¿Cómo demostrar la identidad $\frac{1}{\sin(z)} = \cot(z) + \tan(\frac{z}{2})$?

Question

$$\frac{1}{\sin(z)} = \cot (z) + \tan (\tfrac{z}{2})$$

Hice esto:

Primer intento: $$\displaystyle{\frac{1}{\sin (z)} = \frac{\cos (z)}{\sin (z)} + \frac{\sin (\frac{z}{2})}{ \cos (\frac{z}{2})} = \frac{\cos (z) }{\sin (z)} + \frac{2\sin(\frac{z}{4})\cos(\frac{z}{4})}{\cos^{2}(\frac{z}{4})-\sin^{2}(\frac{z}{4})}} = $ $ $$\frac{\cos (z)(\cos^{2}(\frac{z}{4})-\sin^{2}(\frac{z}{4}))+2\sin z \sin(\frac{z}{4})\cos(\frac{z}{4})}{\sin (z)(\cos^{2}(\frac{z}{4})-\sin^{2}(\frac{z}{4}))}$ $

Pegado.

Segundo intento:

$$\displaystyle{\frac{1}{\sin z} = \left(\frac{1}{2i}(e^{iz}-e^{-iz})\right)^{-1} = 2i\left(\frac{1}{e^{iz}-e^{-iz}}\right)}$$

Pegado.

¿Nadie ve un camino a seguir?

Answer 1

Que $w = \frac{z}{2}$. Entonces $$ \cot(2W) + \tan(w) = \frac{\cos^2(w)-\sin^2(w)} {\sin(w) \cos(w) 2} + \frac{\sin(w)}{\cos(w)} = \frac{1}{\cos(w)} \left (\frac{\cos^2(w)-\sin^2(w) + 2 \sin^2(w)} {\sin(w) 2} \right) $$ El numerador se convierte en 1, y llegamos al resultado $\frac{1}{2 \sin(w) \cos(w)} = \frac{1}{\cos(2w)} = \frac{1}{\cos(z)}$.

Answer 2

$$ \frac{\cos (z)}{\sin (z)} + \frac{\sin (\frac{z}{2})}{ \cos (\frac{z}{2})} =\frac{\cos (z)\cos (\frac{z}{2})+ \sin(z)\sin (\frac{z}{2}) }{\sin (z)\cos (\frac{z}{2})} =\frac{\cos (z-\frac{z}{2})}{\sin (z)\cos (\frac{z}{2})}$$

Answer 3

Empieza con $$ \frac{1-\cos(z)}{\sin(z)}=\frac{2\sin^2(\tfrac{z}{2})}{2\sin(\tfrac{z}{2})\cos(\tfrac{z}{2})} = \tan(\tfrac{z}{2})\tag {1} $$ y $\cot(z)$ a ambos lados: $$ \frac{1}{\sin(z)} = \cot (z) + \tan(\tfrac{z}{2})\tag {2} $$

Answer 4

Voy a ir hacia atrás; Espero que no te importa.

$$\begin{align*}\cot\,z+\tan\frac{z}{2}&=\frac{\cos\,z}{\sin\,z}+\frac{\sin\,z}{1+\cos\,z}\\&=\frac{\sin^2 z+(1+\cos\,z)\cos\,z}{(1+\cos\,z)\sin\,z}\\&=\frac{\cos^2 z+\sin^2 z+\cos\,z}{(1+\cos\,z)\sin\,z}\\&=\frac{1+\cos\,z}{(1+\cos\,z)\sin\,z}\\&=\csc\,z\end{align*}$$