להוכיח כי Cos ^ 6 (x) + חטא ^ 6 (x) = 1/8 (5 + 3cos4x)?

נשתמש

# 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 # ו

# rarr2sin ^ 2x = 1-cos2x #

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

# = (cos ^ 2x) ^ 3 + (חטא ^ 2x) ^ 3 #

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

# 2 * חטא = 2x-cos ^ 2x * sin 2xx # 2x *

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

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

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

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

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

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

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