Ako si overíte identitu tanthetacsc ^ 2theta-tantheta = cottheta?

odpoveď:

Dôkaz nižšie

vysvetlenie:

#tantheta * csc ^ 2theta - tantheta #

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

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

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

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

# = Cos ^ 2theta / (sinthetacostheta) #

# = Costheta / sintheta #

# = Cottheta #

Poznač si to # sin ^ 2theta + cos ^ 2theta = 1 #, preto # cos ^ 2theta = 1- sin ^ 2theta #

# LHS = tantheta * csc ^ 2theta - tantheta #

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

# = tantheta (1 + detská postieľka ^ 2theta - 1) #

# = tantheta * postieľka ^ 2theta #

# = Cottheta = RHS #

Ako si overíte identitu sec ^ 2 (x / 2) = (2secx + 2) / (secx + 2 + cosx)?

Vyžaduje sa preukázať: sec ^ 2 (x / 2) = (2secx + 2) / (secx + 2 + cosx) "Pravá strana" = (2secx + 2) / (secx + 2 + cosx) Pamätajte, že secx = 1 / cosx => (2 * 1 / cosx + 2) / (1 / cosx + 2 + cosx) Teraz vynásobte vrch a spodok cosx => (cosx xx (2 * 1 / cosx + 2)) / (cosx xx (1 / cosx + 2 + cosx) => (2 + 2cosx) / (1 + 2cosx + cos ^ 2x) Faktorizácia dna, => (2 (1 + cosx)) / (1 + cosx) ^ 2 = > 2 / (1 + cosx) Vyvolanie identity: cos2x = 2cos ^ 2x-1 => 1 + cos2x = 2cos ^ 2x Podobne: 1 + cosx = 2cos ^ 2 (x / 2) => "Pravá strana" = 2 / (2cos ^ 2 (x / 2)) = 1

Ako si overíte identitu sec ^ 4theta = 1 + 2tan ^ 2theta + tan ^ 4theta?

Dôkaz nižšie Najprv sa ukážeme 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 = (1 / costheta) ^ 2 1 + tan ^ 2theta = sec ^ 2theta Teraz môžeme dokázať vašu otázku: sec ^ 4theta = (sec ^ 2theta) ^ 2 = (1 + tan ^ 2theta) ^ 2 = 1 + 2tan ^ theta + tan ^ 4theta

Ako si overíte identitu 3sec ^ 2thetan ^ 2theta + 1 = sec ^ 6theta-tan ^ 6theta?

Pozri nižšie 3sec ^ 2thatan ^ 2theta + 1 = sec ^ 6theta-tan ^ 6theta Pravá strana = sek ^ 6theta-tan ^ 6theta = (sec ^ 2theta) ^ 3- (tan ^ 2theta) ^ 3-> použitie rozdielu dvoch kociek vzorec = (sek. 2teta-tan ^ 2theta) (sek ^ 4theta + sec ^ 2tetatán + 2theta + tan4theta) = 1 * (sek ^ 4theta + sec ^ 2tatta + tan 4theta) = sec ^ 4theta + sec 2 2-teta + tan ^ 4theta = 2 ^ teta + 2-teta + sec ^ 2-teta + 2-teta + 2-teta = 2 ^ teta (tan ^ 2teta + 1) + sek. 2theta (sek ^ 2theta-1) = sec ^ 2thetan ^ 2theta + sec ^ 2theta + sec ^ 2theta + 2theta + sec ^ 2thatan ^ 2theta-tan ^ 2theta = sec ^ 2thetatan ^ 2theta + sec ^ 2