lugreget9

2022-01-01

I was trying to find the least value of $|\mathrm{sin}x|+|\mathrm{cos}x|$ and applied the above inequality as :
$|\mathrm{sin}x+\mathrm{cos}x|\le |\mathrm{sin}x|+|\mathrm{cos}x|$
$⇒\sqrt{2}\le |\mathrm{sin}x|+|\mathrm{cos}x|$
But the range of the given function is $\left[1,\sqrt{2}\right]$.

jean2098

It doesn't hold iff x and y have the same sign. As for your problem remember that if $a\le 1$ then ${a}^{2}\le a$ so
$|\mathrm{sin}x|+|\mathrm{cos}x|\ge {|\mathrm{sin}x|}^{2}+{|\mathrm{cos}x|}^{2}=1$
and for upper bound (remember ${\left(a+b\right)}^{2}\le 2\left({a}^{2}+{b}^{2}\right)$):
$|\mathrm{sin}x|+|\mathrm{cos}|x\le \sqrt{2\left({|\mathrm{sin}x|}^{2}+{|\mathrm{cos}x|}^{2}\right)}=\sqrt{2}$

yotaniwc

$|\mathrm{sin}x+\mathrm{cos}x|\le |\mathrm{sin}x|+|\mathrm{cos}x|$ is correct.
But, for example, if x=0 then both sides are 1; thus we cannot conclude $\sqrt{2}\le |\mathrm{sin}x|+|\mathrm{cos}x|$

Vasquez

Let $\left\{\begin{array}{c}f\left(x\right)=\mathrm{sin}x+\mathrm{cos}x\\ g\left(x\right)=|\mathrm{sin}\left(x\right)|+|\mathrm{cos}\left(x\right)|\end{array}$
Notice that $g\left(x+\frac{\pi }{2}\right)=|\mathrm{cos}\left(x\right)|+|\mathrm{sin}\left(x\right)|=g\left(x\right)$ so
g is periodic of period $\frac{\pi }{2}$ and we can study it on $\left[0,\frac{\pi }{2}\right]$.
But on this interval both sin and cos are positive, so we can get rid of absolute values and g(x)=f(x) on $\left[0,\frac{\pi }{2}\right].$
You now can use the addition formulas to transform f to $f\left(x\right)=\sqrt{2}\mathrm{sin}\left(x+\frac{\pi }{4}\right)$ and see that $f\left(x\right)$ has values between 0 and $\sqrt{2}$ while g(x) has values between 1 and $\sqrt{2}$.

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