Com es resol la sin (x) - cos (x) -tan (x) = -1?

Resposta:

# "El conjunt de solucions" = {2kpi} uu {kpi + pi / 4}, k en ZZ #.

Explicació:

Donat que, # sinx-cosx-tanx = -1.

#:. sinx-cosx-sinx / cosx + 1 = 0 #.

#:. (sinx-cosx) - (sinx / cosx-1) = 0 #.

#:. (sinx-cosx) - (sinx-cosx) / cosx = 0 #.

#:. (sinx-cosx) cosx- (sinx-cosx) = 0 #.

#:. (sinx-cosx) (cosx-1) = 0.

#:. sinx = cosx o cosx = 1 #.

# "Cas 1:" sinx = cosx #.

Observeu-ho #cosx! = 0, perquè, "en cas contrari," tanx "esdevé" # "

indefinit.

Per tant, dividint per #cosx! = 0, sinx / cosx = 1 o, tanx = 1 #.

#:. tanx = tan (pi / 4) #.

#:. x = kpi + pi / 4, k en ZZ, "en aquest cas".

# "Cas 2:" cosx = 1 #.

# "En aquest cas," cosx = 1 = cos0,:. x = 2kpi + -0, k en ZZ #.

En conjunt, tenim, # "El conjunt de solucions" = {2kpi} uu {kpi + pi / 4}, k en ZZ #.

Resposta:

# rarrx = 2npi, npi + pi / 4 # on #n en ZZ #

Explicació:

# rarrsinx-cosx-tanx = -1

# rarrsinx-cosx-sinx / cosx + 1 = 0 #

#rarr (sinx * cosx-cos ^ 2x-sinx + cosx) / cosx = 0 #

# rarrsinx * cosx-sinx-cos ^ 2x + cosx = 0 #

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

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

Quan # rarrcosx-1 = 0 #

# rarrcosx = cos0 #

# rarrx = 2npi + -0 = 2npi # on #n en ZZ #

Quan # rarrsinx-cosx = 0 #

#rarrcos (90-x) -cosx = 0 #

# rarr2sin ((90-x + x) / 2) * sin ((x-90 + x) / 2) = 0

#rarrsin (x-pi / 4) = 0 Com #sin (pi / 4)! = 0 #

# rarrx-pi / 4 = npi #

# rarrx = npi + pi / 4 # on #n en ZZ #

Com es resol el sin (x + (π / 4)) + sin (x - (π / 4)) = 1?

X = (- 1) ^ n (pi / 4) + npi "", n en ZZ Utilitzem la identitat (en cas contrari anomenada fórmula de factor): sinA + sinB = 2sin ((A + B) / 2) cos (( AB) / 2) Així: sin (x + (pi / 4)) + sin (x - (pi / 4)) = 2sin (((x + pi / 4) + (x-pi / 4)) / / 2] cos [(x + pi / 4 - + (x-pi / 4)) / 2] = 1 => 2sin ((2x) / 2) cos ((2 * (pi / 4)) / 2) = 1 => 2sin (x) cos (pi / 4) = 1 => 2 * sin (x) * sqrt (2) / 2 = 1 => sin (x) = 1 / sqrt (2) = sqrt (2) / 2 => color (blau) (x = pi / 4) La solució general és: x = pi / 4 + 2pik i x = pi-pi / 4 + 2pik = pi / 4 + (2k + 1) pi "" , k en ZZ Pod

Prova: - sin (7 theta) + sin (5 theta) / sin (7 theta) -sin (5 theta) =?

(sin7x + sin5x) / (sin7x-sin5x) = tan6x * cotx rarr (sin7x + sin5x) / (sin7x-sin5x) = (2in ((7x + 5x) / 2) * cos ((7x-5x) / 2) ) / (2sin ((7x-5x) / 2) * cos ((7x + 5x) / 2) = (sin6x * cosx) / (sinx * cos6x) = (tan6x) / tanx = tan6x * cottx

Sin ^ 2 (45 ^ @) + sin ^ 2 (30 ^ @) + sin ^ 2 (60 ^ @) + sin ^ 2 (90 ^ @) = (- 5) / (4)?

Si us plau mireu més a baix. rarrsin ^ 2 (45 °) + sin ^ 2 (30 °) + sin ^ 2 (60 °) + sin ^ 2 (90 °) = (1 / sqrt (2)) ^ 2+ (1/2) ^ 2 + (sqrt (3) / 2) ^ 2 + (1) ^ 2 = 1/2 + 1/4 + 3/4 + 1 = 1/2 + 2 = 5/2