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 .

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

1 Answer 1


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

The conclusion is now obvious

Method 2 as $0\le |\sin t|,|\cos t|\le 1$ we must have $$|\sin t|+|\cos t|\ge |\sin^2 t|+|\cos^2 t|=\sin^2 t+\cos ^2t=1$$

  • 1
    $\begingroup$ I have always hated squaring ,it makes us loose solution or gains them .why did nothing go wrong here ? $\endgroup$ 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 $\endgroup$ Nov 29, 2020 at 3:33
  • $\begingroup$ In the first method how can we square it ? Do we lose or gain anything?? $\endgroup$ 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 $\endgroup$ Nov 29, 2020 at 3:40

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