Identidades trigonométricas III. Ejercicios con soluciones.

Aguí tienes otra entrada de ejercicios de identidades trigonométricas por si los ejercicios de esta entrada te han parecido difíciles.

La primera, si te das cuenta $a=\alpha +\beta$ y entonces tenemos:

${\displaystyle \frac{\mathrm{tg}(a)-\mathrm{tg}\alpha }{1+\mathrm{tg}(a)\cdot \mathrm{tg}\alpha }}=\mathrm{tg}(a-\alpha )=\mathrm{tg}(\alpha +\beta -\alpha )=\mathrm{tg}(\beta )$

La segunda, desarrollamos la fórmula de la tangente de la suma de dos ángulos:

${\displaystyle \frac{\mathrm{tg}(\alpha +\beta )-\mathrm{tg}\alpha }{1+\mathrm{tg}(\alpha +\beta )\cdot \mathrm{tg}\alpha }}={\displaystyle \frac{{\displaystyle \frac{\mathrm{tg}\alpha +\mathrm{tg}\beta }{1-\mathrm{tg}\alpha \cdot \mathrm{tg}\beta }}-\mathrm{tg}\alpha }{1+{\displaystyle \frac{(\mathrm{tg}\alpha +\mathrm{tg}\beta )\cdot \mathrm{tg}\alpha }{1-\mathrm{tg}\alpha \cdot \mathrm{tg}\beta }}}}={\displaystyle \frac{{\displaystyle \frac{\mathrm{tg}\alpha +\mathrm{tg}\beta -\mathrm{tg}\alpha (1-\mathrm{tg}\alpha \cdot \mathrm{tg}\beta )}{1-\mathrm{tg}\alpha \cdot \mathrm{tg}\beta }}}{{\displaystyle \frac{1-\mathrm{tg}\alpha \cdot \mathrm{tg}\beta +(\mathrm{tg}\alpha +\mathrm{tg}\beta )\cdot \mathrm{tg}\alpha }{1-\mathrm{tg}\alpha \cdot \mathrm{tg}\beta }}}}=$


$={\displaystyle \frac{{\displaystyle \frac{\mathrm{tg}\alpha +\mathrm{tg}\beta -\mathrm{tg}\alpha (1-\mathrm{tg}\alpha \cdot \mathrm{tg}\beta )}{\cancel{1-\mathrm{tg}\alpha \cdot \mathrm{tg}\beta }}}}{{\displaystyle \frac{1-\mathrm{tg}\alpha \cdot \mathrm{tg}\beta +(\mathrm{tg}\alpha +\mathrm{tg}\beta )\cdot \mathrm{tg}\alpha }{\cancel{1-\mathrm{tg}\alpha \cdot \mathrm{tg}\beta }}}}}={\displaystyle \frac{\mathrm{tg}\alpha +\mathrm{tg}\beta -\mathrm{tg}\alpha +{\mathrm{tg}}^2\alpha \cdot \mathrm{tg}\beta }{1-\mathrm{tg}\alpha \cdot \mathrm{tg}\beta +{\mathrm{tg}}^2\alpha +\mathrm{tg}\beta \mathrm{tg}\alpha }}={\displaystyle \frac{\mathrm{tg}\beta \cdot (1+{\mathrm{tg}}^2\alpha )}{1+{\mathrm{tg}}^2\alpha }}=$

$={\displaystyle \frac{\mathrm{tg}\beta \cdot \cancel{(1+{\mathrm{tg}}^2\alpha )}}{\cancel{1+{\mathrm{tg}}^2\alpha }}}=\mathrm{tg}\beta =$

$\mathrm{sen}3\alpha =\mathrm{sen}(2\alpha +\alpha )=\mathrm{sen}2\alpha \cdot \cos \alpha +\cos 2\alpha \cdot \mathrm{sen}\alpha =$

$=2\mathrm{sen}\alpha \cdot {\cos }^2\alpha +({\cos }^2\alpha -{\mathrm{sen}}^2\alpha )\cdot \mathrm{sen}\alpha =$

