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

Odgovor:

glej spodaj

Pojasnilo:

Uporabite lastnost:

# cos2A = 2cos ^ 2A-1 #

Desna stran roke:# = (1 + cos4A) / 2 #

# = (1 + cos2 (2A)) / 2 #

# = (1+ (2cos ^ 2 (2A) -1)) / 2 #

# = (1-1 + 2cos ^ 2 (2A)) / 2 #

# = (cancel1-cancel1 + 2cos ^ 2 (2A)) / 2 #

# = (2cos ^ 2 (2A)) / 2 #

# = (cancel2cos ^ 2 (2A)) / cancel2 #

# = cos ^ 2 (2A) #

#=# Stran leve roke

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

Dokaz pod Ekspanzijo ^ 3 + b ^ 3 = (a + b) (^ 2-ab + b ^ 2) in lahko uporabimo to: (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

Kako preverite posteljico (x) / sin (x) -tan (x) / cos (x) = csc (x) sec (x) 1 / (sin (x) + cos (x))?

"To ni res, zato izpolnite x = 10 ° npr. In videli boste, da enakost ne drži." "Nič več dodati."