Dokazati da Cos ^ 6 (x) + sin ^ 6 (x) = 1/8 (5 + 3cos4x)?

Koristit ćemo

# ^ 3 rarra + b ^ 3 = (a + b) (a ^ 2-ab + b ^ 2) *

# Rarra ^ 2 + b ^ 2 = (a-b) ^ 2 + 2ab #

# Rarrsin ^ 2x + cos ^ 2x = 1 #

# rarr2cos ^ 2x = 1 + # cos2x i

# Rarr2sin ^ 2x = 1 cos2x #

# LHS = cos ^ 6 (x) + sin ^ 6 (x) *

# = (Cos ^ 2 x) ^ 3 + (sin ^ 2 x) ^ 3 #

# = Cos ^ 2x + sin ^ 2x (cos ^ 2 x) ^ 2-cos ^ 2x * sin ^ 2x + sin ^ 2 x) ^ 2 #

# = 1 x (cos ^ 2x-sin ^ 2 x) ^ 2 + 2cos ^ 2x * sin ^ 2x-cos ^ 2x * sin ^ 2x #

# = Cos ^ 2 (2x) + cos ^ 2x * sin ^ 2x #

# = 1/4 4cos ^ 2 (2x) + 4cos ^ 2x * sin ^ 2x #

# = 1/4 2 (1 + cos4x) + sin ^ 2 (2 x) #

# = 2 / (4 * 2) 2 + 2cos4x + sin ^ 2 (2 x) #

# = 1/8 4 + + 4cos4x 2sin ^ 2 (2 x) #

# = 1/8 4 + + 1 4cos4x cos4x #

# = 1/8 5 + 3cos4x = RHS #