<s>[INST] Largest element of the set $\{ \sin{1}, \sin{2}, \sin{3}\}$ I have to find the largest element of the following set $\{ \sin{1}, \sin{2}, \sin{3}\}$.
I converted every element to the first quadrant so I can use the monotony of cosine, the set becomes: 
$$\Big\{ \cos{\frac{\pi-2}{2}}, \cos{\frac{\pi-4}{2}}, \cos{\frac{\pi-6}{2}}\Big\}$$. 
$$\Big|\frac{\pi-6}{2}\Big| > \Big|\frac{\pi-4}{2}\Big|$$
$$\frac{6-\pi}{2} - \frac{\pi-2}{2} > 0  \Rightarrow \frac{6-\pi}{2} > \frac{\pi-2}{2}$$
$$ \frac{4-\pi}{2} - \frac{\pi-2}{2} < 0 \Rightarrow \frac{4-\pi}{2} < \frac{\pi-2}{2}$$
$$ \frac{4-\pi}{2} < \frac{\pi-2}{2} < \frac{6-\pi}{2} \Rightarrow \cos{\frac{\pi-4}{2}} < \cos{\frac{\pi-2}{2}}<  \cos{\frac{\pi-6}{2}}$$ so since cos is monotonically decreesing in the first quadrant, $\cos{\frac{\pi-4}{2}} = \sin 2$ is the biggest element in the set.
Is this correct and is there any easier solution?
 [/INST] (Let's see how rusty my TeX is...)
We could also use trig identities:
$$\sin 2  =  \sin(2·1)  =  2 \sin 1 \cos 1  >  2 \sin 1 \cos \frac{\pi}{3}  =  \sin 1  ,$$
$$\sin 3  =  \sin (2+1)  =  \sin 2 \cos 1  + \cos 2 \sin 1  <  \sin 2 \cos \frac{\pi}{4}  + \cos \frac{\pi}{2} \sin 1  = \frac{\sqrt{2}}{2} \cdot \sin 2 < \sin 2 
$$ 
Thus, sin 2  is the largest of the three values.
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