<s>[INST] How to find : $\lim_{x \to \pi/2}\frac{1-\sin x}{\sin x+\sin 3x}$ How to find : 
$$\lim_{x \to \pi/2}\frac{1-\sin x}{\sin x+\sin 3x}$$

My Try :
$$x-\pi/2=t \to x=t+\frac{\pi}{2}$$
And:
$$ \sin (t+\frac{\pi}{2})=\cos t \\ \sin 3(t+\frac{\pi}{2})=-\cos 3t $$
So :
$$\lim_{t \to 0}\frac{1-\cos t}{\cos t-\cos 3t}=\\\lim_{t \to 0}\frac{1-\cos t}{t^2} \cdot \frac{t^2}{\cos t-\cos 3t}$$
Now what ?
 [/INST] $$\\ \lim _{ t\to 0 } \frac { 1-\cos  t }{ t^{ 2 } } \cdot \frac { t^{ 2 } }{ \cos  t-\cos  3t } =\lim _{ t\to 0 } \frac { \sin ^{ 2 }{ \frac { t }{ 2 }  }  }{ 2{ \left( \frac { t }{ 2 }  \right)  }^{ 2 } } \cdot \frac { t\cdot t }{ 2\sin { t } \sin { 2t }  } =\\ =\lim _{ t\to 0 } \frac { \sin ^{ 2 }{ \frac { t }{ 2 }  }  }{ 2{ \left( \frac { t }{ 2 }  \right)  }^{ 2 } } \cdot \frac { 1 }{ 4\frac { \sin { t }  }{ t } \frac { \sin { 2t }  }{ 2t }  } =\frac { 1 }{ 8 } \\ $$
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