$=2\mathrm{sen}\alpha \cdot (1-{\mathrm{sen}}^2\alpha )+(1-{\mathrm{sen}}^2\alpha -{\mathrm{sen}}^2\alpha )\cdot \mathrm{sen}\alpha =$

$=2\mathrm{sen}\alpha -2{\mathrm{sen}}^3\alpha +\mathrm{sen}\alpha -2{\mathrm{sen}}^3\alpha =$

$=3\mathrm{sen}\alpha -4{\mathrm{sen}}^3\alpha$

$\cos 4\alpha =\cos (2\alpha +2\alpha )=$

$={\cos }^{2}2\alpha -{\mathrm{sen}}^{2}2\alpha =$

$=({\cos }^2\alpha -{\mathrm{sen}}^2\alpha {)}^2-(2\cdot \mathrm{sen}\alpha \cdot \cos \alpha {)}^2=$

$=({\cos }^2\alpha -(1-{\cos }^2\alpha ){)}^2-4\cdot {\mathrm{sen}}^2\alpha \cdot {\cos }^2\alpha =$

$=(2{\cos }^2\alpha -1{)}^2-4\cdot (1-{\cos }^2\alpha )\cdot {\cos }^2\alpha =$

$=4{\cos }^4\alpha -4{\cos }^2\alpha +1-4{\cos }^2\alpha +4\cdot {\cos }^4\alpha =$

$=8{\cos }^4\alpha -8{\cos }^2\alpha +1$

${\displaystyle \frac{\cos \alpha +\mathrm{sen}\alpha }{\cos \alpha -\mathrm{sen}\alpha }}-{\displaystyle \frac{\cos \alpha -\mathrm{sen}\alpha }{\cos \alpha +\mathrm{sen}\alpha }}={\displaystyle \frac{(\cos \alpha +\mathrm{sen}\alpha {)}^2}{(\cos \alpha -\mathrm{sen}\alpha )\cdot (\cos \alpha +\mathrm{sen}\alpha )}}-{\displaystyle \frac{(\cos \alpha -\mathrm{sen}\alpha {)}^2}{(\cos \alpha +\mathrm{sen}\alpha )\cdot (\cos \alpha -\mathrm{sen}\alpha )}}=$

$={\displaystyle \frac{({\cos }^2\alpha +2\cdot \mathrm{sen}\alpha \cdot \cos \alpha +{\mathrm{sen}}^2\alpha )}{{\cos }^2\alpha -{\mathrm{sen}}^2\alpha }}-{\displaystyle \frac{({\cos }^2\alpha -2\cdot \mathrm{sen}\alpha \cdot \cos \alpha +{\mathrm{sen}}^2\alpha )}{{\cos }^2\alpha -{\mathrm{sen}}^2\alpha }}=$

$={\displaystyle \frac{1+2\cdot \mathrm{sen}\alpha \cdot \cos \alpha }{{\cos }^2\alpha -{\mathrm{sen}}^2\alpha }}-{\displaystyle \frac{1-2\cdot \mathrm{sen}\alpha \cdot \cos \alpha }{{\cos }^2\alpha -{\mathrm{sen}}^2\alpha }}=$

$={\displaystyle \frac{\cancel{1}+2\cdot \mathrm{sen}\alpha \cdot \cos \alpha }{{\cos }^2\alpha -{\mathrm{sen}}^2\alpha }}-{\displaystyle \frac{\cancel{1}-2\cdot \mathrm{sen}\alpha \cdot \cos \alpha }{{\cos }^2\alpha -{\mathrm{sen}}^2\alpha }}={\displaystyle \frac{4\cdot \mathrm{sen}\alpha \cdot \cos \alpha }{{\cos }^2\alpha -{\mathrm{sen}}^2\alpha }}=$

