Kako poenostavite f (theta) = sin4theta-cos6theta za trigonometrične funkcije enote theta?

Odgovor:

#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 #

Pojasnilo:

Uporabili bomo naslednji dve identiteti:

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

# = 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 #

Graf funkcije f (x) = abs (2x) se prevede 4 enote navzdol. Kakšna je enačba transformirane funkcije?

F_t (x) = abs (2x) -4 f (x) = abs (2x) Za preoblikovanje f (x) 4 enot navzdol f_t (x) = f (x) -4 f_t (x) = abs (2x) - 4 Graf f_t (x) je prikazan spodaj: graf {abs (2x) -4 [-18.02, 18.03, -9.01, 9.01]}

Kako poenostavite f (theta) = csc2theta-sec2theta-3tan2theta za trigonometrične funkcije enote theta?

F (theta) = (cos ^ 2theta-sin ^ 2theta-2costhetasintheta-4sin ^ 2thetacos ^ 2theta) / (2sinthetacos ^ 3theta-sin ^ 3thetacostheta) Najprej prepišemo kot: f (theta) = 1 / sin (2theta) -1 / cos (2theta) -sin (2theta) / cos (2theta) Potem kot: f (theta) = 1 / sin (2theta) - (1-sin (2theta)) / cos (2theta) = (cos (2theta) - sin (2theta) -sin ^ 2 (2theta)) / (sin (2theta) cos (2theta)) Uporabili bomo: cos (A + B) = cosAcosB-sinAsinB sin (A + B) = sinAcosB + cosAsinB Torej get: f (theta) = (cos ^ 2theta-sin ^ 2theta-2costhetasintheta-4sin ^ 2thetacos ^ 2theta) / ((2sinthetacostheta) (cos ^ 2theta-sin ^ 2theta)) f (theta) = (cos

Produkt pozitivnega števila dveh števk in števila na mestu njegove enote je 189. Če je številka na desetem mestu dvakrat večja kot na mestu enote, kakšna je številka na mestu enote?

3. Upoštevajte, da sta dve števki. izpolnjevanje drugega pogoja (cond.) so, 21,42,63,84. Med temi, od 63xx3 = 189, sklepamo, da dvomestni št. je 63 in želena številka v enoti je 3. Za metodično reševanje problema predpostavimo, da je številka mesta deset enaka x in enota enote, y. To pomeni, da dvomestni št. je 10x + y. "1" (st) "cond." RArr (10x + y) y = 189. "The" 2 ^ (nd) "cond." RArr x = 2y. Sub.ing x = 2y v (10x + y) y = 189, {10 (2y) + y} = 189. :. 21y ^ 2 = 189 rArr y ^ 2 = 189/21 = 9 rArr y = + - 3. Jasno je, da je y = -3 nedopustno. :. y = 3, je želena številka, kot prej! Už