Bagaimana Anda menyederhanakan f (theta) = sin4theta-cos6theta menjadi fungsi trigonometri dari unit theta?

Menjawab:

#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 #

Penjelasan:

Kami akan menggunakan dua identitas berikut:

#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) + dosa (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) 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 #

Bagaimana Anda dapat menggunakan fungsi trigonometri untuk menyederhanakan 12 e ^ ((19 pi) / 12 i) menjadi bilangan kompleks non-eksponensial?

3sqrt6-3sqrt2-i (3sqrt6 + 3sqrt2) Kita dapat berubah menjadi re ^ (itheta) menjadi bilangan kompleks dengan melakukan: r (costheta + isintheta) r = 12, theta = (19pi) / 12 12 (cos ((19pi)) / 12) + isin ((19pi) / 12)) 3sqrt6-3sqrt2-i (3sqrt6 + 3sqrt2)

Bagaimana Anda menyederhanakan f (theta) = csc2theta-sec2theta-3tan2theta menjadi fungsi trigonometrik dari unit theta?

F (theta) = (cos ^ 2 theta-sin ^ 2theta-2costhetasintheta-4sin ^ 2thacac ^ 2theta) / (2sinthetacos ^ 3theta-sin ^ 3thetacostheta) Pertama, tulis ulang sebagai: f (theta) = 1 / sin (2theta) -1 / cos (2theta) -sin (2theta) / cos (2theta) Kemudian sebagai: f (theta) = 1 / sin (2theta) - (1-sin (2theta)) / cos (2theta) = (cos (2theta) - sin (2theta) -sin ^ 2 (2theta)) / (sin (2theta) cos (2theta)) Kita akan menggunakan: cos (A + B) = cosAcosB-sinAsinB sin (A + B) = sinAcosB + cosAsinB Jadi, kami dapatkan: f (theta) = (cos ^ 2 theta-sin ^ 2theta-2costhetasintheta-4sin ^ 2thacac ^ 2theta) / ((2sinthetacostheta) (cos ^ 2theta-sin

Bagaimana Anda dapat menggunakan fungsi trigonometri untuk menyederhanakan 4 e ^ ((pi) / 4 i) menjadi bilangan kompleks non-eksponensial?

Gunakan rumus Moivre. Rumus Moivre memberi tahu kita bahwa e ^ (itheta) = cos (theta) + isin (theta). Terapkan ini di sini: 4e ^ (i (5pi) / 4) = 4 (cos ((5pi) / 4) + isin ((5pi) / 4)) Pada lingkaran trigonometri, (5pi) / 4 = (-3pi) / 4. Mengetahui bahwa cos ((- 3pi) / 4) = -sqrt2 / 2 dan sin ((- 3pi) / 4) = -sqrt2 / 2, kita dapat mengatakan bahwa 4e ^ (i (5pi) / 4) = 4 (- sqrt2 / 2 -i (sqrt2) / 2) = -2sqrt2 -2isqrt2.