Expresiones, identidades y ecuaciones trigonométricas

\[\frac{\cos x+\text{sen}\,x}{\cos x-\text{sen}\,x}-\frac{\cos x-\text{sen}\,x}{\cos x+\text{sen}\,x}=\]

\[=\frac{\left( \cos x+\text{sen}\,x \right)\left( \cos x+\text{sen}\,x \right)}{\left( \cos x-\text{sen}\,x \right)\left( \cos x+\text{sen}\,x \right)}-\frac{\left( \cos x-\text{sen}\,x \right)\left( \cos x-\text{sen}\,x \right)}{\left( \cos x-\text{sen}\,x \right)\left( \cos x+\text{sen}\,x \right)}=\]

\[=\frac{{{\cos }^{2}}x+2\cos x\,\text{sen}\,x+\text{se}{{\text{n}}^{2}}x}{\left( \cos x-\text{sen}\,x \right)\left( \cos x+\text{sen}\,x \right)}-\frac{{{\cos }^{2}}x-2\cos x\,\text{sen}\,x+\text{se}{{\text{n}}^{2}}x}{\left( \cos x-\text{sen}\,x \right)\left( \cos x+\text{sen}\,x \right)}=\]

\[=\frac{2\cos x\,\text{sen}\,x+2\cos x\,\text{sen}\,x}{\left( \cos x-\text{sen}\,x \right)\left( \cos x+\text{sen}\,x \right)}=\]

\[=\frac{\text{sen}\,2x+\text{sen}\,2x}{{{\cos }^{2}}x-\text{se}{{\text{n}}^{2}}x}=\frac{2\,\text{sen}\,2x}{\cos 2x}=2\,\text{tg}\,2x\]

\[\frac{1+\text{t}{{\text{g}}^{2}}x}{\text{cotg}\,x}=\frac{1+\displaystyle\frac{\text{se}{{\text{n}}^{2}}\,x}{{{\cos }^{2}}x}}{\displaystyle\frac{\cos x}{\text{sen}\,x}}=\frac{\displaystyle\frac{{{\cos }^{2}}x+\text{se}{{\text{n}}^{2}}\,x}{{{\cos }^{2}}x}}{\displaystyle\frac{\cos x}{\text{sen}\,x}}=\frac{\displaystyle\frac{1}{{{\cos }^{2}}x}}{\displaystyle\frac{\cos x}{\text{sen}\,x}}=\]

\[=\frac{\text{sen}\,x}{\cos x\cdot {{\cos }^{2}}x}=\frac{\text{sen}\,x}{\cos x}\cdot \frac{1}{\cos {{x}^{2}}}=\frac{\text{tg}\,x}{{{\cos }^{2}}x}\]

\[\frac{\text{sen}\,\alpha +\text{cotg}\,\alpha }{\text{tg}\,\alpha +\text{cosec}\,\alpha }=\frac{\text{sen}\,\alpha +\displaystyle\frac{\cos \alpha }{\text{sen}\,\alpha }}{\displaystyle\frac{\text{sen}\,\alpha }{\cos \alpha }+\frac{1}{\text{sen}\,\alpha }}=\frac{\displaystyle\frac{\text{se}{{\text{n}}^{2}}\,\alpha +\cos \alpha }{\text{sen}\,\alpha }}{\displaystyle\frac{\text{se}{{\text{n}}^{2}}\,\alpha +\cos \alpha }{\cos \alpha \,\text{sen}\,\alpha }}=\]

\[=\frac{\left( \text{se}{{\text{n}}^{2}}\,\alpha +\cos \alpha \right)\cos \alpha \,\text{sen}\,\alpha }{\left( \text{se}{{\text{n}}^{2}}\,\alpha +\cos \alpha \right)\,\text{sen}\,\alpha }=\cos \alpha\]

\[2\,\text{tg}\,\alpha \cdot {{\cos }^{2}}\frac{\alpha }{2}-\text{sen}\,\alpha =2\frac{\text{sen}\,\alpha }{\cos \alpha }\cdot \frac{1+\cos \alpha }{2}-\text{sen}\,\alpha =\]

