A trigonometric inequality

We can do it by trigonometry We know for a series of type Sin (a), sin (a+b), sin (a+2b), sin (a+3b),..............., sin (a+(n-1) b)

The sum of terms is given by

Sin (nb/2) ×sin (a+(n-1)b/2) /sin (b/2)

In this let the sum till sin nth terme be Sn Here value of a=1/100, b=1/100

Sn=sin (n/200)×sin (1/100+(n-1)/200) /sin (1/200)

=) Sn=sin (n/200)×sin ((n+1)/200) /sin (1/200)

We know for small value of x , sin (x) approx equal to x So we can write

sin(1/200) as 1/200

=) Sn=200 sin (n/200)×sin ((n+1)/200)

=) Sn=100 (2sin (n/200)×sin ((n+1)/200)

Using 2sin (a)×sin (b)=cos ((a-b)/2)-cos ((a+b)/2)

We write Sn=100 (cos (1/2)-cos ((2n+1)/400)

Maximum cos (1/2)-cos (2n+1)/400

where cos (1/2) will be approx 1 will be when cos (2n+1/400)=-1

So it is at max 2Sn=100×(approx 2)

And the result will always be less than 2 So N=200

It's tricky because $sin \frac{n}{100}$ has a period of $\infty$ .

I have another solution:

We can see that the above-given sum of sines will simply cancel out to approximately 0 after 2n\pi for all integer n. Hence, we only need to look at $\sin x$ for $0 < x < \pi$ The sum will simply become 100*(area under graph of $\sin x$ for $0 < x < \pi$ ), which can easily be evaluated to 200 ny integration, giving the required value of N. :)

There is a typo in the second last line:- $f''(n)= \frac{ \cos (2n+1)/200 }{100}$ . The $n$ was replaced by the $x$ . Also sorry for poor latex formatting. :(

Here is a potential solution (it uses some of the info here ). It isn't quite rigorous, but I think it might be close:

The given expression looks like a Riemann Sum. Let $S$ be our desired sum. It is clear that, where $k$ is a whole number $n=\left\lfloor\dfrac{(2k+1)\pi}{100}\right\rfloor$ This is simply because we want $n$ to be at the last point where $\sin(n)$ is still positive for that period. The integral from 0 to $(2k+1)\pi$ of $\sin(x)$ is $2$ . The Riemann sum from 0 to $(2k+1)\pi$ of $\sin(x)$ is $\dfrac{1}{100}\left(\sin\dfrac{1}{100}+\sin\dfrac{2}{100}+\cdots+\sin\dfrac{n}{100}\right)=\dfrac{1}{100}S$ Now, using the bounds in the linked article (we have the left Riemann sum), $2-\dfrac{S}{100}\le 2\cdot\dfrac{1}{100}\cdot \dfrac{\pi}{2}=\dfrac{\pi}{100}$ I split the area into two parts, the area from 0 to $\dfrac{\pi}{2}$ , and the area from $\dfrac{\pi}{2}$ to $\pi$ .

Therefore, $2-\dfrac{\pi}{100}\le S\le 2$ $200-\pi\le S\le 200$ Now, if we choose a $k$ , $(2k+1)\pi$ can be arbitrarily close to a multiple of $\dfrac{1}{100}$ (this is probably a problem). Therefore, we can get $S$ very close to $\boxed{200}$ (also probably a problem).

The ending is not quite rigorous; can anyone fix it?

Reply to Challenge Master note
I found out later that $|S|$ could be maximized in a much simpler method. Note that $f(n)= \frac{1}{2} (\cos \frac{1}{200} - \cos \frac{2n+1}{200})$ . To maximize this, we have to minimize $\cos \frac{2n+1}{200}$ , whose minimum is $-1$ , and attained when $n= 100 \pi (2c \pm 1) - \frac{1}{2}$ ( $c \epsilon Z$ ).

the summation of the series will be maximum when

n/100 =3.14

now applying summation of sine series

difference= 0.01 no.of terms = 314

Now,

{sin(100 3.14 0.5* .01) * sin([3.14+.01]/2)} / sin(0.02)

taking pi & 3.14 as interchangeable

we get,

{1*1} / 0.02

= 200(approx)