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

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

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Answer 1 · 2018-06-26T13:04:01

impossible equation

Explanation:

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

Answer 2 · 2018-06-26T16:31:14

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

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

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