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

Antworten:

#f (theta) = (cos ^ 2theta-sin ^ 2theta-2costhetasintheta-4sin ^ 2thetacos ^ 2theta) / (2sinthetacos ^ 3theta-sin ^ 3thetacostheta) #

Erläuterung:

Zuerst schreiben Sie als:#f (theta) = 1 / sin (2 theta) -1 / cos (2 theta) -sin (2 theta) / cos (2 theta) #

Dann als:

#f (theta) = 1 / sin (2 theta) - (1-sin (2 theta)) / cos (2 theta) = (cos (2 theta) -sin (2 theta) -sin ^ 2 (2 theta)) / (sin (2 theta)) cos (2theta)) #

Wir werden verwenden:

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

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

Also bekommen wir:

#f (theta) = (cos ^ 2theta-sin ^ 2theta-2costhetasintheta-4sin ^ 2thetacos ^ 2theta) / ((2sinthetacostheta) (cos ^ 2theta-sin ^ 2theta)) #

#f (theta) = (cos ^ 2theta-sin ^ 2theta-2costhetasintheta-4sin ^ 2thetacos ^ 2theta) / (2sinthetacos ^ 3theta-sin ^ 3thetacostheta) #