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

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 .

• $(|\sin 2x|+|\cos 2x|)^2=(\sin 2x)^2 + (\cos 2x)^2 + 2|\sin 2x||\cos 2x|=1+2|\sin 2x||\cos 2x|\ge 1$ 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. Nov 29, 2020 at 3:41
• squaring could sometimes introduce extraneous results, but in this case the expression must be non-negative Nov 29, 2020 at 3:50

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