Nyatakan kos4x sebagai kuasa cosx. ?

Jawapan:

# cos4x = cos2 (2x) = warna (merah) 2cos ^ 2 (2x) -1 #

Penjelasan:

# cos2 (2x) = cos ^ 2 (2x) -sin ^ 2 (2x) #

# = cos ^ 2 (2x) -1 + cos ^ 2 (2x) = warna (merah) 2cos ^ 2 (2x) -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 #

Jawapan:

# rarrcos4x = 8cos ^ 4x-8cos ^ 2x + 1 #

Penjelasan:

# rarrcos4x #

# = cos2 * (2x) #

# = cos ^ 2 (2x) -in ^ 2 (2x) #

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

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

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

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

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