Wie vereinfacht man f (theta) = sin4theta-cos6theta zu trigonometrischen Funktionen einer Einheitstheta?

Antworten:

#sin (Theta) ^ 6-15cos (Theta) ^ 2sin (Theta) ^ 4-4cos (Theta) sin (Theta) ^ 3 + 15cos (Theta) ^ 4sin (Theta) ^ 2 + 4cos (Theta) ^ 3sin (Theta) -cos (Theta) ^ 6 #

Erläuterung:

Wir werden die folgenden zwei Identitäten verwenden:

#sin (A + -B) = sinAcosB + -cosAsinB #

#cos (A + -B) = cosAcosB sinAsinB #

#sin (4theta) = 2sin (2theta) cos (2theta) = 2 (2sin (theta) cos (theta)) (cos ^ 2 (theta) -sin ^ 2 (theta)) = 4sin (theta) cos ^ 3 (theta) -4sin ^ 3 (theta) cos (theta) #

#cos (6theta) = cos ^ 2 (3theta) -sin ^ 2 (3theta) #

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

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

# = (cos ^ 3 (theta) -sin ^ 2 (theta) cos (theta) -2sin ^ 2 (theta) cos (theta)) ^ 2- (2cos ^ 2 (theta) sin (theta) + cos ^ 2 (Theta) sin (Theta) -sin ^ 3 (Theta)) ^ 2 #

# = (cos ^ 3 (theta) -3sin ^ 2 (theta) cos (theta)) ^ 2- (3cos ^ 2 (theta) sin (theta) -sin ^ 3 (theta)) ^ 2 #

# = cos ^ 6 (theta) -6sin ^ 2 (theta) cos ^ 4 (theta) + 9sin ^ 4 (theta) cos ^ 2 (theta) -9sin ^ 2 (theta) cos ^ 4 (theta) + 6sin ^ 4 (Theta) cos ^ 2 (Theta) -sin ^ 6 (Theta) #

# sin (4theta) -cos (6 theta) = 4 sin (theta) cos ^ 3 (theta) -4sin 3 (theta) cos (theta) - (cos ^ 6 (theta) -6sin ^ 2 theta) cos ^ 4 (theta) + 9sin ^ 4 (theta) cos ^ 2 (theta) -9sin ^ 2 (theta) cos ^ 4 (theta) + 6sin ^ 4 (theta) cos ^ 2 (theta) -sin ^ 6 (theta)) #

# = 4sin (theta) cos ^ 3 (theta) -4sin ^ 3 (theta) cos (Theta) -cos ^ 6 (theta) + 6sin ^ 2 (theta) cos ^ 4 (theta) -9sin ^ 4 (theta) cos ^ 2 (theta) + 9sin ^ 2 (theta) cos ^ 4 (theta) -6sin ^ 4 (theta) cos ^ 2 (theta) + sin ^ 6 (theta) #

# = sin (theta) ^ 6-15cos (theta) ^ 2sin (theta) ^ 4-4cos (theta) sin (Theta) ^ 3 + 15cos (Theta) ^ 4sin (Theta) ^ 2 + 4cos (Theta) ^ 3sin (Theta) -cos (Theta) ^ 6 #

Wie vereinfacht man f (theta) = csc2theta-sec2theta-3tan2theta zu trigonometrischen Funktionen einer Einheit Theta?

F (theta) = (cos ^ 2theta-sin ^ 2theta-2costhetasintheta-4sin ^ 2thetacos ^ 2theta) / (2sinthetacos ^ 3theta-sin ^ 3thetacostheta) Zuerst schreibt man wie folgt: f (theta) = 1 / sin (2theta) -1 / cos (2 theta) -sin (2 theta) / cos (2 theta) Dann dann: f (theta) = 1 / sin (2 theta) - (1-sin (2 theta)) / cos (2 theta) = (cos (2 theta) - sin (2theta) -sin ^ 2 (2theta)) / (sin (2theta) cos (2theta)) Wir werden verwenden: cos (A + B) = cosAcosB-sinAsinB sin (A + B) = sinAcosB + cosAsinB Also, wir get: f (theta) = (cos ^ 2theta-sin ^ 2theta-2costhetasintheta-4sin ^ 2thetacos ^ 2theta) / ((2sinthetacostheta) (cos ^ 2theta-sin ^ 2

Was ist Kinderbett (Theta / 2) in Bezug auf die trigonometrischen Funktionen einer Einheit Theta?

Tut mir leid, falsch gelesen, Kinderbett ( theta / 2) = sin ( theta) / {1-cos ( theta)}, das Sie erhalten können, wenn Sie tan ( theta / 2) = {1-cos ( theta)} wenden. / sin ( theta), Beweis kommt. theta = 2 * arctan (1 / x) Wir können das nicht ohne rechte Seite lösen, also gehe ich einfach mit x. Torumstellung, Kinderbett ( theta / 2) = x für theta. Da die meisten Taschenrechner oder andere Hilfsmittel keine Schaltfläche "Kinderbett" oder Kinderbett ^ {- 1} oder Bogenbett ODER Wechselkorb "" ^ 1 haben (anderes Wort für die inverse Cotangens-Funktion, Kinderbett rückw&

Wie drückt man f (theta) = sin ^ 2 (theta) + 3cot ^ 2 (theta) -3csc ^ 2 theta aus, wenn man die nichtexponentiellen trigonometrischen Funktionen berücksichtigt?

Siehe unten f (theta) = 3sin ^ 2theta + 3cot ^ 2theta-3csc ^ 2theta = 3sin ^ 2theta + 3cot ^ 2theta-3csc ^ 2theta = 3sin ^ 2theta + 3 (csc ^ 2theta-1) -3csc ^ 2theta = 3sin ^ 2theta + annullieren (3csc ^ 2theta) -cancel3csc ^ 2theta-3 = 3sin ^ 2theta-3 = -3 (1-sin ^ 2theta) = -3cos ^ 2theta