Zdarzają się zadania, w których mamy podane dwa kąty sumujące się do 90∘90^{\circ}90∘, na przykład α= \alpha=\ α= 32∘32^{\circ}32∘ i β= \beta=\ β= 58∘58^{\circ}58∘:
α+β= \alpha+\beta=\ α+β= 32∘+58∘= 32^{\circ}+58^{\circ}=\ 32∘+58∘= 90∘90^{\circ}90∘
Zatem:
58∘= 58^{\circ}=\ 58∘= 90∘−32∘90^{\circ}-32^{\circ}90∘−32∘
32∘= 32^{\circ}=\ 32∘= 90∘−58∘90^{\circ}-58^{\circ}90∘−58∘
Możemy wtedy korzystać ze wzorów:
sin58∘= \sin 58^{\circ}=\ sin58∘= sin(90∘−32∘)= \sin(90^{\circ}-32^{\circ})=\ sin(90∘−32∘)= cos32∘\cos 32^{\circ}cos32∘
sin32∘= \sin 32^{\circ}=\ sin32∘= sin(90∘−58∘)= \sin(90^{\circ}-58^{\circ})=\ sin(90∘−58∘)= cos58∘\cos 58^{\circ}cos58∘
cos58∘= \cos 58^{\circ}=\ cos58∘= cos(90∘−32∘)= \cos(90^{\circ}-32^{\circ})=\ cos(90∘−32∘)= sin32∘\sin 32^{\circ}sin32∘
cos32∘= \cos 32^{\circ}=\ cos32∘= cos(90∘−58∘)= \cos(90^{\circ}-58^{\circ})=\ cos(90∘−58∘)= sin58∘\sin 58^{\circ}sin58∘
tg58∘= \tg 58^{\circ}=\ tg58∘= tg(90∘−32∘)= \tg(90^{\circ}-32^{\circ})=\ tg(90∘−32∘)= ctg32∘\ctg 32^{\circ}ctg32∘
tg32∘= \tg 32^{\circ}=\ tg32∘= tg(90∘−58∘)= \tg(90^{\circ}-58^{\circ})=\ tg(90∘−58∘)= ctg58∘\ctg 58^{\circ}ctg58∘
ctg58∘= \ctg58^{\circ}=\ ctg58∘= ctg(90∘−32∘)= \ctg(90^{\circ}-32^{\circ})=\ ctg(90∘−32∘)= tg32∘\tg 32^{\circ}tg32∘
ctg32∘= \ctg32^{\circ}=\ ctg32∘= ctg(90∘−58∘)= \ctg(90^{\circ}-58^{\circ})=\ ctg(90∘−58∘)= tg58∘\tg 58^{\circ}tg58∘
