Proving Linear Dependence of Trig Functions on the Real Line

[misterau]

Homework Statement

show {cos x ,sin x , cos 2x , sin 2x , (cos x − sin x)^2 − 2*sin^2( x)} is not a linearly independent set of real valued functions on the real line R.

The Attempt at a Solution

Not linearly independent = linearly dependent?
So if
f(x) = cos (x)
g(x) = sin (x)
m(x) = cos (2*x) = 1 - 2sin^2(x)
k(x) = sin (2*x) = 2sin(x)cos(x)
h(x) = (cos (x) − sin (x))^2 − 2*sin^2(x)

z(x) = (a*f(x)) + (b*g(x)) + (c*m(x)) + (d*k(x)) + (e*h(x))

To prove it is linearly dependence we need scalars a,b,c,d,e that work with ANY x that make the equation z(x) equal to zero? Some of the scalars can be zero correct? Just not all of them then it becomes an non trivial answer.

For instance:

0=1 * 2sin(x)cos(x) + -1 * (1 - 2sin^2(x) ) + 1 *( cos^2(x) + sin^2(x) - 2sin(x)cos(x) - 2sin^2(x) ) + 0 * cos(x) + 0 * sin(x)

 

[HallsofIvy]

Homework Statement

show {cos x ,sin x , cos 2x , sin 2x , (cos x − sin x)^2 − 2*sin^2( x)} is not a linearly independent set of real valued functions on the real line R.

The Attempt at a Solution

Not linearly independent = linearly dependent?

Yes, that is correct.

So if
f(x) = cos (x)
g(x) = sin (x)
m(x) = cos (2*x) = 1 - 2sin^2(x)
k(x) = sin (2*x) = 2sin(x)cos(x)
h(x) = (cos (x) − sin (x))^2 − 2*sin^2(x)

?? m, k, and h are not among the functions you give initially. Is this a different example?

z(x) = (a*f(x)) + (b*g(x)) + (c*m(x)) + (d*k(x)) + (e*h(x))

To prove it is linearly dependence we need scalars a,b,c,d,e that work with ANY x that make the equation z(x) equal to zero? Some of the scalars can be zero correct? Just not all of them then it becomes an non trivial answer.

Yes.

For instance:

0=1 * 2sin(x)cos(x) + -1 * (1 - 2sin^2(x) ) + 1 *( cos^2(x) + sin^2(x) - 2sin(x)cos(x) - 2sin^2(x) ) + 0 * cos(x) + 0 * sin(x)