พิสูจน์ว่า 32sin ^ 4x.cos ^ 2x = cos6x-2cos4x-cos 2x + 2 หรือไม่

# RHS = cos6x-2cos4x-cos2x + 2 #

# = cos6x-cos2x + 2 (1-cos4x) #

# = - 2sin ((6x + 2x) / 2) * บาป ((6x-2x) / 2) + 2 * 2sin ^ 2 (2x) #

# = 4sin ^ 2 (2x) -2sin4x * sin2x = 4sin ^ 2 (2x) -2 * 2 * * * * * * * * sin2x cos2x * sin2x #

# = 4sin ^ 2 (2x) -4sin ^ 2 (2x) * cos2x #

# = 4sin ^ 2 (2x) 1 cos2x #

# = 4 * (* 2sinx cosx) ^ 2 * 2sin ^ 2x #

# = 4 * 4sin ^ 2x * cos ^ 2x * 2sin ^ 2x = ^ 32sin 4x * cos ^ 2x = LHS #