Odgovor:
Identiteta mora biti resnična za katero koli številko
Pojasnilo:
Preverite secx • cscx + cotx = tanx + 2cosx • cscx?
RHS = tanx + 2cosx * cscx = sinx / cosx + (2cosx) / sinx = (sin ^ 2x + 2cos ^ 2x) / (sinx * cosx) = (sin ^ 2x + cos ^ 2x + cos ^ 2x) / (sinx * cosx) = (1 + cos ^ 2x) / (sinx * cosx) = 1 / (sinx * cosx) + (cos ^ 2x) / (sinx * cosx) = cscx * secx + cotx = LHS
Kako dokazati to identiteto? sin ^ 2x + tan ^ 2x * sin ^ 2x = tan ^ 2x
Spodaj prikazano ... Uporabite naše trigonomske lastnosti ... 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 faktor leva stran vaše težave ... => 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
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