How do you verify #cos4x = 8cos(^4)x - 8cos²x + 1#?

1 Answer
Nov 27, 2015

To verify:

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

We will need the following trigonometrical identities:

  1. # sin^2x + cos^2 x = 1 color(white)(xxx)<=> sin^2(x) = 1 - cos^2(x)#
  2. # sin(x + y) = sin x cos y + cos x sin y #
  3. # cos(x + y) = cos x cos y - sin x sin y#

So, let's start. :-)

#cos (4x) = cos(2x + 2x) #

# color(white) (xxxxx) = cos(2x) cos(2x) - sin(2x) sin(2x)#

# color(white) (xxxxx) = (cos(2x))^2 - (sin(2x))^2#

# color(white) (xxxxx) = (cos(x +x ))^2 - (sin(x + x))^2#

# color(white) (xxxxx) = ( cos x cos x - sin x sin x )^2 - ( sin x cos x + sin x cos x )^2#

# color(white) (xxxxx) = ( cos^2 x - sin^2 x )^2 - ( 2 sin x cos x )^2#

# color(white) (xxxxx) = ( cos^2 x - sin^2 x )^2 - 4 sin^2 x cos^2 x #

# color(white) (xxxxx) = ( cos^2 x - (1 - cos^2 x ) )^2 - 4 (1 - cos^2 x ) cos^2 x #

# color(white) (xxxxx) = ( 2 cos^2 x - 1 )^2 - 4 cos^2 x + 4 cos^4 x #

# color(white) (xxxxx) = 4 cos^4 x - 4 cos^2 x + 1 - 4 cos^2 x + 4 cos^4 x #

# color(white) (xxxxx) = 8 cos^4 x - 8 cos^2 x + 1#

Hope that this helped!