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For every $x \in \left[\pi/2,\pi\right]\,,\ \sin\left(x\right) − \cos\left(x\right) \geq 1$.

I have drawn the graph and can clearly see that A is true however how do I prove it correctly.

Hanul Jeon
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Tk706
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2 Answers2

1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \sin\pars{x} - \cos\pars{x}& =\sin\pars{x} - \tan\pars{\pi/4}\cos\pars{x} ={\sin\pars{x}\cos\pars{\pi/4} - \sin\pars{\pi/4}\cos\pars{x}\over \cos\pars{\pi/4}} \\[5mm]&=\root{2}\sin\pars{x - {\pi \over 4}} \end{align}

Also, with $\quad\ds{\xi\ \in\ \bracks{{\pi \over 4},{3\pi \over 4}}}$: $$ \sin\pars{\xi} \geq \sin\pars{\pi \over 4} = {1 \over \root{2}} $$

Then, $$ \color{#66f}{\large\sin\pars{x} - \cos\pars{x}} \geq \root{2}\,{1 \over \root{2}} =\color{#66f}{\Large 1}\,,\qquad x\ \in\ \bracks{{\pi \over 2},\pi} $$

Felix Marin
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Let’s assume the contrary. For simplicity, I have changed the region of x to an open interval.

That is, there exists an $x ∈ (π/2, π)$, such that $\sin x − \cos x \lt 1$.

[Please skip the following line and continue from its next.]

Then, try some values for x in that range to see the contradiction. An example is x = 3π/4.

Then, $\sin x \lt 1 + \cos x$

For x belongs to the given region, $0 \lt \sin x$.

∴ $0 \lt sin x \lt 1 + \cos x$

Squaring, we have $\sin ^2 x \lt 1 + 2\cos x + \cos ^2 x$

Simplifying, we have $0 \lt \cos x(\cos + 1)$

Then, $0 \lt \cos x(\cos + 1) = (negative)(positive)$; for x lying in the given region

i.e. , $0 \lt \cos x(\cos + 1) \lt 0$, which is a contradiction.

Mick
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