Tangente (trigonometria)

Identitate trigonometriko hau bi angeluren batuketaren identitatetik abiatzen da, dagoeneko sinu eta kosinuarentzat ezagutzen dena.

• ${\displaystyle \phi ,\theta \ }$ angeluak izanda:

${\displaystyle \operatorname {tg} \left(\phi +\theta \right)={\cfrac {\operatorname {sen}(\phi +\theta )}{\cos(\phi +\theta )}}}$

• Lehengo identitateengatik ordezkatuta:

${\displaystyle \operatorname {tg} \left(\phi +\theta \right)={\cfrac {\operatorname {sen} \phi \cos \theta +\cos \phi \operatorname {sen} \theta }{\cos \phi \cos \theta -\operatorname {sen} \phi \operatorname {sen} \theta }}}$

• Zatikiaren aldeak ebatziz ${\displaystyle \cos \phi \cos \theta \,}$:

${\displaystyle \operatorname {tg} \left(\phi +\theta \right)={\cfrac {\cfrac {\operatorname {sen} \phi \cos \theta +\cos \phi \operatorname {sen} \theta }{\cos \phi \cos \theta }}{\cfrac {\cos \phi \cos \theta -\operatorname {sen} \phi \operatorname {sen} \theta }{\cos \phi \cos \theta }}}}$

• ${\displaystyle \operatorname {tg} \left(\phi +(-\theta )\right)={\frac {\operatorname {tg} \phi +\operatorname {tg} (-\theta )}{1-\operatorname {tg} \phi \operatorname {tg} (-\theta )}}}$

• Funtzi bakoitia denez, hau lortzen da:

${\displaystyle \operatorname {tg} \left(\phi -\theta \right)={\frac {\operatorname {tg} \phi -\operatorname {tg} \theta }{1+\operatorname {tg} \phi \operatorname {tg} \theta }}}$

${\displaystyle \operatorname {tg} \left(\phi \pm \theta \right)={\frac {\operatorname {tg} \phi \pm \operatorname {tg} \theta }{1\mp \operatorname {tg} \phi \operatorname {tg} \theta }}}$

Hemendik abiatuz

${\displaystyle \operatorname {tg} \left(\phi +\theta \right)={\frac {\operatorname {tg} \phi +\operatorname {tg} \theta }{1-\operatorname {tg} \phi \operatorname {tg} \theta }}}$

eta ${\displaystyle \phi =\theta \,}$ eginez, orduan:

${\displaystyle \operatorname {tg} \left(2\phi \right)={\frac {2\operatorname {tg} \phi }{1-\operatorname {tg} ^{2}\phi }}}$

ψ angeluaren tangentea ezagututa, 3ψ tangentea aurkitu:

${\displaystyle \operatorname {tg} \left(3\psi \right)={\frac {3\operatorname {tg} \psi -\operatorname {tg} ^{3}\psi }{1-3\operatorname {tg} ^{2}\psi }}}$

θ angeluaren erdiaren tangentea aurkitu:

${\displaystyle \operatorname {tg} {\frac {\theta }{2}}={\frac {\operatorname {sen} \theta }{1+\cos \theta }}}$ [2]