Prokázat, že 32sin ^ 4x.cos ^ 2x = cos6x-2cos4x-cos 2x + 2?

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

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

# = - 2sin ((6x + 2x) / 2) * sin ((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 * 4s ^ 2x * cos ^ 2x * 2sin ^ 2x = 32sin ^ 4x * cos ^ 2x = LHS #