\[=\frac{\text{sen}\,\alpha \left( 1+\cos \alpha \right)}{\cos \alpha }-\text{sen}\,\alpha =\frac{\text{sen}\,\alpha +\text{sen}\,\alpha \cos \alpha }{\cos \alpha }-\frac{\text{sen}\,\alpha \cos \alpha }{\cos \alpha }=\]

\[=\frac{\text{sen}\,\alpha }{\cos \alpha }=\text{tg}\,\alpha\]

\[\text{tg}\,x+2\text{sen}\,x=0\Rightarrow \frac{\text{sen}\,x}{\cos x}+2\text{sen}\,x=0\Rightarrow\]

\[\Rightarrow\text{sen}\,x+2\text{sen}\,x\cos x=0\Rightarrow \text{sen}\,x\left( 1+2\cos x \right)=0\Rightarrow\]

\[\Rightarrow \left\{ \begin{align} & \text{sen}\,x=0\Rightarrow x=0{}^\text{o}\,\,;\,\,x=180{}^\text{o} \\ & \cos x=-\frac{1}{2}\Rightarrow x=120{}^\text{o}\,\,;\,\,x=240{}^\text{o} \\ \end{align} \right. \]

\[\text{sen }x\cdot \text{tg}\,x=\frac{\sqrt{3}}{6}\Rightarrow \text{sen}\,x\cdot \frac{\text{sen}\,x}{\cos x}=\frac{\sqrt{3}}{6}\Rightarrow\]

\[\Rightarrow 6\,\text{se}{{\text{n}}^{2}}\,x=\sqrt{3}\cos x\Rightarrow 6\left( 1-{{\cos }^{2}}x \right)=\sqrt{3}\cos x\Rightarrow\]

\[\Rightarrow 6-6{{\cos }^{2}}x=\sqrt{3}\cos x\Rightarrow 6{{\cos }^{2}}x+\sqrt{3}\cos x-6=0\]

El discriminante de la ecuación anterior es \(\sqrt{3}^2-4\cdot6\cdot(-6)=3+144=147\). Por tanto:

\[\cos x=\frac{-\sqrt{3}\pm \sqrt{147}}{12}=\frac{-\sqrt{3}\pm \sqrt{{{7}^{2}}\cdot 3}}{12}=\]

\[=\frac{-\sqrt{3}\pm 7\sqrt{3}}{12}=\left\{ \begin{align} & \frac{6\sqrt{3}}{12}=\frac{\sqrt{3}}{2} \\ & \frac{-8\sqrt{3}}{12}=\frac{-2\sqrt{3}}{3} \\ \end{align} \right. \]

Si \(\cos x=\dfrac{\sqrt{3}}{2}\), entonces \($x=30{}^\text{o}\,\,;\,\,x=330{}^\text{o}$\).

Si \(\cos x=\frac{-2\sqrt{3}}{3}\), entonces no existe solución para \(x\) pues \(\dfrac{-2\sqrt{3}}{3}\cong -1,15\), y el coseno no puede ser un número menor que \(-1\).

Dividiendo ambas ecuaciones tenemos:

\[\frac{\text{sen}\,x}{\cos x}=\frac{\displaystyle\frac{\sqrt{2}}{4}}{\displaystyle\frac{\sqrt{6}}{4}}\Rightarrow \text{tg}\,x=\frac{4\sqrt{2}}{4\sqrt{6}}\Rightarrow \text{tg}\,x=\sqrt{\frac{2}{6}}\Rightarrow\]

\[\Rightarrow\text{tg}\,x=\sqrt{\frac{1}{3}}\Rightarrow \text{tg}\,x=\frac{\sqrt{3}}{3}\Rightarrow x=30{}^\text{o}\]

Sustituyendo en la primera ecuación:

\[\text{sen}\,30{}^\text{o}\cdot \text{sen}\,y=\frac{\sqrt{2}}{4}\Rightarrow \frac{1}{2}\text{sen}\,y=\frac{\sqrt{2}}{4}\Rightarrow \text{sen}\,y=\frac{2\sqrt{2}}{4}\Rightarrow\]

\[\Rightarrow\text{sen}\,y=\frac{\sqrt{2}}{2}\Rightarrow y=45{}^\text{o}\]