MHB Find K in Radians: Integer Solution

[anemone]

Find $K$, an integer and is in radians, such that $\sin K>\sin \left(\dfrac{11\pi}{60}\right)$.

 

[Nono713]

Spoiler

Note that for all integers $k$, we have:
$$\sin \frac{(4k + 1)\pi}{2} = 1$$
Therefore to solve the challenge it suffices to find an integer $K$ sufficiently close to $(4k + 1) \pi / 2$ for some $k \in \mathbb{Z}$. A straightforward way to do this is to obtain a good rational approximation of $\pi / 2$ as follows:
$$\frac{\pi}{2} \approx \frac{p}{q}$$
Then, if $q \equiv 1 \pmod{4}$ we can use the continuity of $\sin$ to deduce that:
$$\frac{q \pi}{2} \approx p ~ ~ ~ \implies ~ ~ ~ \sin p \approx \sin \frac{q \pi}{2} = 1$$
The successive convergents of $\pi/2$ can be computed through the continued fraction of $\pi/2$, and the first few are given below:
$$1, 2, \frac{3}{2}, \frac{11}{7}, \frac{355}{226}, \frac{51819}{32989}, \cdots$$
Now $51819 / 32989$ is the first nontrivial convergent such that $32989 \equiv 1 \pmod{4}$ and in fact we find that (in radians):
$$\sin 51819 \approx 0.999999999696513$$
Which satisfies the challenge's inequality with considerable overkill.
​

 

[anemone]

Spoiler

Note that for all integers $k$, we have:
$$\sin \frac{(4k + 1)\pi}{2} = 1$$
Therefore to solve the challenge it suffices to find an integer $K$ sufficiently close to $(4k + 1) \pi / 2$ for some $k \in \mathbb{Z}$. A straightforward way to do this is to obtain a good rational approximation of $\pi / 2$ as follows:
$$\frac{\pi}{2} \approx \frac{p}{q}$$
Then, if $q \equiv 1 \pmod{4}$ we can use the continuity of $\sin$ to deduce that:
$$\frac{q \pi}{2} \approx p ~ ~ ~ \implies ~ ~ ~ \sin p \approx \sin \frac{q \pi}{2} = 1$$
The successive convergents of $\pi/2$ can be computed through the continued fraction of $\pi/2$, and the first few are given below:
$$1, 2, \frac{3}{2}, \frac{11}{7}, \frac{355}{226}, \frac{51819}{32989}, \cdots$$
Now $51819 / 32989$ is the first nontrivial convergent such that $32989 \equiv 1 \pmod{4}$ and in fact we find that (in radians):
$$\sin 51819 \approx 0.999999999696513$$
Which satisfies the challenge's inequality with considerable overkill.
​

Very well done, Bacterius!

You're right, we can find other value(s) of $K$ using purely elementary method, therefore, anyone who wants to try that is welcome to post your solution here, hehehe...(Wink)

 

[anemone]

I just realized I posted a wrong problem, since the problem should read $\sin K>\sin 33 \,\text{radians}$. Sorry...

Here is the solution of other that I wanted to share with MHB:

Spoiler

Recall that $\sin 3x=3\sin x-4\sin^3 x$.

Thus we have $\sin 33-\sin(-11)=4\sin 11-4\sin^3 11=4\sin 11 \cos^2 11<0$, since the angle 11 in radians is in the fourth quadrant.

Thus, $\sin 11<0$ and hence $\sin(-11)>\sin 33$.

So, $K=-11$ will do.