Ejercicio resuelto de Razones trigonométricas · Matemáticas I

Si $\mathrm{tg}a={\displaystyle \frac{3}{4}}$, hallar $\mathrm{tg}(a+{30}^{\circ })$ y $\mathrm{tg}({45}^{\circ }-a)$.

Se utilizan las fórmulas de la tangente de la suma y de la diferencia de dos ángulos:

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

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

Como dato se tiene $\mathrm{tg}a={\displaystyle \frac{3}{4}}$, y son conocidos los valores $\mathrm{tg}{30}^{\circ }={\displaystyle \frac{\sqrt{3}}{3}}$ y $\mathrm{tg}{45}^{\circ }=1$.

Cálculo de $\mathrm{tg}(a+{30}^{\circ })$:

$\mathrm{tg}(a+{30}^{\circ })={\displaystyle \frac{\mathrm{tg}a+\mathrm{tg}{30}^{\circ }}{1-\mathrm{tg}a\mathrm{tg}{30}^{\circ }}}={\displaystyle \frac{{\displaystyle \frac{3}{4}}+{\displaystyle \frac{\sqrt{3}}{3}}}{1-{\displaystyle \frac{3}{4}}\cdot {\displaystyle \frac{\sqrt{3}}{3}}}}$

Se opera el numerador: ${\displaystyle \frac{3}{4}}+{\displaystyle \frac{\sqrt{3}}{3}}={\displaystyle \frac{9+4\sqrt{3}}{12}}$

Se opera el denominador: $1-{\displaystyle \frac{3}{4}}\cdot {\displaystyle \frac{\sqrt{3}}{3}}=1-{\displaystyle \frac{\sqrt{3}}{4}}={\displaystyle \frac{4-\sqrt{3}}{4}}$

Al dividir ambas expresiones:

$\mathrm{tg}(a+{30}^{\circ })={\displaystyle \frac{{\displaystyle \frac{9+4\sqrt{3}}{12}}}{{\displaystyle \frac{4-\sqrt{3}}{4}}}}={\displaystyle \frac{9+4\sqrt{3}}{12}}\cdot {\displaystyle \frac{4}{4-\sqrt{3}}}={\displaystyle \frac{9+4\sqrt{3}}{3(4-\sqrt{3})}}$

Se racionaliza multiplicando numerador y denominador por $4+\sqrt{3}$:

$\mathrm{tg}(a+{30}^{\circ })={\displaystyle \frac{(9+4\sqrt{3})(4+\sqrt{3})}{3(4-\sqrt{3})(4+\sqrt{3})}}={\displaystyle \frac{36+9\sqrt{3}+16\sqrt{3}+12}{3(16-3)}}={\displaystyle \frac{48+25\sqrt{3}}{39}}$

Cálculo de $\mathrm{tg}({45}^{\circ }-a)$:

$\mathrm{tg}({45}^{\circ }-a)={\displaystyle \frac{\mathrm{tg}{45}^{\circ }-\mathrm{tg}a}{1+\mathrm{tg}{45}^{\circ }\mathrm{tg}a}}={\displaystyle \frac{1-{\displaystyle \frac{3}{4}}}{1+{\displaystyle \frac{3}{4}}}}={\displaystyle \frac{{\displaystyle \frac{1}{4}}}{{\displaystyle \frac{7}{4}}}}={\displaystyle \frac{1}{7}}$