MHB Trigonometric inequality: sin (1/(n+1934))<1/1994

[anemone]

Find the smallest natural number $n$ for which $\sin \left(\dfrac{1}{n+1934}\right)<\dfrac{1}{1994}$.

 

[Olinguito]

Spoiler

As $\sin x<x$ for all $x>0$ – in particular $\sin\left(\dfrac1{1994}\right)<\dfrac1{1994}$ – it suffices to show that $\sin\left(\dfrac1{1993}\right)>\dfrac1{1994}$.

By Taylor’s theorem, $\sin x=x-\dfrac{x^2}2\sin\xi$ for some $0<\xi<x$ (using the Lagrange form of the remainder). Thus:

$\begin{array}{rcl}\sin x &=& x-\dfrac{x^2}2\sin{\xi} \\\\ {} &>&x-\dfrac{x^2}2\xi \\\\ {} &>& x-\dfrac{x^3}2.\end{array}$

Hence $\sin\left(\dfrac1{1993}\right)>\dfrac1{1993}-\dfrac{\left(\frac1{1993}\right)^3}2 = \dfrac{7944097}{15832587314}>\dfrac1{1994}$, as required. So the smallest natural number is $\boxed{n=60}$.

 

[Joppy]

Spoiler

    for (unsigned int n = 0; n <= 60; n++)
        if ((sin(1/(n + 1934.0))) < 1/1994.0)
            std::cout << n;

(Bigsmile)

 

[castor28]

Find the smallest natural number $n$ for which $\sin \left(\dfrac{1}{n+1934}\right)<\dfrac{1}{1994}$.

[sp]
For $0<x<1$, we have $\sin(x)<x$. This shows that the inequality is satisfied for $n=60$. Taking $n=59$, we find:
$$
\sin\left(\frac{1}{1993}\right)\approx 0.00050176> \frac{1}{1994}\approx 0.00050150
$$
As $\sin\left(\dfrac{1}{n+1934}\right)$ is a decreasing function of $n$ for $n>0$, this shows that the smallest $n$ that satifies the inequality is $60$.
[/sp]