Izrazite cos4x kao moć cosxa. ?

Odgovor:

# Cos4x = cos2 (2 x) = boja (crvena) 2cos ^ 2 (2 x) -1 #

Obrazloženje:

# Cos2 (2 x) = cos ^ 2 (2 x) -sin ^ 2 (2 x) *

# = Cos ^ 2 (2 x) -1 + cos ^ 2 (2 x) = boja (crvena) 2cos ^ 2 (2 x) -1 #

# = 2 cos2x * cos2x -1-2 (cos ^ 2x-sin ^ 2x) + (cos ^ 2x-sin ^ 2x) - 1 #

# = 2 cos ^ 4x-sin ^ 2x * cos ^ 2x-sin ^ 2x * cos ^ 2x + sin ^ 4x -1 #

# = 2cos ^ 4x-4sin ^ 2x * cos ^ 2x + 2sin ^ 4x -1 #

Odgovor:

# Rarrcos4x = 8cos ^ 4x-8cos ^ 2x + 1 #

Obrazloženje:

# Rarrcos4x #

# = Cos2 * (2x) #

# = Cos ^ 2 (2 x) -sin ^ 2 (2 x) *

# = Cos ^ 2x-sin ^ 2x ^ 2- 2sinx * cosx ^ 2 #

# = Cos ^ 4x-2cos ^ 2x * grijeh ^ 2x + (grijeh ^ 2x) ^ 2-4sin ^ 2x * cos ^ 2x #

# = Cos ^ 4x-2cos ^ 2x (1-cos ^ 2x) + (1-cos ^ 2 x) ^ 2-4 (1-cos ^ 2x) + cos ^ 2x #

# = Cos ^ 4x-2cos ^ 2x + 2cos ^ 4x + 1-2cos ^ 2x + cos ^ 4x-4cos ^ 2x + 4cos ^ 4x #

# = 8cos ^ 4x-8cos ^ 2x + 1 #