Miten ilmaisette cos (4theta) cos (2theta): n suhteen?

Vastaus:

#cos (4theta) = 2 (cos (2theta)) ^ 2-1 #

Selitys:

Aloita korvaamalla # 4theta # kanssa # 2theta + 2theta #

#cos (4theta) = cos (2theta + 2theta) #

Sen tietäen #cos (a + b) = cos (a) cos (b) -sin (a) sin (b) # sitten

#cos (2theta + 2theta) = (cos (2theta)) ^ 2- (sin (2theta)) ^ 2 #

Sen tietäen # (cos (x)) ^ 2+ (sin (x)) ^ 2 = 1 # sitten

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

#rarr cos (4theta) = (cos (2theta)) ^ 2- (1- (cos (2theta)) ^ 2) #

# = 2 (cos (2theta)) ^ 2-1 #

Miten osoitat cos ^ 4theta-sin ^ 4theta = cos2theta?

Käytämme rarrsin ^ 2x + cos ^ 2x = 1, a ^ 2-b ^ 2 = (a + b) (a-b) ja cos ^ 2x-sin ^ 2x = cos2x. LHS = cos ^ 4x-sin ^ 4x = (cos ^ 2x) ^ 2- (sin ^ 2x) ^ 2 = (cos ^ 2x + sin ^ 2x) * (cos ^ 2x-sin ^ 2x) = 1 * cos2x = cos2x = RHS

Miten ilmaisette cos ((15 pi) / 8) * cos ((5 pi) / 8) ilman trigonometristen toimintojen tuotteita?

Cos ((15pi) / 8) cos ((5pi) / 8) = 1/2 cos ((5pi) / 2) +1/2 cos ((5pi) / 4) = - sqrt2 / 2 2cos A cos B = cos (A + B) + cos (AB) cosAcos B = 1/2 (cos (A + B) + cos (AB)) A = (15pi) / 8, B = (5pi) / 8 => cos (( 15pi) / 8) cos ((5pi) / 8) = 1/2 (cos ((15pi) / 8 + (5pi) / 8) + cos ((15pi) / 8- (5pi) / 8)) = 1 / 2 (cos ((20pi) / 8) + cos ((10pi) / 8)) = 1/2 cos ((5pi) / 2) +1/2 cos ((5pi) / 4) = 0 + -sqrt2 / 2 = -sqrt2 / 2 cos ((15pi) / 8) cos ((5pi) / 8) = 1/2 cos ((5pi) / 2) +1/2 cos ((5pi) / 4) = - sqrt2 / 2

Miten vahvistat identiteetin sec ^ 4theta = 1 + 2tan ^ 2theta + tan ^ 4theta?

Alla oleva todistus Aluksi osoitamme 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 Nyt voimme todistaa kysymyksesi: sek ^ 4theta = (sek ^ 2theta) ^ 2 = (1 + tan ^ 2theta) ^ 2 = 1 + 2tan ^ theta + tan ^ 4theta