Rewrite 2sin ^ 6 (x) sa mga tuntunin ng isang expression na naglalaman lamang ng mga cosines sa kapangyarihan ng isa?

Sagot:

# 2sin ^ 6x = (10-cos (6x) + 6cos (4x) -15cos (2x)) / 16 #

Paliwanag:

Kami ay binigyan # 2sin ^ 6x #

Paggamit ng Teorem ni De Moivre alam natin na:

# (2isin (x)) ^ n = (z-1 / z) ^ n # kung saan # z = cosx + isinx #

# (2isin (x)) ^ 6 = -64sin ^ 6x = z ^ 6-6z ^ 4 + 15z ^ 2-20 + 15 / z ^ 2-6 / z ^ 4 + 1 / z ^ 6 #

Una naming isinaayos ang lahat ng bagay upang makakuha ng:

# -20 + (z + 1 / z) ^ 6-6 (z + 1 / z) ^ 4 + 15 (z + 1 / z) ^ 2 #

Gayundin, alam natin iyan # (z + 1 / z) ^ n = 2cos (nx) #

# -64sin ^ 6x = -20 + (2cos (6x)) - 6 (2cos (4x)) + 15 (2cos (2x)) #

# -64sin ^ 6x = -20 + 2cos (6x) -12cos (4x) + 30cos (2x) #

# sin ^ 6x = (- 20 + 2cos (6x) -12cos (4x) + 30cos (2x)) / - 64 #

# 2sin ^ 6x = 2 * (- 20 + 2cos (6x) -12cos (4x) + 30cos (2x)) / - 64 = (- 20 + 2cos (6x) -12cos (4x) + 30cos (2x)) / -32 = (10-cos (6x) + 6cos (4x) -15cos (2x)) / 16 #

Sagot:

# rarr2sin ^ 6x = 1/16 10-15cos2x + 6cos4x-cos6x #

Paliwanag:

# rarr2sin ^ 6x #

# = 1/4 (2sin ^ 2x) ^ 3 #

# = 1/4 (1-cos2x) ^ 3 #

# = 1/4 1-3cos2x + 3cos ^ 2 (2x) -cos ^ 3 (2x) #

# = 4 / (4 * 4) 1-3cos2x + 3cos ^ 2 (2x) -cos ^ 3 (2x) #

# = 1/16 4-12cos2x + 3 * 2 * {2cos ^ 2 (2x)} - 4cos ^ 3 (2x) #

# = 1/16 4-12cos2x + 3 * 2 * {1 + cos4x} -cos6x-3cos2x #

# = 1/16 4-15cos2x + 6 + 6cos4x-cos6x #

# = 1/16 10-15cos2x + 6cos4x-cos6x #