Question #fcb2e

#cosxcos3x=0.707#

#f(x)=cosxcos3x-0.707=0#

Use Newton method to find a zero of #f#

#x_(n+1)=x_n-(f(x_n))/(f'(x_n))#

#f'(x)=-2sin2x-2sin4x#

#0.707=cos(1/4pi)#, so a good #x_0# would be #(1/4pi)/3=0.262#

#x_1=0.262-(cos(0.262)cos(3(0.262))-0.707)/(2sin(2(0.262)-2sin(4(0.262))=0.253#

#x_2=0.251#

#x_3=0.251#

#x_4=0.251#

The iterations approach #0.251# to #0.251# is a solution to #cosxcos3x=0.707#

cos x.cos 3x = 0.707
Use trig identity:
#cos a.cos b = (1/2)(cos (a - b) + cos (a + b))#
In this case:
#cos x.cos 3x = (1/2)(cos 2 x + cos 4x)#-->
cos 2x + cos 4x = 2(0.707) = 1.414
Call 2x = X,
cos X + cos 2X - 1.414 = 0.
Replace cos 2X by (2cos^2 X - 1):
#cos X + (2cos^2 X - 1) - 1.414 = 0#
2cos^2 X + cos X - 2.414 = 0.
Solve this quadratic equation for cos X.
#D = d^2 = b^2 - 4ac = 1 + 19.28 = 20.28# --> #d = +- 4.50#
There are 2 real roots:
#cos X = -b/(2a) +- d/(2a) = - 1/4 +- (4.50)/4 = - 0.25 +- 1.125#
a. cos X = - 1.375 (rejected as < - 1)
b. cos 2x = cos X = 0.875
Calculator and unit circle give:
#2x = +- 28^@96# --> #x = +- 14^@48#
Check by calculator:
#x = 14^@48# --> #cos x = 0.97# --> #3x = 43^@44# --> #cos 3x = 0.73#.
cos x.cos 3x = 0.97(0.73) = 0.71. Proved.