Some question about sine functions Let $x \in [0, \frac \pi 2]$ and $I_n$ be a length of intervals in which $\sin 4nx \leq \sin x$. Find a limit of $I_n$ as $n \rightarrow \infty$.
I tried to represent intersecting points of two sine functions implicitly but it is too hard for me.
 A: First a little result: Suppose we have $c \in [0,1]$ and the function $f_n(x) = \sin (4nx)$. Now compute the length of the set $J=\{t \in [0, {\pi \over 2 n}] | f_n(t) \le c \}$; it is straightforward to get $m(J) = 2 {1 \over 4n} \arcsin c + { \pi \over 4n}$. Note that this length is the same if we take $t \in [k {\pi \over 2 n}, (k+1){\pi \over 2 n}]$ instead.
Let $J_{k,n}(c) = \{t \in [k {\pi \over 2 n}, (k+1){\pi \over 2 n}] | f_n(t) \le c \}$, and note that if $c \le c'$, then $J_{k,n}(c) \subset J_{k,n}(c')$. Also, as above, note that $m(J_{k,n}(c)) = 2 {1 \over 4n} \arcsin c + { \pi \over 4n}$, which is increasing as a function of $c$.
Let $I_{k,n} = \{ t \in [k {\pi \over 2 n}, (k+1){\pi \over 2 n}] | f_n(t) \le  \sin t\} $, and note that
$J_{k,n}(\sin(k {\pi \over 2 n})) \subset I_{k,n} \subset J_{k,n}(\sin((k+1) {\pi \over 2 n}))$, and hence
$2 {1 \over 4n} \arcsin (\sin(k {\pi \over 2 n})) + { \pi \over 4n} \le m(I_{k,n}) \le 2 {1 \over 4n} \arcsin (\sin((k+1) {\pi \over 2 n})) + { \pi \over 4n}$, or, simplifying a little, we have:
${1 \over 2} {\pi \over 2n}({k \over n}+1) \le m(I_{k,n}) \le {1 \over 2} {\pi \over 2n}({k+1 \over n}+1) $.
Now take sums to get
$\sum_{k=0}^{n-1} {1 \over 2} {\pi \over 2n}({k \over n}+1) \le m(I_n) \le \sum_{k=0}^{n-1} {1 \over 2} {\pi \over 2n}({k+1 \over n}+1)$, and note that if we let $\phi(x) = {1 \over 2} ({2 \over \pi}x+1)$, we can rewrite this as
$  L(\phi,P_k) \le m(I_n) \le   U(\phi,P_k)$, where $L,U$ are the lower and upper Riemann sums respectively, and $P_k$ is the partition $(0,{\pi \over 2n},...,k {\pi \over 2n},...,{\pi \over 2})$.
Since $\phi$ is Riemann integrable, we have $\lim_n m(I_n) =   \int_0^{\pi \over 2} \phi(t)dt = { 3 \pi \over 8}$.
A: Use the definition of the sine on the unit circle to find that $\sin(A)=\sin(B)$ if either $B=A+2\pi k$ or $B=\pi-A+2\pi k$.
A second, more complicated form of this result can be obtained using trig. identities
$$0=\sin(A)-\sin(B)=2\sin\frac{A-B}2\cos\frac{A+B}2$$
But in the end it is really the same.
