How to prove: cos^2(2x) - sin^2(x) = cos(x).cos(3x)?

2 Answers
Nghi N.
Jun 26, 2018

impossible equation

Explanation:

#cos^2 x - sin^2 x = cos x.cos 3x#
Trig identity:
#cos^2 x - sin^2 x = cos 2x#
#cos 2x = cos x.cos 3x#.(1)
This equation is impossible.
If #x = pi/6# --> #2x = pi/3# --> #cos 2x = cos (pi/3) = 1/2# -->
#cos 3x = cos (pi/2) = 0#.
Equation (1) becomes:
1/2 = (sqrt3/2)(zero)
This is impossible.

Nghi N.
Jun 26, 2018

First, develop the left side:
#f(x) = cos^2 (2x) - sin^2 x = (cos 2x - sin x)(cos 2x + sin x)#
Use trig identities:
#cos a + cos b = 2cos ((a + b)/2).cos ((a - b)/2)#
#cos a - cos b = - 2sin ((a +b)/2)sin ((a - b)/2)#
Note that:
#cos 2x - sin x = cos 2x - cos (pi/2 - x) = -2sin(x/2 + pi/4)sin ((3x)/2 - pi/4)#.
#(cos 2x + sin x) = cos 2x + cos (pi/2 - x) = = 2cos(x/2 + pi/4)cos ((3x)/2 - pi/4)#
Also note that:
a. #2sin (x/2 + pi/4)(cos (x/2 + pi/4) = sin (x + pi/2) = cos x#
b. #-2cos((3x)/2 - pi/4)sin ((3x)/2 - pi/4) = -sin (3x - pi/2) = cos 3x#
Finally:
#f(x) = cos^2 2x - sin^2 x = cos x.cos 3x#. Proved

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