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 ?

Adding mine to the pile:
$$\require{cancel}\begin{aligned}\frac{1-\cos t}{\cos t\color{blue}{-\cos 3t}}
&=\frac{1-\cos t}{\color{purple}{\cos t}\color{blue}{-\cos^3 t+3\sin^2 t\cos t}}
\\&=\frac{1-\cos t}{\cos t\left(\color{purple}{\sin^2 t+\cancel{\cos^2 t}}-\cancel{\cos^2 t}+3\sin^2 t\right)}
\\&=\frac{1-\cos t}{\color{green}{4\cos t}\sin^2 t}\times\frac{1+\cos t}{\color{orange}{1+\cos t}}
\\&=\frac{1}{\color{green}{4\cos t}\color{orange}{(1+\cos t)}}
\end{aligned}
$$
