Razones trigonométricas de ángulos compuestos
Caso seno:
$sen(\alpha + \beta )=sen(\alpha)cos(\beta)+cos(\alpha)sen(\beta)$$sen(\alpha - \beta )=sen(\alpha)cos(\beta)-cos(\alpha)sen(\beta)$
Caso coseno:
$cos(\alpha + \beta )=cos(\alpha)cos(\beta)-sen(\alpha)sen(\beta)$$cos(\alpha - \beta )=cos(\alpha)cos(\beta)+sen(\alpha)sen(\beta)$
Caso tangente:
$$tan(\alpha + \beta )=\frac{tan(\alpha )+ tan(\beta )}{1-tan(\alpha ).tan(\beta )}$$
$$tan(\alpha - \beta )=\frac{tan(\alpha )- tan(\beta )}{1+tan(\alpha ).tan(\beta )}$$
Ángulos dobles
De estas identidades se desprende el caso del ángulo doble:Caso seno:
$sen(\alpha + \alpha )=sen(\alpha)cos(\alpha )+cos(\alpha )sen(\alpha )=2sen(\alpha)cos(\alpha )$Es decir:
$$sen(2\alpha )=2sen(\alpha)cos(\alpha )$$
Caso coseno:
$cos(\alpha + \alpha )=cos(\alpha)cos(\alpha)-sen(\alpha)sen(\alpha)=cos^{2}(\alpha)-sen^{2}(\alpha )$Es decir:
$$cos(2\alpha)=cos^{2}(\alpha)-sen^{2}(\alpha )$$
Caso tangente:
$tan(\alpha + \alpha )=\frac{tan(\alpha )+ tan(\alpha )}{1-tan(\alpha ).tan(\alpha )}=\frac{2tan(\alpha )}{1-tan^{2}(\alpha )}$Es decir:
$$tan(2\alpha)=\frac{2tan(\alpha )}{1-tan^{2}(\alpha )}$$
