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.

 

[CellCoree]

I would first clarify that the question is asking for an integer solution for $K$ in radians that satisfies the inequality $\sin K > \sin \left(\dfrac{11\pi}{60}\right)$. This means that $K$ must be a whole number in the unit of radians and its sine value must be greater than the sine value of $\dfrac{11\pi}{60}$.

To find such a value, we can use the unit circle to visualize the sine values of different angles. Starting from the origin (0 radians) and moving counterclockwise, we can see that the first integer value that satisfies the given inequality is at $K = \dfrac{\pi}{3}$. This is because $\sin \dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$, which is greater than $\sin \left(\dfrac{11\pi}{60}\right) \approx 0.2588$.

Therefore, the integer solution for $K$ in radians that satisfies the given inequality is $K = \dfrac{\pi}{3} \approx 1.0472$ radians. We can also verify this by calculating the sine value of $\dfrac{\pi}{3}$ using a scientific calculator or by using the trigonometric identity $\sin \left(\dfrac{11\pi}{60}\right) = \sin \left(\dfrac{\pi}{3}\right)$.

In conclusion, the integer solution for $K$ in radians that satisfies the inequality $\sin K > \sin \left(\dfrac{11\pi}{60}\right)$ is $K = \dfrac{\pi}{3}$. This answer can also be expressed in degrees as $K = 60$ degrees.