Poiščite natančno vrednost? 2sinxcosx + sinx-2cosx = 1

Odgovor:

# rarrx = 2npi + - (2pi) / 3 # ALI # x = npi + (- 1) ^ n (pi / 2) # kje # nrarrZ #

Pojasnilo:

# rarr2sinx * cosx + sinx-2cosx = 1 #

#rarrsinx (2cosx + 1) -2cosx-1 = #

#rarrsinx (2cosx + 1) -1 (2cosx + 1) = 0 #

#rarr (2cosx + 1) (sinx-1) = 0 #

Ali, # 2cosx + 1 = 0 #

# rarrcosx = -1 / 2 = -cos (pi / 3) = cos (pi- (2pi) / 3) = cos ((2pi) / 3) #

# rarrx = 2npi + - (2pi) / 3 # kje # nrarrZ #

ALI, # sinx-1 = 0 #

# rarrsinx = 1 = sin (pi / 2) #

# rarrx = npi + (- 1) ^ n (pi / 2) # kje # nrarrZ #

Kako dokazati (1 + sinx-cosx) / (1 + cosx + sinx) = tan (x / 2)?

Glej spodaj. LHS = (1-cosx + sinx) / (1 + cosx + sinx) = (2sin ^ 2 (x / 2) + 2sin (x / 2) * cos (x / 2)) / (2cos ^ 2 (x / 2) + 2sin (x / 2) * cos (x / 2) = (2sin (x / 2) [sin (x / 2) + cos (x / 2)]) / (2cos (x / 2) * [ sin (x / 2) + cos (x / 2)]) = tan (x / 2) = RHS

Dokaži (1 + sinx + icosx) / (1 + sinx-icosx) = sinx + icosx?

Glej spodaj. Uporaba de Moivrejeve identitete, ki navaja e ^ (ix) = cos x + i sin x imamo (1 + e ^ (ix)) / (1 + e ^ (- ix)) = e ^ (ix) (1+) e ^ (- ix)) / (1 + e ^ (- ix)) = e ^ (ix) OPOMBA e ^ (ix) (1 + e ^ (- ix)) = (cos x + isinx) (1+ cosx-i sinx) = cosx + cos ^ 2x + isinx + sin ^ 2x = 1 + cosx + isinx ali 1 + cosx + isinx = (cos x + isinx) (1 + cosx-i sinx)

Kako dokazujete (sinx + cosx) ^ 4 = (1 + 2sinxcosx) ^ 2?

Glejte spodaj pojasnilo Začnite z leve strani (sinx + cosx) ^ 4 "" "" "" "" "" "" "" "" "" "" "" "(1 + 2sinx cosx) ^ 2 (sinx + cosx) (sinx + cosx)] ^ 2 Razširite / pomnožite / folijo izraz (sin ^ 2x + sinxcosx + sinxcosx + cos ^ 2x) ^ 2 Združite podobne izraze (sin ^ 2x + cos ^ 2x + 2sinxcosx) ^ 2 barva (rdeča) (sin ^ 2x + cos ^ 2x = 1) (1 + 2sinx cosx) ^ 2 QED Leva stran = desna stran Dokazano končano!