Rescrieți sin ^ 4 (x) tan ^ 2 (x) în ceea ce privește prima putere a cosinusului?

Răspuns:

(X) + cosc (x) / cos ^ 2 (x) # #

Explicaţie:

# Păcat ^ 4 (x) tan ^ 2 (x) #

= (1-cos ^ 2 (x)) ^ 2 (sin ^ 2 (x)) / cos ^

(X) = cos (2) cos 2 (x) + cos ^ 4 (x)

(x) cos sin ^ 2 (x) -2sin ^ 2 (x) cos ^ 2 (x) + sin ^

= ((1-cos ^ 2 (x)) -2 (1-cos ^ 2 (x)) cos ^ / cos ^ 2 (x) #

(X) + 2 cos ^ 2 (x) + 2cos ^ 4 (x) + cos ^ (x) -cos ^

(X) + cosc (x) / cos ^ 2 (x) # #

Răspuns:

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

Explicaţie:

# Păcat ^ 4xtan ^ 2x = sin ^ 6x / cos ^ 2x #

#cos (2x) = cos ^ 2x-sin ^ 2x #

#color (alb) (cos (2x)) = cos ^ 2x- (1-cos ^ 2x) #

#color (alb) (cos (2x)) = 2cos ^ 2x-1 #

# cos ^ 2x = (cos (2x) +1) / 2 #

Folosind Theoreomul lui De Moivre, putem evalua # Păcat ^ # 6x:

# 2isin (x) = z 1 / z # (Unde # Z = cosx + isinx #)

# (2isin (x)) ^ 6 = (z-1 / z) ^ 6 #

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

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

# (Z ^ n-1 / z ^ n) = 2cos (nx) #

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

# ((- 20 + 2cos (6x) -12cos (4x) + 30cos (2x)) / - 64) / ((cos (2x) +1) / 2) = - (2cos (6x) -12cos (4x) + 30cos (2x) -20) / (32cos (2x) +32) #

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

Răspuns:

# Păcat ^ 4x * tan ^ 2x = 1/16 (10-15cos2x + 6cos4x-cos6x) / (1 + cos2x) #

Explicaţie:

Vom folosi, # Rarrsin ^ 2x = (1-cos2x) / 2 #

# rarrcos ^ 2x = (1 + cos2x) / 2 #

# rarr4cos ^ 3x = cos3x + 3cosx #

Acum, # RArrtan ^ 2 x * sin ^ # 4x

# = Sin ^ 2x / cos ^ 2x * sin ^ # 4x

# = (Sin ^ 2x) ^ 3 / cos ^ 2x #

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

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

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

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

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

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

# = 1/16 (4-12cos2x + 6 + 6cos4x-cos6x-3cos2x) / (1 + cos2x) #

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