Sine function

So one way of solving is the one which Khang Nguyen Thanh has pointed out which is quite marvelous. And if you are wondering for a non generalized version then you might be tempted to do this -> $sin(a+b)=sin(2015x)$ for all x , then take three integer values of x $1,2,3$ then Add what you get(You must have got $a+b,a+2b,a+3b$ as arguments for sine on LHS And similarly you must have got $2015,4030,6045$ on the RHS as arguments of sine . You add them using the formula for addition of sines. You'll see some cancellations . Subsequently you would convert triple angle formula to single angle and again we'll see some cancellations.In the end convert the answer to the general solution for sine. And you must have got $\frac{a}{2}$ as some integral multiple of pi plus or minus $\frac{2015}{2}$ Then it is done choose such value of integral multiple such that it does not make your value of a negative . After getting the answer try to generalize .

First, since $\sin(b) = \sin(0) = 0$ , we have $b = n\pi$ for some integer $n$ .

Since $\sin$ function has period $2\pi$ , we need only consider the cases when $b = 0$ and $b = \pi$ .

Now let $b \in \{0, \pi\}$ and $a$ be any real number.

If for all integers $x$ , $\sin(ax + b) = \sin(2015x)$ , then for any integer $n$ , $\sin((a + 2\pi n)x + b) = \sin(ax + b + 2\pi nx) = \sin(ax + b) = \sin(2015x)$ for all integers $x$ as well.

Conversely, assume for some $a$ and $c$ that for all integers $x$ , $\sin(ax+b) = \sin(cx + b) = \sin(2015x)$ .

Then, for all integers $x$ ,

$\sin(ax) = \dfrac{\sin(ax) \cos(b) + \cos(ax) \sin(b)}{\cos(b)} = \dfrac{\sin(ax + b)}{\cos(b)}$

$\qquad\quad\;\;= \dfrac{\sin(cx + b)}{\cos(b)} =\dfrac{\sin(cx)\cos(b) + \cos(cx) \sin(b)}{\cos(b)} = \sin(cx)$ ,

since $\sin(0) = \sin(\pi) = 0$ and $\cos(0),\cos(\pi) \ne 0$ .

But then, $\sin(a) = \sin(c)$ and $2\sin(a)\cos(a) = \sin(2a) = \sin(2c) = 2\sin(c)\cos(c)$ implies $\cos(a) = \cos(c)$ since $\sin(a) = \sin(c) = \dfrac{\sin(2015)}{\cos(b)}\ne 0$ .

Hence, $a$ and $c$ are the same angle, modulo integer multiples of $2\pi$ . Now, we consider the two cases concretely.

If $b = 0$ , one valid assignment of $a$ is $a =2015$ , so all possible ones are $a = 2015 + 2\pi n$ for integers $n$ .

The smallest positive number we can make this is $2015 - 640\pi$ .

Meanwhile, if $b = \pi$ , one valid assignment of $a$ is $a = -2015$ , since $\sin(-2015x+\pi) = \sin(-2015x) \cos(\pi) + \cos(-2015x) \sin(\pi) = - \sin(-2015x) = \sin(2015x)$ . So, all possible ones are $a = -2015 + 2\pi n$ for integers $n$ .

The smallest positive number we can make this is $642\pi - 2015$ .

Because $642\pi-2015<2015-640\pi$ so the answer is $642\pi-2015\approx \boxed{1.902}$ .