Kako preverite identiteto tanthetacsc ^ 2theta-tantheta = cottheta?

Odgovor:

Dokaz spodaj

Pojasnilo:

#tantheta * csc ^ 2theta - tantheta #

# = sintheta / costheta * (1 / sintheta) ^ 2 - sintheta / costheta #

# = sintheta / costheta * 1 / sin ^ 2tea - sintheta / costheta #

# = 1 / (sinthetacostheta) - sintheta / costheta #

# = (1-sin ^ 2theta) / (sinthetacostheta) #

# = cos ^ 2theta / (sinthetacostheta) #

# = costheta / sintheta #

# = cottheta #

Upoštevajte, da # sin ^ 2theta + cos ^ 2theta = 1 #zato # cos ^ 2theta = 1- sin ^ 2theta #

# LHS = tanteta * csc ^ 2tea - tantheta #

# = tantheta (csc ^ 2theta - 1) #

# = tantheta (1 + cot ^ 2theta - 1) #

# = tantheta * cot ^ 2theta #

# = cottheta = RHS #

Kako preverite identiteto sec ^ 2 (x / 2) = (2secx + 2) / (secx + 2 + cosx)?

Potrebno je dokazati: sec ^ 2 (x / 2) = (2secx + 2) / (secx + 2 + cosx) "Desna stran" = (2secx + 2) / (secx + 2 + cosx) Ne pozabite, da je sekx = 1 / cosx => (2 * 1 / cosx + 2) / (1 / cosx + 2 + cosx) Zdaj, pomnožimo vrh in dno s cosx => (cosx xx (2 * 1 / cosx + 2)) / (cosx xx) (1 / cosx + 2 + cosx)) => (2 + 2cosx) / (1 + 2cosx + cos ^ 2x) Faktoriziraj dno, => (2 (1 + cosx)) / (1 + cosx) ^ 2 = > 2 / (1 + cosx) Recall identiteta: cos2x = 2cos ^ 2x-1 => 1 + cos2x = 2cos ^ 2x Podobno: 1 + cosx = 2cos ^ 2 (x / 2) => "Desna stran" = 2 / (2cos ^ 2 (x / 2)) = 1 / cos ^ 2 (x / 2) = barva (m

Kako preverite identiteto sec ^ 4theta = 1 + 2tan ^ 2theta + tan ^ 4theta?

Dokaz spodaj Najprej bomo dokazali 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 Zdaj lahko dokažemo vaše vprašanje: sec ^ 4theta = (sec ^ 2theta) ^ 2 = (1 + tan ^ 2theta) ^ 2 = 1 + 2tan ^ theta + tan ^ 4theta

Kako preverite identiteto 3sec ^ 2thetatan ^ 2theta + 1 = sec ^ 6theta-tan ^ 6theta?

Glej spodaj 3sec ^ 2thetatan ^ 2theta + 1 = sec ^ 6theta-tan ^ 6theta desna stran = sec ^ 6theta-tan ^ 6theta = (sec ^ 2theta) ^ 3- (tan ^ 2theta) ^ 3-> uporabi razliko dveh kock formula = (sec ^ 2teta-tan ^ 2theta) (sec ^ 4theta + sec ^ 2thetatan ^ 2theta + tan ^ 4theta) = 1 * (sec ^ 4theta + sec ^ 2thetatan ^ 2theta + tan ^ 4theta) = sec ^ 4theta + sec ^ 2thetatan ^ 2theta + tan ^ 4theta = sek ^ 2tea sek ^ 2 theta + sec ^ 2tetatan ^ 2theta + tan ^ 2theta tan ^ 2 theta = sek ^ 2tea (tan ^ 2theta + 1) + sec ^ 2thetatan ^ 2theta + tan ^ 2theta (sec ^ 2theta-1) = sec ^ 2tetatan ^ 2theta + sec ^ 2theta + sec ^ 2tetatan ^ 2