Αποδείξτε ότι 32sin ^ 4x.cos ^ 2x = cos6x-2cos4x-cos 2x + 2?

Αποδείξτε ότι 32sin ^ 4x.cos ^ 2x = cos6x-2cos4x-cos 2x + 2?
Anonim

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

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

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

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

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

# = 4sin ^ 2 (2χ) 1-cos2x #

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

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