$={\displaystyle \frac{2\cdot 2\cdot \mathrm{sen}\alpha \cdot \cos \alpha }{{\cos }^2\alpha -{\mathrm{sen}}^2\alpha }}={\displaystyle \frac{2\cdot 2\mathrm{sen}2\alpha }{\cos 2\alpha }}=2\cdot \mathrm{tg}2\alpha$

${\displaystyle \frac{3}{8}}-{\displaystyle \frac{1}{2}}\cdot \cos 2\alpha +{\displaystyle \frac{1}{8}}\cdot \cos 4\alpha =$ \ \ $={\displaystyle \frac{3}{8}}-{\displaystyle \frac{1}{2}}\cdot ({\cos }^2\alpha -{\mathrm{sen}}^2\alpha )+{\displaystyle \frac{1}{8}}\cdot ({\cos }^2\left(2\alpha \right)-{\mathrm{sen}}^2\left(2\alpha \right))=$

$={\displaystyle \frac{3}{8}}-{\displaystyle \frac{1}{2}}\cdot (1-2{\mathrm{sen}}^2\alpha )+{\displaystyle \frac{1}{8}}\cdot {\cos }^2\left(2\alpha \right)-{\displaystyle \frac{1}{8}}\cdot {\mathrm{sen}}^2\left(2\alpha \right)=$

$={\displaystyle \frac{3}{8}}-{\displaystyle \frac{1}{2}}+{\mathrm{sen}}^2\alpha +{\displaystyle \frac{1}{8}}\cdot {({\cos }^2\alpha -{\mathrm{sen}}^2\alpha )}^2-{\displaystyle \frac{1}{8}}\cdot 4{\mathrm{sen}}^2\alpha {\cos }^2\cdot \alpha =$

$={\displaystyle \frac{3}{8}}-{\displaystyle \frac{1}{2}}+{\mathrm{sen}}^2\alpha +{\displaystyle \frac{1}{8}}\cdot {(1-2\cdot {\mathrm{sen}}^2\alpha )}^2-{\displaystyle \frac{1}{8}}\cdot 4{\mathrm{sen}}^2\alpha \cdot (1-{\mathrm{sen}}^2\alpha )=$

$={\displaystyle \frac{3}{8}}-{\displaystyle \frac{1}{2}}+{\mathrm{sen}}^2\alpha +{\displaystyle \frac{1}{8}}\cdot (1-4\cdot {\mathrm{sen}}^2\alpha +4\cdot {\mathrm{sen}}^4\alpha )-{\displaystyle \frac{1}{2}}\cdot {\mathrm{sen}}^2\alpha +{\displaystyle \frac{1}{2}}\cdot {\mathrm{sen}}^4\alpha =$

$={\displaystyle \frac{3}{8}}-{\displaystyle \frac{1}{2}}+{\mathrm{sen}}^2\alpha +{\displaystyle \frac{1}{8}}-{\displaystyle \frac{1}{2}}\cdot {\mathrm{sen}}^2\alpha +{\displaystyle \frac{1}{2}}\cdot {\mathrm{sen}}^4\alpha -{\displaystyle \frac{1}{2}}\cdot {\mathrm{sen}}^2\alpha +{\displaystyle \frac{1}{2}}\cdot {\mathrm{sen}}^4\alpha =$

$={\displaystyle \frac{3}{8}}-{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{8}}+{\mathrm{sen}}^2\alpha -{\displaystyle \frac{1}{2}}\cdot {\mathrm{sen}}^2\alpha -{\displaystyle \frac{1}{2}}\cdot {\mathrm{sen}}^2\alpha +{\displaystyle \frac{1}{2}}\cdot {\mathrm{sen}}^4\alpha +{\displaystyle \frac{1}{2}}\cdot {\mathrm{sen}}^4\alpha =$

$={\displaystyle \frac{3}{8}}-{\displaystyle \frac{4}{8}}+{\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{2}}\cdot {\mathrm{sen}}^4\alpha +{\displaystyle \frac{1}{2}}\cdot {\mathrm{sen}}^4\alpha =$

