How to find : $\lim_{x \to \pi/2}\frac{1-\sin x}{\sin x+\sin 3x}$

Question

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 ?

Answer 1

$$\lim_{x\to \pi/2}\frac{1-\sin x}{\sin x + \sin 3x}=\lim_{x\to \pi/2}\frac{1-\sin x}{2\sin 2x\cos x}$$ $$=\lim_{x\to \pi/2}\frac{1-\sin x}{4\sin x \cos^2 x}$$ $$=\lim_{x\to \pi/2}\frac{1-\sin x}{4\sin x (1+\sin x)(1-\sin x)}$$ $$=\lim_{x\to \pi/2}\frac{1}{4\sin x(1+\sin x)}$$ $$=\frac{1}{8}$$

Answer 2

$$\\ \lim _{ t\to 0 } \frac { 1-\cos t }{ t^{ 2 } } \cdot \frac { t^{ 2 } }{ \cos t-\cos 3t } =\lim _{ t\to 0 } \frac { \sin ^{ 2 }{ \frac { t }{ 2 } } }{ 2{ \left( \frac { t }{ 2 } \right) }^{ 2 } } \cdot \frac { t\cdot t }{ 2\sin { t } \sin { 2t } } =\\ =\lim _{ t\to 0 } \frac { \sin ^{ 2 }{ \frac { t }{ 2 } } }{ 2{ \left( \frac { t }{ 2 } \right) }^{ 2 } } \cdot \frac { 1 }{ 4\frac { \sin { t } }{ t } \frac { \sin { 2t } }{ 2t } } =\frac { 1 }{ 8 } \\ $$

Answer 3

Adding mine to the pile:

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

Answer 4

Some elementary trigonometry will suffice: all you have to know is $$\sin 3x=3\sin x-4\sin^3x.$$ Indeed, this relation yields $$\frac{1-\sin x}{\sin x-\sin 3x}=\frac{1-\sin x}{4\sin x-4\sin^3x}=\frac{1-\sin x}{4\sin x(1-\sin^2x)}=\frac1{4\sin x(1+\sin x)}\to\frac18$$