Prueba $ \frac{\sin^3(x)-\cos^3(x)}{\sin(x)+\cos(x)} = \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $

Question

Pregunta: Prueba $$ \frac{\sin^3(x)-\cos^3(x)}{\sin(x)+\cos(x)} = \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $$

RHS: $$ \frac{\csc^2(x) -\cot(x) -2\cos^2(x)}{1-\cot^2(x)} $$

$$ \frac{\frac{1}{\sin^2(x)} +\frac{\cos(x)}{\sin(x)}-2\cos^2(x)}{1-\frac{\cos^2(x)}{\sin^2(x)}} $$

$$ \frac{\frac{1}{\sin^2(x)} +\frac{\cos(x)\sin(x)}{\sin^2(x)}-\frac{2\cos^2(x)\sin^2(x)}{\sin^2(x)}}{\frac{\sin^2(x)-\cos^2(x)}{\sin^2(x)}} $$

$$ \frac{\frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{\sin^2(x)}}{\frac{\sin^2(x)-\cos^2(x)}{\sin^2(x)}} $$

$$ \frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{\sin^2(x)} \times {\frac{\sin^2(x)}{\sin^2(x)-\cos^2(x)}} $$

$$ \frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{\sin^2(x)-\cos^2(x)} $$

$$ \frac{1+\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{(\sin(x)-\cos(x))(\sin(x)+\cos(x))} $$

Ahora estoy atascado

Answer 1

Está usted en lo cierto excepto en un signo : $$\frac{1\color{red}{-}\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)}{(\sin(x)-\cos(x))(\sin(x)+\cos(x))}$$ (El error comienza al principio).

Ahora usando $1=\cos^2(x)+\sin^2(x)$ , $$\begin{align}&1-\cos(x)\sin(x)-2\cos^2(x)\sin^2(x)\\&=(1-2\cos(x)\sin(x))(1+\cos(x)\sin(x))\\&=(\cos^2(x)+\sin^2(x)-2\sin(x)\cos(x))(\cos^2(x)+\sin^2(x)+\cos(x)\sin(x))\\&=(\sin(x)-\cos(x))(\sin(x)-\cos(x))(\sin^2(x)+\sin(x)\cos(x)+\cos^2(x))\\&=(\sin(x)-\cos(x))(\sin^3(x)-\cos^3(x))\end{align}$$

Answer 2

SUGERENCIA:

Escribir $\cos x\sin x=u$

$$1-u-2u^2=(1-2u)(1+u)$$

$1-2u=(\sin x-\cos x)^2$