How to prove this inequality in interval: (0,\frac{\pi}{2}) \ \

oliviayychengwh 2022-01-02 Answered
How to prove this inequality in interval: \(\displaystyle{\left({0},{\frac{{\pi}}{{{2}}}}\right)}\ \ {2}{\cos{{x}}}+{{\sec}^{{2}}{x}}-{3}\geq{0}\)

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Expert Answer

Thomas Nickerson
Answered 2022-01-03 Author has 6020 answers
Restrict the domain to some meaningful interval such as \(\displaystyle{x}\in{\left(-{\frac{{\pi}}{{{2}}}},{\frac{{\pi}}{{{2}}}}\right)}\) so that \(\displaystyle{\cos{{x}}}{>}{0}\). In that case, using the AM-GM inequality you have:
\(\displaystyle{L}{H}{S}={\cos{{x}}}+{\cos{{x}}}+{{\sec}^{{2}}{x}}-{3}\geq{3}{\sqrt[{{3}}]{{{\cos{{x}}}\cdot{\cos{{x}}}\cdot{{\sec}^{{2}}{x}}}}}-{3}={3}-{3}={0}={R}{H}{S}\)
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Papilys3q
Answered 2022-01-04 Author has 3156 answers
We have
\(\displaystyle{2}{\cos{{x}}}+{{\sec}^{{2}}{x}}-{3}={2}{\left({\cos{{x}}}-{1}\right)}+{\left({\frac{{{1}}}{{{{\cos}^{{2}}{x}}}}}-{1}\right)}={\frac{{{\left({2}{\cos{{x}}}+{1}\right)}{\left({\cos{{x}}}-{1}\right)}^{{2}}}}{{{{\cos}^{{2}}{x}}}}}\)
Thefore the inequality does not hold.
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Vasquez
Answered 2022-01-09 Author has 9499 answers

By Fermat's theorem, we examine zeroes of derivative of the function \(f(x)=2\cos x+\sec^2 x-3\) to find the extrema, that is, to solve the following:
\(f′(x)=2\tan x \sec^2 x−2\sin x=0\)
The solution is \(x=\{k\pi | k \in \mathbb{Z}\}\). When \(x=\pi\), we get
\(2\cos(x)+\sec^2(x)−3=-4\)
Thus \(2\cos x+\sec^2 x-3 \geq 0\) doesn't hold.
Since the interval is added, we can prove the claim is true. For x=0, f(x)=0. And when \(x \in (0,\frac{\pi}{2})\), it is trivial that \(f′(x) \geq 0\), thus f(x) monotonously increases on \((0,\frac{\pi}{2})\). Hence \(f(x) \geq f(0)=0 \text{ for all } x\in (0,\frac{\pi}{2})\) and we finished the proof.

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Prove that \(\displaystyle{\sin{{x}}}+{\cos{{x}}}=\sqrt{{2}}{\sin{{\left({x}+{\frac{{\pi}}{{{4}}}}\right)}}}\)
\(\displaystyle\sqrt{{2}}{\left({x}+{\frac{{\pi}}{{{4}}}}\right)}=\sqrt{{2}}{\left({\sin{{x}}}{\cos{{\frac{{\pi}}{{{4}}}}}}+{\cos{{x}}}{\sin{{\frac{{\pi}}{{{4}}}}}}\right)}={\sin{{x}}}+{\cos{{x}}}\)
Could you solve it from opposite?
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I want to solve this integral :
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How to prove \(\displaystyle{\frac{{{e}^{{{j}{x}}}-{j}{e}^{{-{j}{x}}}}}{{{j}{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}}}}={\frac{{{\tan{{x}}}-{1}}}{{{\tan{{x}}}+{1}}}}\)
where \(\displaystyle{j}=\sqrt{{-{1}}}\)
I've tried to prove it from the right hand
\(\displaystyle{\tan{{x}}}={\frac{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}}}{{{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}}\) (2)
and thus
\(\displaystyle{\tan{{x}}}-{1}={\frac{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}-{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}{{{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}}\) (3)
and
\(\displaystyle{\tan{{x}}}+{1}={\frac{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}+{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}{{{j}{\left({e}^{{{j}{x}}}+{e}^{{-{j}{x}}}\right)}}}}\) (4)
The right side of Eq. (1) can be written as
\(\displaystyle{\frac{{{\tan{{x}}}-{1}}}{{{\tan{{x}}}+{1}}}}={\frac{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}-{j}{e}^{{{j}{x}}}-{j}{e}^{{-{j}{x}}}}}{{{e}^{{{j}{x}}}-{e}^{{-{j}{x}}}+{j}{e}^{{{j}{x}}}+{j}{e}^{{-{j}{x}}}}}}\) (5)
Compared with the left side of Eq. (1), the numerator has the same part \(\displaystyle{e}^{{{j}{x}}}−{j}{e}^{{−{j}{x}}}\), and the denominator has the same part \(\displaystyle{j}{e}^{{{j}{x}}}−{e}^{{-{j}{x}}}\), but the other parts don't seem to be zero constantly.
I've checked Eqs. (1) and (5), they are all correct numerically. Can anyone help to find where I am wrong? Thanks in advance!
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