Kako dokazujete (tanx + sinx) / (2tanx) = cos ^ 2 (x / 2)?

Za dokončanje dokaza bomo potrebovali ti dve identiteti:

# tanx = sinx / cosx #

#cos (x / 2) = + - sqrt ((1 + cosx) / 2) #

Začel bom z desno stranjo in nato manipuliral, dokler ne bo videti kot na levi strani:

# RHS = cos ^ 2 (x / 2) #

#barva (bela) (RHS) = (cos (x / 2)) ^ 2 #

#color (bel) (RHS) = (+ - sqrt ((1 + cosx) / 2)) ^ 2 #

#barva (bela) (RHS) = (1 + cosx) / 2 #

#barva (bela) (RHS) = (1 + cosx) / 2barva (rdeča) (* sinx / sinx) #

#barva (bela) (RHS) = (sinx + sinxcosx) / (2sinx) #

#barva (bela) (RHS) = (sinx + sinxcosx) / (2sinx) barva (rdeča) (* (1 / cosx) / (1 / cosx)) #

#barva (bela) (RHS) = (sinx / cosx + (sinxcosx) / cosx) / (2sinx / cosx) #

#barva (bela) (RHS) = (tanx + sinx) / (2tanx) #

#barva (bela) (RHS) = LHS #

To je dokaz. Upam, da je to pomagalo!

Prizadevamo si dokazati identiteto:

# (tanx + sinx) / (2tanx) - = cos ^ 2 (x / 2) #

Upoštevajte LHS izraza in uporabite definicijo tangente:

# LHS = (tanx + sinx) / (2tanx) #

# (sinx / cosx + sinx) / (2 (sinx / cosx)) # t

# (cosx / sinx) ((sinx / cosx + sinx) / 2) # t

# (cosx / sinx * sinx / cosx + cosx / sinx * sinx) / 2 # t

# (1 + cosx) / 2 # t

Zdaj pa razmislite o RHS in uporabite identiteto:

# cos2A - = 2cos ^ 2A - 1 #

Dajemo:

# cosx - = 2cos ^ 2 (x / 2) - 1 => 1 + cosx - = 2cos ^ 2 (x / 2) #

#:. cos ^ 2 (x / 2) = (1 + cosx) / 2 = RHS #

Tako:

# LHS = RHS => (tanx + sinx) / (2tanx) - = cos ^ 2 (x / 2) t QED

# LHS = (tanx + sinx) / (2tanx) #

# = (prekliči (tanx) (1 + sinx / tanx)) / (2zaključi (tanx)) #

# = (1 + cosx) / 2 = (2cos ^ 2 (x / 2)) / 2 = cos ^ 2 (x / 2) = RHS #

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

Preverjeno spodaj (cotx + cscx) / (sinx + tanx) = (cotx) (cscx) (cosx / sinx + 1 / sinx) / (sinx + sinx / cosx) = (cotx) (cscx) ((cosx + 1) / ((sinxcosx) / cosx + sinx / cosx) = (cotx) (cscx) ((cosx + 1) / sinx) / ((sinx (cosx + 1)) / cosx) = (cotx) (cscx) ) (preklic (cosx + 1) / sinx) * (cosx / (sinxcancel ((cosx + 1)))))) (cotx) (cscx) (cosx / sinx * 1 / sinx) = (cotx) (cscx) ( cotx) (cscx) = (cotx) (cscx)

Kako dokazujete (cosx / (1 + sinx)) + ((1 + sinx) / cosx) = 2secx?

Pretvorite levo stran v izraze s skupnim imenovalcem in dodajte (pretvorite cos ^ 2 + sin ^ 2 v 1 na poti); poenostavimo in se sklicujemo na definicijo sek = 1 / cos (cos (x) / (1 + sin (x))) + ((1 + sin (x)) / cos (x)) = (cos ^ 2 (x) + 1 + 2sin (x) + sin ^ 2 (x)) / (cos (x) (1 + sin (x) = (2 + 2sin (x)) / (cos (x) (1 + sin (x)) ) = 2 / cos (x) = 2 * 1 / cos (x) = 2sek (x)

Kako dokazujete: secx - cosx = sinx tanx?

S pomočjo definicij sekx in tanx, skupaj z identiteto sin ^ 2x + cos ^ 2x = 1, imamo secx-cosx = 1 / cosx-cosx = 1 / cosx-cos ^ 2x / cosx = (1-cos ^ 2x) / cosx = sin ^ 2x / cosx = sinx * sinx / cosx = sinxtanx