How to find the maximum and minimum values of $\left|\sin 2x\right|+\left|\cos 2x\right|$

Question

The expression is :$$\left|\sin2x\right|+\left|\cos2x\right|$$

For the maximum value I can ignore the absolute value and can get $$\sqrt2$$

But finding the minimum is what stumps me

I can understand it is $1$ by guessing but is there a rigorous approach ?

I didn't try differentiation because it can not be used on absolute value function .

Thank you in advance .

$(|\sin 2x|+|\cos 2x|)^2=(\sin 2x)^2 + (\cos 2x)^2 + 2|\sin 2x||\cos 2x|=1+2|\sin 2x||\cos 2x|\ge 1$ — J. W. Tanner, Nov 29, 2020 at 3:37
@J. W. Tanner how are we not getting a wrong answer by squaring? I have learnt that squaring gives wrong results. — Glowingbluejuicebox, Nov 29, 2020 at 3:41
squaring could sometimes introduce extraneous results, but in this case the expression must be non-negative — J. W. Tanner, Nov 29, 2020 at 3:50

Albus Dumbledore · Accepted Answer · 2020-11-29 03:32:52Z

5

Method 1 $2x=t$ we have $${(|\sin t|+|\cos t|)}^2=1+|\sin 2t|\ge 1$$

The conclusion is now obvious

answered Nov 29, 2020 at 3:25

Albus Dumbledore

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1

$\begingroup$ I have always hated squaring ,it makes us loose solution or gains them .why did nothing go wrong here ? $\endgroup$
– Glowingbluejuicebox
Nov 29, 2020 at 3:29
3

$\begingroup$ I always wonder why, if something is obvious, that it needs to be pointed out? $\endgroup$
– copper.hat
Nov 29, 2020 at 3:30
$\begingroup$ @Glowingbluejuicebox another method is put $\endgroup$
– Albus Dumbledore
Nov 29, 2020 at 3:33
$\begingroup$ In the first method how can we square it ? Do we lose or gain anything?? $\endgroup$
– Glowingbluejuicebox
Nov 29, 2020 at 3:36
$\begingroup$ @Glowingbluejuicebox i started out with a known inequality ,it we take square roots we only have to take pesitive value as sum of modulus is always greater than or equal to zero $\endgroup$
– Albus Dumbledore
Nov 29, 2020 at 3:40

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