Kako dokazujete (cotx + cscx / sinx + tanx) = (cotx) (cscx)?

Odgovor:

Preverjeno spodaj

Pojasnilo:

# (cotx + cscx) / (sinx + tanx) = (cotx) (cscx) #

# (cosx / sinx + 1 / sinx) / (sinx + sinx / cosx) = (cotx) (cscx) #

# ((cosx + 1) / sinx) / ((sinxcosx) / cosx + sinx / cosx) = (cotx) (cscx) #

# ((cosx + 1) / sinx) / ((sinx (cosx + 1)) / cosx) = (cotx) (cscx) #

# (prekliči (cosx + 1) / sinx) * (cosx / (sinxcancel ((cosx + 1)))) = (cotx) (cscx) #

# (cosx / sinx * 1 / sinx) = (cotx) (cscx) #

# (cotx) (cscx) = (cotx) (cscx) #

To poskušamo dokazati # (cotx + cscx) / (sinx + tanx) = cotxcscx #. Tu so identitete, ki jih potrebujete:

# tanx = sinx / cosx #

# cotx = cosx / sinx #

# cscx = 1 / sinx #

Začel bom z levo stranjo in manipuliral z njo, dokler ne bo enaka desni strani:

#barva (bela) = (cotx + cscx) / (sinx + tanx) #

# = (qquadcosx / sinx + 1 / sinxqquad) / (qquadsinx / 1 + sinx / cosxqquad) #

# = (qquad (cosx + 1) / sinxqquad) / (qquad (sinxcosx) / cosx + sinx / cosxqquad) #

# = (qquad (cosx + 1) / sinxqquad) / (qquad (sinxcosx + sinx) / cosxqquad) #

# = (cosx + 1) / sinx * cosx / (sinxcosx + sinx) #

# = (cosx + 1) / sinx * cosx / (sinx (cosx + 1)) #

# = (cosx (cosx + 1)) / (sin ^ 2x (cosx + 1)) #

# = (cosxcolor (rdeča) cancelcolor (črna) ((cosx + 1))) / (sin ^ 2xcolor (rdeča) preklicna barva (črna) ((cosx + 1))) #

# = cosx / sin ^ 2x #

# = cosx / sinx * 1 / sinx #

# = cotx * cscx #

# = RHS #

To je dokaz. Upam, da je to pomagalo!

# LHS = (cotx + cscx) / (sinx + tanx) #

# = (cotx + cscx) / (sinx + tanx) * ((cotx * cscx) / (cotx * cscx)) #

# = cotx * cscx (cotx + cscx) / ((sinx + tanx) * cotx * cscx) #

# = cotx * cscx (cotx + cscx) / ((sinx * cscx * cotx + tanx * cotx * cscx)) #

# = cotx * cscxcancel ((cotx + cscx) / (cotx + cscx)) = cotx * cscx = RHS #

Preverite secx • cscx + cotx = tanx + 2cosx • cscx?

RHS = tanx + 2cosx * cscx = sinx / cosx + (2cosx) / sinx = (sin ^ 2x + 2cos ^ 2x) / (sinx * cosx) = (sin ^ 2x + cos ^ 2x + cos ^ 2x) / (sinx * cosx) = (1 + cos ^ 2x) / (sinx * cosx) = 1 / (sinx * cosx) + (cos ^ 2x) / (sinx * cosx) = cscx * secx + cotx = LHS

Kako preverite (1 + tanx) / (sinx) = cscx + secx?

Uporabite naslednja pravila: tanx = sinx / cosx 1 / sinx = cscx 1 / cosx = secx Začnite z leve strani ("LHS"): => "LHS" = (1 + tanx) / sinx = 1 / sinx + tanx / sinx = cscx + tanx xx1 / sinx = cscx + preklic (sinx) / cosx xx1 / preklic (sinx) = cscx + 1 / cosx = barva (modra) (cscx + secx) QED

Kako dokazati to identiteto? (cosxcotx-tanx) / cscx = cosx / secx-sinx / cotx

Identiteta mora biti resnična za katero koli število x, ki preprečuje delitev na nič. (cosxcotx-tanx) / cscx = {cos x (cos x / sin x) - sin x / cos x} / (1 / sin x) = cos ^ 2x - sin ^ 2 x / cos x = cos x / (1 / cos x) - sin x / (cos x / sin x) = cosx / secx-sinx / cotx