$={\mathrm{sen}}^4\alpha$

$\cos 3\alpha =\cos (2\alpha +\alpha )=$

$=\cos 2\alpha \cdot \cos \alpha -\mathrm{sen}2\alpha \cdot \mathrm{sen}\alpha =$

$=({\cos }^2\alpha -{\mathrm{sen}}^2\alpha )\cdot \cos \alpha -2\cdot {\mathrm{sen}}^2\alpha \cdot \cos \alpha =$

$=(2\cdot {\cos }^2\alpha -1)\cdot \cos \alpha -2\cdot (1-{\cos }^2\alpha )\cdot \cos \alpha =$

$=2\cdot {\cos }^3\alpha -\cos \alpha -2\cdot \cos \alpha +2\cdot {\cos }^3\alpha =$

$=4\cdot {\cos }^3\alpha -3\cdot \cos \alpha$

$\mathrm{sen}\alpha =2\cdot \mathrm{sen}{\displaystyle \frac{\alpha }{2}}\cdot \cos {\displaystyle \frac{\alpha }{2}}={\displaystyle \frac{2\cdot \mathrm{sen}{\displaystyle \frac{\alpha }{2}}\cdot \cos {\displaystyle \frac{\alpha }{2}}}{1}}={\displaystyle \frac{2\cdot \mathrm{sen}{\displaystyle \frac{\alpha }{2}}\cdot \cos {\displaystyle \frac{\alpha }{2}}}{{\cos }^2{\displaystyle \frac{\alpha }{2}}+{\mathrm{sen}}^2{\displaystyle \frac{\alpha }{2}}}}=$

$={\displaystyle \frac{{\displaystyle \frac{2\cdot \mathrm{sen}{\displaystyle \frac{\alpha }{2}}\cdot \cos {\displaystyle \frac{\alpha }{2}}}{{\cos }^2{\displaystyle \frac{\alpha }{2}}}}}{{\displaystyle \frac{{\cos }^2{\displaystyle \frac{\alpha }{2}}+{\mathrm{sen}}^2{\displaystyle \frac{\alpha }{2}}}{{\cos }^2{\displaystyle \frac{\alpha }{2}}}}}}={\displaystyle \frac{{\displaystyle \frac{2\cdot \mathrm{sen}{\displaystyle \frac{\alpha }{2}}\cdot \cancel{\cos {\displaystyle \frac{\alpha }{2}}}}{{\cos }^{\cancel{2}}{\displaystyle \frac{\alpha }{2}}}}}{{\displaystyle \frac{{\cos }^2{\displaystyle \frac{\alpha }{2}}+{\mathrm{sen}}^2{\displaystyle \frac{\alpha }{2}}}{{\cos }^2{\displaystyle \frac{\alpha }{2}}}}}}=$

$={\displaystyle \frac{{\displaystyle \frac{2\cdot \mathrm{sen}{\displaystyle \frac{\alpha }{2}}}{\cos {\displaystyle \frac{\alpha }{2}}}}}{1+{\mathrm{tg}}^2{\displaystyle \frac{\alpha }{2}}}}={\displaystyle \frac{2\cdot \mathrm{tg}{\displaystyle \frac{\alpha }{2}}}{1+{\mathrm{tg}}^2{\displaystyle \frac{\alpha }{2}}}}$

Primero vamos a desarrollar $\mathrm{sen}3\alpha$ y $\cos 3\alpha$:

$$\mathrm{sen}3\alpha =\mathrm{sen}(\alpha +2\alpha )=\mathrm{sen}\alpha \cdot \cos 2\alpha +\cos \alpha \cdot \mathrm{sen}2\alpha$$ $$\cos 3\alpha =\cos (\alpha +2\alpha )=\cos \alpha \cdot \cos 2\alpha -\mathrm{sen}\alpha \cdot \mathrm{sen}2\alpha$$
${\displaystyle \frac{\mathrm{sen}3\alpha }{\mathrm{sen}\alpha }}-{\displaystyle \frac{\cos 3\alpha }{\cos \alpha }}=$

