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

Uporabili bomo

# rarra ^ 3 + 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 # in

# rarr2sin ^ 2x = 1-cos2x #

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

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

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

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

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

# = 1/8 4 + 4os4x + 2sin ^ 2 (2x) #

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

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