Kako preverite [sin ^ 3 (B) + cos ^ 3 (B)] / [sin (B) + cos (B)] = 1-sin (B) cos (B)?

Odgovor:

Dokaz spodaj

Pojasnilo:

Širitev # a ^ 3 + b ^ 3 = (a + b) (a ^ 2-ab + b ^ 2) #in to lahko uporabimo:

# (sin ^ 3B + cos ^ 3B) / (sinB + cosB) = ((sinB + cosB) (sin ^ 2B-sinBcosB + cos ^ 2B)) / (sinB + cosB) #

# = sin ^ 2B-sinBcosB + cos ^ 2B #

# = sin ^ 2B + cos ^ 2B-sinBcosB # (identiteta: # sin ^ 2x + cos ^ 2x = 1 #)

# = 1-sinBcosB #

Pokažite, da cos²π / 10 + cos²4π / 10 + cos² 6π / 10 + cos²9π / 10 = 2. Malo sem zmeden, če naredim Cos²4π / 10 = cos² (π-6π / 10) & cos²9π / 10 = cos² (π-π / 10), bo postal negativen kot cos (180 ° - theta) = - costheta v drugi kvadrant. Kako naj dokazujem vprašanje?

Glej spodaj. LHS = cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) + cos ^ 2 ((6pi) / 10) + cos ^ 2 ((9pi) / 10) = cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) + cos ^ 2 (pi- (4pi) / 10) + cos ^ 2 (pi- (pi) / 10) = cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) + cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) = 2 * [cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10)] = 2 * [cos ^ 2 (pi / 2- (4pi) / 10) + cos ^ 2 ((4pi) / 10)] = 2 * [sin ^ 2 ((4pi) / 10) + cos ^ 2 ((4pi) / 10)] = 2 * 1 = 2 = RHS

Kako preverite 1/8 [3 + 4cos2x + cos4x] = cos ^ 4x?

RHS = cos ^ 4x = [(2cos ^ 2x) / 2] ^ 2 = 1/4 [1 + cos2x] ^ 2 = 2 / (4 * 2) [1 + 2cos2x + cos ^ 2 (2x)] = 1 / 8 [2 + 4cos2x + 2cos ^ 2 (2x)] = 1/8 [2 + 4cos2x + 1 + cos4x] = 1/8 [3 + 4cos2x + cos4x] = LHS

Kako preverite cos ^ 2A = (1 + cos4A) / 2?

Glej spodaj Uporabite lastnost: cos2A = 2cos ^ 2A-1 Desna stran: = (1 + cos4A) / 2 = (1 + cos2 (2A)) / 2 = (1+ (2cos ^ 2 (2A) -1)) / 2 = (1-1 + 2cos ^ 2 (2A)) / 2 = (prekliči1-prekliči1 + 2kz ^ 2 (2A)) / 2 = (2cos ^ 2 (2A)) / 2 = (cancel2cos ^ 2 (2A) )) / cancel2 = cos ^ 2 (2A) = Leva stran