How can one show that [math](\sin x)^2 < \sin(x^2)[/math] for [math] 0 < x < { \left( \frac { \pi  }{ 2 }  \right)  }^{ \frac { 1 }{ 2 }  } [/math] ?
How one can prove that [math]\sin\left(\dfrac{\pi}{2}-x\right)=\sin \left(x+\dfrac{\pi}{2}\right)[/math], from [math]\sin(-x)=\sin x[/math]?