Alisha Pitts

2022-01-23

How do I prove $\mathrm{tan}1<\frac{\pi }{2}$?
Prove that the equation
$\mathrm{sin}x\mathrm{sin}\left(\mathrm{sin}x\right)=\frac{\pi }{2}\mathrm{cos}\left(\mathrm{sin}x\right)$
Let $t=\mathrm{sin}x,-1\le t\le 1$. Then the expression above is equvalent to $t\mathrm{sin}t=\frac{\pi }{2}\mathrm{cos}t$. As the function $f\left(t\right)=t\mathrm{sin}t-\frac{\pi }{2}\mathrm{cos}t$ is even, and t=0 is not a solution, I have to prove that f(t) has no positive roots (t>0). So, for the left side $0 and $0<\mathrm{sin}t\le \mathrm{sin}1$, then $t\mathrm{sin}t\le \mathrm{sin}1$. For the right side $\mathrm{cos}t\ge \mathrm{cos}1$, so $\frac{\pi }{2}\mathrm{cos}t\ge \frac{\pi }{2}\mathrm{cos}1$. The objective is to prove that $\mathrm{sin}1<\frac{\pi }{2}\mathrm{cos}1$, or, equivalently, $\mathrm{tan}1<\frac{\pi }{2}$

Tapanuiwp

Expert

We can use the Taylor series and alternating series theorem to say
$\mathrm{sin}1<1-\frac{1}{3!}+\frac{1}{5!}=\frac{101}{120}$
$\mathrm{cos}1>1-\frac{1}{2!}+\frac{1}{4!}-\frac{1}{6!}=1-\frac{12}{+}\frac{1}{24}-\frac{1}{720}=\frac{389}{720}$
$\mathrm{tan}1=\frac{\mathrm{sin}1}{\mathrm{cos}1}<\frac{606}{389}<1.56<\frac{\pi }{2}$

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