \cos x \le \cos\frac\pi4, & x\in [\frac \pi4,\frac{5\pi}4]\\ \sin x\le \sin\frac\pi4, & x \in[0,\frac \pi4]\cup [\frac{5\pi}4,2\pi] \end{array}$$

Since $\sin\frac\pi4 =\cos\frac\pi4 < \frac\pi4$, there is no solution.

edited Jan 21 '20 at 04:52

answered Jan 21 '20 at 03:58

Quanto

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score 0 · Answer 4 · answered Jan 21 '20 at 04:25

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Note: $\sin x = \pm \sqrt{1 - \cos^2 x}$ and vice versa.

If $\sin x > \frac 1{\sqrt 2}$ then $\cos x <\sqrt{ 1- (\frac 1{\sqrt 2})^2} = \frac 1{\sqrt 2}$ (and vice versa).

So $\min(\sin x, \cos x)\le \frac 1{\sqrt 2}$. Always.

And $\frac \pi 4 > \frac 1{\sqrt 2}$ so there are no solutions.

I have to wonder. Did you write the question correctly.

answered Jan 21 '20 at 04:25

fleablood

124,253

I think I did write it correctly. I am apparently not even the first one to ask it. It was answered before here but the person who solved it used plots to see that there are no solutions. – Jan 21 '20 at 17:29
The thing that should be clear is that if $\sin x$ "goes up" then $\cos y$ goes down (assuming both are positive in the first place). So the very maximum $\min (\sin x, \cos x)$ can possibly be is when $\sin x = \cos x>0$ and that should be well known to be when $x=\frac \pi 4$ and $\sin x =\cos x = \frac 1{\sqrt 2}$. Now having $\sin x|\cos x = \frac \pi 4$ should be ... wierd.. as usually expression of $\pi$ are input; not output. But ... what the heck... so just compare. $\frac \pi 4 > \frac 1{\sqrt 2}$ and that's that.... – fleablood Jan 21 '20 at 17:49

Solve the equation $\min \{ \sin x, \cos x \} = \frac{\pi}{4}$ in $[0, 2\pi]$.

4 Answers4