Trigonometry and Proportions Numbers are given $a = {\sin1\over \sin2}$, $b = {\sin2\over \sin3}$ and $c = {\sin3\over \sin4}$. Then: 


*

*$a < b < c$

*$c < b < a$

*$c < a < b$


What is the solution and can someone explain me why or just give me a hint. This seems to me that I need to use addition formulas and double angle formulas but I do not know how to find that out?
NOTICE : Angles are not expressed in degrees!
 A: Think of the sine curve:

Recall that $\pi\approx 3.14$ and $\pi/2\approx 1.57$. So $x=1$ lies to the left of $\pi/2$ while $x=2$ lies to the right but before $x=\pi$. However the graph is symmetric about $\pi/2$ between $x=0$ and $x=\pi$ and since $2$ is a little closer to $\pi/2$ than $1$ is, the value of $\sin 2$ is going to be a bit larger than than the value of $\sin 1$. Since both are positive, we can see that $\sin 1/\sin2$ is somewhere between $0$ and $1$. 
Similarly, $x=2$ and $x=3$ are both between $\pi/2$ and $\pi$ and the sine curve is decreasing on this interval. So $\sin2$ is bigger than $\sin3$ which implies $\sin2/\sin3>1$. 
Finally, $\sin3$ is positive and you can see that $\sin4$ is negative since $\pi<4<2\pi$. Therefore $\sin3/\sin4<0$. 
So the answer is $c<a<b$. 
A: $$\sin(\alpha)=\sin(\pi-\alpha)$$and
$$-\frac\pi2<\pi-4<0<\pi-3<1<\pi-2<\dfrac\pi2$$
As the sine is growing in that range,
$$-1<\sin(4)<0<\sin(3)<\sin(1)<\sin(2)<1$$then
$$\sin(1)\sin(3)<\sin^2(2)$$and
$$\frac{\sin(1)}{\sin(2)}<\frac{\sin(2)}{\sin(3)}.$$
The order is $c<a<b$.
A: Assuming $x+1$ is not of the form $n\pi$ ($n\in\Bbb Z$),
$$\displaystyle{f'(x)=\left(\frac{\sin x}{\sin (x+1)}\right)'=\frac{\sin 1}{\sin^2(x+1)}>0}$$(simple to calculate, or see WolframAlpha).   
$f(x)$ is strictly increasing in $(-1,\pi-1)$. Note $f(x)=f(x-\pi)$.  
Since $-1<3-\pi<1<2<\pi-1$, we have $f(3)<f(1)<f(2)$.
A: I want to show that if $1, 2, 3,4$  are in degrees, then $\,a<b<c$ with a pure trigonometric solution.
More generally, if $\alpha,\beta, \gamma,\delta$ is an arithmetic progression  such that $\sin\alpha,\sin\beta, \sin\gamma,\sin\delta>0 $, if we  set: 
$$a=\frac{\sin\alpha}{\sin\beta}, \quad b=\frac{\sin\beta}{\sin\gamma},\quad c=\frac{\sin\gamma}{\sin\delta}, $$
then $ a<b<c$.
Indeed, as $a,b,c>0$, it is enough to prove $\,\dfrac ab,\: \dfrac bc<1$. Let $$\beta=\alpha+r,\enspace \gamma=\alpha+2r=\beta+r,\enspace\delta=\beta+2r.$$
As $\,\dfrac ab=\dfrac{\sin\alpha\sin(\alpha+2r)}{\sin^2(
\alpha+r)}\;$ and $\;\dfrac bc=\dfrac{\sin\beta\sin(\beta+2r)}{\sin^2(
\beta+r)}$, it enough to prove $\dfrac ab <1$.
With the linearisation formulae, we get:
\begin{align*}
\dfrac{\sin\alpha\sin(\alpha+2r)}{\sin^2(\alpha+r)}&=\frac{\frac12(\cos 2r-\cos2(\alpha+r))}{\sin^2(\alpha+r)}=\frac{\frac12(\cos 2r-1)+\sin^2(\alpha+r)}{\sin^2(\alpha+r)}\\&=1+\frac{\cos 2r-1}{2\sin^2(\alpha+r)}<1\quad\text{since}\enspace r\not\equiv0\bmod\pi.
\end{align*}
