show that cos^2(x) - sin^2(x) = cos(2x)?

Answer 1

You can show this using the angle sum identity, cos(x + y) = cos(x)*cos(y) - sin(x)*sin(y). Since cos(2x) = cos(x + x), the identity reduces to cos^2(x) - sin^2(x).

Answer 2

since cos^2 (x) = 1-sin^2(x),
you get 1-2*sin^2(x).

But recall that sin^2(x) = [1-cos(2x)]/2
therefore 2*sin^2(x) = 1-cos(2x)

plug back in and get
1-((1-cos(2x))) = cos(2x)
identity.

Answer 3

Use addition identity: (put x's in for a & b)

cos (a + b) = (cos a)(cos b) - (sin a)(sin b)


cos^2(x) - sin^2(x) = cos(2x)?
(cos x)(cos x) - (sin x)(sin x) = cos(2x)

Answer 4

LHS:

cos^2(x) - (1 - cos^2(x)) ...................{sin^2(x)+cos^2(x)=1}
2cos^2(x) - 1
=cos(2x)........................................{axiom : 2cos^(2x) - 1 =
1 - 2sin^(2x) =
cos(2x)}
hope tht explains it :D

Answer 5

a pretty demonstration: with i (where i^2=-1)
exp(ix)=cosx + i*sinx
so : cosx = Re(exp(ix)) and sinx=Im(exp(ix))
cos(2x) = Re(exp(i2x))
=Re((exp(ix)^2)
=Re((cosx+isinx)^2)
=Re( cos^2(x) + 2i cosx sinx - sin^2(x))
= cos^2(x) - sin^2(x)