I need a proof for the following trigonometric inequality $$\frac{|\sin x| + |\cos x|}{\sqrt2} \leq 1- \frac{\cos^2(2x)}{8}$$ Can someone please help me with this?
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$\begingroup$ How about: write $a=|\sin x|$ and $b=|\cos x|$, then write $\cos^2(2x)$ in terms of $a$ and $b$ to get an inequality to be proved for all $a,b \ge 0$ with $a^2+b^2=1$. $\endgroup$– GEdgarSep 23, 2013 at 17:06
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$\begingroup$ I don't see how the inequality $$\frac{a+b}{\sqrt 2} \leq 1- \frac{(a^2-b^2)^2}{8}$$ is less complicated. Am I missing something? $\endgroup$– MartinSep 23, 2013 at 18:25
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1 Answer
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Let$$y=f(x)=\frac{|\sin x| + |\cos x|}{\sqrt2}+\frac{\cos^22x}{8}\quad(x\in \Bbb R).$$The shift $x\leftarrow x+{\frac12}n\pi\;(n\in \Bbb Z)$ leaves $f(x)$ invariant; so we may take $x\in [0,{\frac12}\pi].$ Write $x={\frac14}\pi+\theta.$ Then $$y=\cos\theta+{\frac18}\sin^22\theta=u+{\frac12}u^2-{\frac12}u^4,$$ where $u=\cos\theta$ with $\theta\in[-{\frac14}\pi,{\frac14}\pi].$ Now$$\frac{\mathrm dy}{\mathrm du}=1+u-2u^3=(1-u)[u^2+(1+u)^2],$$ which is clearly positive for $1/\sqrt2< u< 1.$ Hence the extrema of $y$ are attained at $u=1/\sqrt2$ and $u=1,$ where $y=1/\sqrt2+1/8$ and $y=1$ respectively. Therefore $y\leqslant 1.$
answered Sep 23, 2013 at 22:55