$={\displaystyle \frac{\mathrm{sen}\alpha \cdot \cos 2\alpha +\cos \alpha \cdot \mathrm{sen}2\alpha }{\mathrm{sen}\alpha }}-{\displaystyle \frac{\cos \alpha \cdot \cos 2\alpha -\mathrm{sen}\alpha \cdot \mathrm{sen}2\alpha }{\cos \alpha }}=$

$={\displaystyle \frac{\mathrm{sen}\alpha \cdot \cos 2\alpha }{\mathrm{sen}\alpha }}+{\displaystyle \frac{\cos \alpha \cdot \mathrm{sen}2\alpha }{\mathrm{sen}\alpha }}-{\displaystyle \frac{\cos \alpha \cdot \cos 2\alpha }{\cos \alpha }}+{\displaystyle \frac{\mathrm{sen}\alpha \cdot \mathrm{sen}2\alpha }{\cos \alpha }}=$

$={\displaystyle \frac{\cancel{\mathrm{sen}\alpha }\cdot \cos 2\alpha }{\cancel{\mathrm{sen}\alpha }}}+{\displaystyle \frac{\cos \alpha \cdot \mathrm{sen}2\alpha }{\mathrm{sen}\alpha }}-{\displaystyle \frac{\cancel{\cos \alpha }\cdot \cos 2\alpha }{\cancel{\cos \alpha }}}+{\displaystyle \frac{\mathrm{sen}\alpha \cdot \mathrm{sen}2\alpha }{\cos \alpha }}=$

$=\cos 2\alpha +\mathrm{cotg}\alpha \cdot \mathrm{sen}2\alpha -\cos 2\alpha +\mathrm{tg}\alpha \cdot \mathrm{sen}2\alpha =$

$=\cancel{\cos 2\alpha }+\mathrm{cotg}\alpha \cdot \mathrm{sen}2\alpha -\cancel{\cos 2\alpha }+\mathrm{tg}\alpha \cdot \mathrm{sen}2\alpha =$

$=\mathrm{cotg}\alpha \cdot \mathrm{sen}2\alpha +\mathrm{tg}\alpha \cdot \mathrm{sen}2\alpha =$

$=\mathrm{sen}2\alpha \cdot (\mathrm{cotg}\alpha +\mathrm{tg}\alpha )=$

$=\mathrm{sen}2\alpha \cdot ({\displaystyle \frac{1}{\mathrm{tg}\alpha }}+\mathrm{tg}\alpha )=$

$=\mathrm{sen}2\alpha \cdot \left({\displaystyle \frac{1+{\mathrm{tg}}^2\alpha }{\mathrm{tg}\alpha }}\right)=$

$=2\cdot \mathrm{sen}\alpha \cdot \cos \alpha \left({\displaystyle \frac{{\displaystyle \frac{1}{{\cos }^{\cancel{2}}\alpha }}}{{\displaystyle \frac{\mathrm{sen}\alpha }{\cancel{\cos \alpha }}}}}\right)=$

$=2\cdot \mathrm{sen}\alpha \cdot \cos \alpha \cdot \left({\displaystyle \frac{1}{\cos \alpha \cdot \mathrm{sen}\alpha }}\right)=$

$=2\cdot \cancel{\mathrm{sen}\alpha \cdot \cos \alpha }\cdot \left({\displaystyle \frac{1}{\cancel{\cos \alpha \cdot \mathrm{sen}\alpha }}}\right)=2$

En un triángulo sabemos que $A+B+C=\pi$ Luego

${\mathrm{sen}}^{2}A+{\mathrm{sen}}^{2}B+{\mathrm{sen}}^2[\pi -(A+B)]=2$
$$\text{si los ángulos son suplementarios }\mathrm{sen}[\pi -(A+B)]=\mathrm{sen}(A+B)$$ ${\mathrm{sen}}^{2}A+{\mathrm{sen}}^{2}B+{\mathrm{sen}}^2(A+B)=2$

