Riješite cos2A = sqrt (2) (cosA-sinA)?

Odgovor:

Pogledajte odgovor u nastavku …

Obrazloženje:

# Cos2A = sqrt2 (cosa-sina) #

# => Cos2A (cosa + sina) = sqrt2 (cos ^ 2A-sin ^ 2A) #

# => cos2A (cosA + sinA) = sqrt2 cdot cos2A #

# => otkazati (cos2A) (cosA + sinA) = sqrt2 otkazivanje cdot-a (cos2A #

# => (Cosa + sina) = sqrt2 #

# => Grijeh ^ 2A + cos ^ 2A + 2sinAcosA = 2 #na obje strane

# => 1 + sin2A = 2 #

# => Sin2A = 1 = sin90 ^ '#

# => 2A = 90 ^ '#

# => A = 45 ^ '#

NADA NAMJERAVAM ODGOVOR …

HVALA VAM…

# Cos2A = sqrt2 (cosa-sina) #

# => Cos ^ 2A-sin ^ 2A-sqrt2 (cosa-sina) = 0 #

# => (Cosa-sina) (+ cosa sina) -sqrt2 (cosa-sina) = 0 #

# => (Cosa-sina) (+ cosa sina-sqrt2) = 0 #

Kada

# Cosa + sina = 0 #

# => Tana = 1 = tan (pi / 4) *

# => A = npi + pi / 4 "gdje" n u ZZ #

# Cosa + sina = sqrt2 #

# => 1 / sqrt2cosA + 1 / sqrt2sinA = 1 #

# => Cos (pi / 4) cosa + sin (pi / 4) sina = 1 #

# => Cos (A-pi / 4) = 1 #

# => A = 2mpi + pi / 4 "gdje" m u ZZ #

Što je (sqrt (5+) sqrt (3)) / (sqrt (3+) sqrt (3+) sqrt (5)) - (sqrt (5-) sqrt (3)) / (sqrt (3+) sqrt) (3) sqrt (5))?

2/7 Primamo, A = (sqrt5 + sqrt3) / (sqrt3 + sqrt3 + sqrt5) - (sqrt5-sqrt3) / (sqrt3 + sqrt3-sqrt5) = (sqrt5 + sqrt3) / (2sqrt3 + sqrt5) - (sqrt5) -sqrt3) / (2sqrt3-sqrt5) = (sqrt5 + sqrt3) / (2sqrt3-sqrt5) = ((sqrt5 + sqrt3) (2sqrt3-sqrt5) - (sqrt5-sqrt3) (2sqrt3 + sqrt5) (2sqrt3 + sqrt5) / ((2sqrt15-5 + 2 * 3-sqrt15) - (2sqrt15 + 5-2 * 3-sqrt15)) / ((2sqrt3) ^ 2- (sqrt5) ^ 2) = (poništi (2sqrt15) -5 + 2 * 3kkazati (-sqrt15) - otkazati (2sqrt15) -5 + 2 * 3 + otkazati (sqrt15)) / (12-5) = ( Imajte na umu da, ako su u nazivnicima (sqrt3 + sqrt (3 + sqrt5)) i (sqrt3 + sqrt (3-sqrt5)), odgovor će biti promijenjen.

Kako pojednostavljujete (1 / sqrt (a-1) + sqrt (a + 1)) / (1 / sqrt (a + 1) -1 / sqrt (a-1)) div sqrt (a + 1) / ( (a-1) sqrt (a + 1) - (a + 1) sqrt (a-1)), a> 1?

Ogromno formatiranje matematike ...> boja (plava) (((1 / sqrt (a-1) + sqrt (a + 1)) / (1 / sqrt (a + 1) -1 / sqrt (a-1)) ) / (sqrt (a + 1) / ((a-1) sqrt (a + 1) - (a + 1) sqrt (a-1))) = boja (crvena) (((1 / sqrt (a- 1) + sqrt (a + 1)) / ((sqrt (a-1) -sqrt (a + 1)) / (sqrt (a + 1) cdot sqrt (a-1)))) / (sqrt (a +1) / (sqrt (a-1) cdot sqrt (a-1) cdot sqrt (a + 1) -sqrt (a + 1) cdot sqrt (a + 1) sqrt (a-1))) = boja ( plava) (((1 / sqrt (a-1) + sqrt (a + 1)) / ((sqrt (a-1) -sqrt (a + 1)) / (sqrt (a + 1) cdot sqrt (a -1)))) / (sqrt (a + 1) / (sqrt (a 1) cdot sqrt (a-1) (sqrt (a-1) -sqrt (a + 1))) = boja (crvena) ((1 / sqrt (a

Riješite sljedeći sustav jednadžbe: [((1), sqrt (2) x + sqrt (3) y = 0), ((2), x + y = sqrt (3) -sqrt (2))]?

{(x = (3sqrt (2) -2sqrt (3)) / (sqrt (6) -2)), (y = (sqrt (6) -2) / (sqrt (2) -sqrt (3))) :} Iz (1) imamo sqrt (2) x + sqrt (3) y = 0 Podijeliti obje strane sa sqrt (2) daje x + sqrt (3) / sqrt (2) y = 0 "(*)" Ako oduzmemo "(*)" iz (2) dobijamo x + y- (x + sqrt (3) / sqrt (2) y) = sqrt (3) -sqrt (2) - 0 => (1-sqrt) (3) / sqrt (2)) y = sqrt (3) -sqrt (2) => y = (sqrt (3) -sqrt (2)) / (1-sqrt (3) / sqrt (2)) = (sqrt (6) -2) / (sqrt (2) -sqrt (3)) Ako zamijenimo vrijednost koju smo pronašli za y natrag u "(*)" dobivamo x + sqrt (3) / sqrt (2) * (sqrt (6) -2) / (sqrt (2) -sqrt (3)) = 0 =>