Hoe deze identiteit te bewijzen? sin ^ 2x + tan ^ 2x * sin ^ 2x = tan ^ 2x

Antwoord:

Hieronder weergegeven…

Uitleg:

Gebruik onze trig-identiteiten …

# sin ^ 2 x + cos ^ 2 x = 1 #

# => sin ^ 2 x / cos ^ 2 x + cos ^ 2 x / cos ^ 2 x = 1 / cos ^ 2 x #

# => tan ^ 2 x + 1 = 1 / cos ^ 2 x #

Factor de linkerkant van uw probleem …

# => sin ^ 2 x (1 + tan ^ 2 x) #

# => sin ^ 2 x (1 / cos ^ 2 x) = sin ^ 2 x / cos ^ 2 x #

# => (sinx / cosx) ^ 2 = tan ^ 2 x #

Gegeven, # sin ^ 2 x + tan ^ 2x sin ^ 2x #

# = sin ^ 2 x (1 + tan ^ 2 x) #

# = sin ^ 2 x sec ^ 2x # (zoals,# sec ^ 2x - tan ^ 2 x = 1) #

# = sin ^ 2x (1 / (cos ^ 2x)) #

# = tan ^ 2 x #

bewezen

Hoe te bewijzen (1 + sinx-cosx) / (1 + cosx + sinx) = tan (x / 2)?

Zie onder. 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

Hoe verifieer je de identiteit sec ^ 4theta = 1 + 2tan ^ 2theta + tan ^ 4theta?

Bewijs hieronder Eerst zullen we bewijzen 1 + tan ^ 2theta = sec ^ 2theta: sin ^ 2theta + cos ^ 2theta = 1 sin ^ 2theta / cos ^ 2theta + cos ^ 2theta / cos ^ 2theta = 1 / cos ^ 2theta tan ^ 2theta + 1 = (1 / costheta) ^ 2 1 + tan ^ 2theta = sec ^ 2theta Nu kunnen we je vraag bewijzen: sec ^ 4theta = (sec ^ 2theta) ^ 2 = (1 + tan ^ 2theta) ^ 2 = 1 + 2tan ^ theta + tan ^ 4theta

Hoe verifieer je de identiteit 3sec ^ 2thetatan ^ 2theta + 1 = sec ^ 6theta-tan ^ 6theta?

Zie onder 3sec ^ 2thetatan ^ 2theta + 1 = sec ^ 6theta-tan ^ 6theta Rechterkant = sec ^ 6theta-tan ^ 6theta = (sec ^ 2theta) ^ 3- (tan ^ 2theta) ^ 3-> gebruik het verschil van twee kubussen formula = (sec ^ 2theta-tan ^ 2theta) (sec ^ 4theta + sec ^ 2thetatan ^ 2theta + tan ^ 4theta) = 1 * (sec ^ 4theta + sec ^ 2thetatan ^ 2theta + tan ^ 4theta) = sec ^ 4theta + sec ^ 2thetatan ^ 2theta + tan ^ 4theta = sec ^ 2theta sec ^ 2 theta + sec ^ 2thetatan ^ 2theta + tan ^ 2theta tan ^ 2 theta = sec ^ 2theta (tan ^ 2theta + 1) + sec ^ 2thetatan ^ 2theta + tan ^ 2theta (sec ^ 2theta-1) = sec ^ 2thetatan ^ 2theta + sec ^ 2theta +