I was trying to find the least value of |\sin

lugreget9 2022-01-01 Answered
I was trying to find the least value of \(\displaystyle{\left|{\sin{{x}}}\right|}+{\left|{\cos{{x}}}\right|}\) and applied the above inequality as :
\(\displaystyle{\left|{\sin{{x}}}+{\cos{{x}}}\right|}\leq{\left|{\sin{{x}}}\right|}+{\left|{\cos{{x}}}\right|}\)
\(\displaystyle\Rightarrow\sqrt{{2}}\leq{\left|{\sin{{x}}}\right|}+{\left|{\cos{{x}}}\right|}\)
But the range of the given function is \(\displaystyle{\left[{1},\sqrt{{2}}\right]}\).

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Expert Answer

jean2098
Answered 2022-01-02 Author has 2912 answers
It doesn't hold iff x and y have the same sign. As for your problem remember that if \(\displaystyle{a}\leq{1}\) then \(\displaystyle{a}^{{2}}\leq{a}\) so
\(\displaystyle{\left|{\sin{{x}}}\right|}+{\left|{\cos{{x}}}\right|}\geq{\left|{\sin{{x}}}\right|}^{{2}}+{\left|{\cos{{x}}}\right|}^{{2}}={1}\)
and for upper bound (remember \(\displaystyle{\left({a}+{b}\right)}^{{2}}\leq{2}{\left({a}^{{2}}+{b}^{{2}}\right)}\)):
\(\displaystyle{\left|{\sin{{x}}}\right|}+{\left|{\cos}\right|}{x}\leq\sqrt{{{2}{\left({\left|{\sin{{x}}}\right|}^{{2}}+{\left|{\cos{{x}}}\right|}^{{2}}\right)}}}=\sqrt{{2}}\)
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yotaniwc
Answered 2022-01-03 Author has 4798 answers
\(\displaystyle{\left|{\sin{{x}}}+{\cos{{x}}}\right|}\leq{\left|{\sin{{x}}}\right|}+{\left|{\cos{{x}}}\right|}\) is correct.
But, for example, if x=0 then both sides are 1; thus we cannot conclude \(\displaystyle\sqrt{{2}}\leq{\left|{\sin{{x}}}\right|}+{\left|{\cos{{x}}}\right|}\)
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Vasquez
Answered 2022-01-08 Author has 8850 answers

Let \(\displaystyle{\left\lbrace\begin{matrix} f{{\left({x}\right)}}= \sin{{x}}+ \cos{{x}}\\ g{{\left({x}\right)}}={\left| \sin{{\left({x}\right)}}\right|}+{\left| \cos{{\left({x}\right)}}\right|}\end{matrix}\right.}\)
Notice that \(g(x+\frac{\pi}{2})=|\cos(x)|+|\sin(x)|=g(x)\) so
g is periodic of period \(\frac{\pi}{2}\) and we can study it on \([0,\frac{\pi}{2}]\).
But on this interval both sin and cos are positive, so we can get rid of absolute values and g(x)=f(x) on \([0,\frac{\pi}{2}].\)
You now can use the addition formulas to transform f to \(f(x)=\sqrt2 \sin(x+\frac{\pi}{4})\) and see that \(f(x)\) has values between 0 and \(\sqrt2\) while g(x) has values between 1 and \(\sqrt2\).

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