Relaciones trigonométricas del ángulo doble y ángulo mitad
Conocidas todas las razones de un ángulo, podemos encontrar todas la razones del doble y de la mitad de dicho ángulo a partir de las siguientes relaciones:
- $$\sin (2\alpha)=2 \cdot \sin \alpha \cdot \cos \alpha$$
- $$\cos (2\alpha)=\cos^2 \alpha - \sin^2\alpha$$
- $$\tan (2\alpha)=\displaystyle \frac{2\cdot \tan\alpha }{1-\tan^2\alpha}$$
- $$\sin \displaystyle \Big(\frac{\alpha}{2}\Big)=\pm \sqrt{\frac{1-\cos \alpha}{2}}$$
- $$\cos \displaystyle \Big(\frac{\alpha}{2}\Big)=\pm \sqrt{\frac{1+\cos \alpha}{2}}$$
- $$\tan \displaystyle \Big(\frac{\alpha}{2}\Big)=\pm \sqrt{\frac{1-\cos \alpha}{1+\cos \alpha}}$$
Dado $$\alpha$$ del que ya conocemos sus razones trigonométricas, ahora podremos calcular las razones del ángulo doble y mitad. Teniendo en cuenta que $$\alpha=30^\circ$$, calcularemos las razones de $$2\alpha=60^\circ$$ y $$\displaystyle \frac{\alpha}{2}=15^\circ$$.
Tenemos:
$$\displaystyle \sin 2\alpha = 2\cdot \sin \alpha \cdot \cos \alpha= 2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2}= \frac{\sqrt{3}}{2}$$
$$\displaystyle \cos (2\alpha)=\cos^2\alpha -\sin^2\alpha=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}$$
$$\displaystyle \tan (2\alpha) = \frac{2 \cdot \tan \alpha }{1-\tan^2\alpha}=\frac{2 \cdot \displaystyle \frac{\sqrt{3}}{3}}{1-\displaystyle \frac{1}{3}}=\sqrt{3}$$
$$\sin \displaystyle \frac{\alpha}{2}= \displaystyle\sqrt{\displaystyle\frac{1-\cos \alpha}{2}}=\displaystyle\sqrt{\displaystyle\frac{1-\displaystyle\frac{\sqrt{3}}{2}}{2}}=\displaystyle\sqrt{\displaystyle\frac{2-\sqrt{3}}{4}}=\displaystyle\frac{\displaystyle\sqrt{2-\displaystyle\sqrt{3}}}{2}$$
$$\cos \displaystyle \frac{\alpha}{2}= \displaystyle\sqrt{\displaystyle\frac{1+\cos \alpha}{2}}=\displaystyle\sqrt{\displaystyle\frac{1+\displaystyle\frac{\sqrt{3}}{2}}{2}}=\displaystyle\sqrt{\displaystyle\frac{2+\sqrt{3}}{4}}=\displaystyle\frac{\displaystyle\sqrt{2+\displaystyle\sqrt{3}}}{2}$$
$$\tan \displaystyle \frac{\alpha}{2}=\displaystyle\sqrt{ \frac{1- \cos \alpha}{1+\cos \alpha}}=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{1+\frac{\sqrt{3}}{2}}}=\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}=$$
$$=\displaystyle\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}\cdot \sqrt{\frac{2+\sqrt{3}}{2+\sqrt{3}}}=\frac{\sqrt{4-3}}{2+\sqrt{3}}=\frac{1}{2+\sqrt{3}} \cdot \frac{2-\sqrt{3}}{2-\sqrt{3}}=$$
$$=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$$
Relaciones trigonométricas de la suma y diferencia de ángulos
- $$\sin (A+B)=\sin A \cdot \cos B+\cos A \cdot \sin B$$
- $$\sin (A-B)=\sin A \cdot \cos B-\cos A \cdot \sin B$$
- $$\cos (A+B)=\cos A \cdot \cos B-\sin A \cdot \sin B$$
- $$\cos (A-B)=\cos A \cdot \cos B+\sin A \cdot \sin B$$
- $$\displaystyle \tan (A+B) =\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}$$
- $$\displaystyle \tan (A-B) =\frac{\tan A-\tan B}{1+\tan A \cdot \tan B}$$
Ahora podemos calcular las razones trigonométricas de $$45^\circ=60^\circ-15^\circ$$.
$$\sin (60-15)=\sin 60 \cdot \cos 15-\cos 60 \cdot \sin 15=\displaystyle \frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2+\sqrt{3}}}{2}-\frac{1}{2}\cdot \frac{\sqrt{2-\sqrt{3}}}{2}=$$
$$\displaystyle =\frac{1}{4}\Big(\sqrt{6+3\sqrt{3}}-\sqrt{2-\sqrt{3}}\Big)= \frac{\sqrt{2}}{2} \\ $$
$$\cos (60-15)=\cos 60 \cdot \cos 15+\sin 60 \cdot \sin 15=\displaystyle \frac{1}{2}\cdot \frac{\sqrt{2+\sqrt{3}}}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{2-\sqrt{3}}}{2}=$$
$$\displaystyle =\frac{1}{4}\Big(\sqrt{2+\sqrt{3}}+\sqrt{6-3\sqrt{3}}\Big)=\frac{\sqrt{2}}{2} \\$$
$$\displaystyle \tan (60-15) =\frac{\tan 60-\tan 15}{1+\tan 60 \cdot \tan 15}=\frac{\sqrt{3}-(2-\sqrt{3})}{1+\sqrt{3}\cdot (2-\sqrt{3}}=\frac{2\sqrt{3}-2}{1+2\sqrt{3}-3}=\frac{2\sqrt{3}-2}{2\sqrt{3}-2}=1$$