${\mathrm{sen}}^{2}A+{\mathrm{sen}}^{2}B+{[\mathrm{sen}A\cos B+\cos A\mathrm{sen}B]}^2=2$

${\mathrm{sen}}^{2}A+{\mathrm{sen}}^{2}B+{\mathrm{sen}}^{2}A{\cos }^{2}B+{\cos }^{2}A{\mathrm{sen}}^{2}B+2\cos A\cos B\mathrm{sen}A\mathrm{sen}B=2$

${\mathrm{sen}}^{2}A+{\mathrm{sen}}^{2}B+{\mathrm{sen}}^{2}A(1-{\mathrm{sen}}^{2}B)+{\mathrm{sen}}^{2}B(1-{\mathrm{sen}}^{2}A)+2\cos A\cos B\mathrm{sen}A\mathrm{sen}B=2$

${\mathrm{sen}}^{2}A+{\mathrm{sen}}^{2}B+{\mathrm{sen}}^{2}A-{\mathrm{sen}}^{2}A{\mathrm{sen}}^{2}B+{\mathrm{sen}}^{2}B-{\mathrm{sen}}^{2}B{\mathrm{sen}}^{2}A+2\cos A\cos B\mathrm{sen}A\mathrm{sen}B=2$

$2{\mathrm{sen}}^{2}A+2{\mathrm{sen}}^{2}B-2{\mathrm{sen}}^{2}A{\mathrm{sen}}^{2}B+2\cos A\cos B\mathrm{sen}A\mathrm{sen}B=2$
$$\text{simplificando por 2}$$ ${\mathrm{sen}}^{2}A+{\mathrm{sen}}^{2}B-{\mathrm{sen}}^{2}A{\mathrm{sen}}^{2}B+\cos A\cos B\mathrm{sen}A\mathrm{sen}B=1$

${\mathrm{sen}}^{2}A+{\mathrm{sen}}^{2}B-{\mathrm{sen}}^{2}A{\mathrm{sen}}^{2}B+\cos A\cos B\mathrm{sen}A\mathrm{sen}B={\cos }^{2}A+{\mathrm{sen}}^{2}A$

${\mathrm{sen}}^{2}B-{\mathrm{sen}}^{2}A{\mathrm{sen}}^{2}B+\cos A\cos B\mathrm{sen}A\mathrm{sen}B={\cos }^{2}A$

${\mathrm{sen}}^{2}B(1-{\mathrm{sen}}^{2}A)+\cos A\cos B\mathrm{sen}A\mathrm{sen}B={\cos }^{2}A$

${\mathrm{sen}}^{2}B{\cos }^{2}A+\cos A\cos B\mathrm{sen}A\mathrm{sen}B={\cos }^{2}A$

$\cos A\cos B\mathrm{sen}A\mathrm{sen}B={\cos }^{2}A(1-{\mathrm{sen}}^{2}B)$

$\cos A\cos B\mathrm{sen}A\mathrm{sen}B={\cos }^{2}A{\cos }^{2}B$

$\cos A\cos B(\mathrm{sen}A\mathrm{sen}B-\cos A\cos B)=0$

$\cos A\cos B[-\cos (A+B)]=0$

Luego pueden ocurrir tres cosas:

$1^{\underset{―}{a}}\cos A=0\Rightarrow A={\displaystyle \frac{\pi }{2}}$, el triángulo es rectángulo.

$2^{\underset{―}{a}}\cos B=0\Rightarrow B={\displaystyle \frac{\pi }{2}}$, el triángulo es rectángulo.

$3^{\underset{―}{a}}\cos (A+B)=0\Rightarrow A+B={\displaystyle \frac{\pi }{2}}\Rightarrow C={\displaystyle \frac{\pi }{2}}$, el triángulo es rectángulo.

Iremos actualizando esta sección con los ejercicios que nos pidáis, para ello podéis mandar un correo a profesor.maties@gmail.com