We want to prove that the functions {1,(((Cos[t]))^2),(((Cos[t]))^4),(((Cos[t]))^6)} are linearly independent. We need to prove that when the linear combination

a1*1+a2*(((Cos[t]))^2)+a3*(((Cos[t]))^4)+a4*(((Cos[t]))^6)

equals to 0 for all t∈ℝ, then all the coefficients a1, a2, a3, a4 must equal to 0.

Since the equality a1*1+a2*(((Cos[t]))^2)+a3*(((Cos[t]))^4)+a4*(((Cos[t]))^6) =0 for all t∈ℝ, we can substitute t with for carefully selected special values:

(a1*1+a2*(((Cos[t]))^2)+a3*(((Cos[t]))^4)+a4*(((Cos[t]))^6))/.{t->Pi/2}

(a1*1+a2*(((Cos[t]))^2)+a3*(((Cos[t]))^4)+a4*(((Cos[t]))^6))/.{t->0}

(a1*1+a2*(((Cos[t]))^2)+a3*(((Cos[t]))^4)+a4*(((Cos[t]))^6))/.{t->Pi/4}

(a1*1+a2*(((Cos[t]))^2)+a3*(((Cos[t]))^4)+a4*(((Cos[t]))^6))/.{t->Pi/6}

Now we have four equations with four unknowns. We collect these four equations in a list of equations and ask Mathematica to solve them, as follows:

Solve[{a1==0,a1+a2+a3+a4==0,a1+(a2/2)+(a3/4)+(a4/8)==0,a1+((3 a2)/4)+((9 a3)/16)+((27 a4)/64)==0},{a1,a2,a3,a4}]

Mathematica gives us the only solution.

Using linear algebra methods we would form the matrix of the homogeneous system and row reduce that matrix:

MatrixForm[RowReduce[{{1,0,0,0},{1,1,1,1},{1,(1/2),(1/4),(1/8)},{1,(3/4),(9/16),(27/64)}}]]

Thus the only solution of the above homogenous system is the trivial solution.

Next we explore some linear combinations of the given four functions:

{1,(((Cos[t]))^2),(((Cos[t]))^4),(((Cos[t]))^6)}

The above is the list of the given functions.

The next command makes a specific linear combination of the given functions:

{1,-8,8,0}.{1,(((Cos[t]))^2),(((Cos[t]))^4),(((Cos[t]))^6)}

Next plot this linear combination:

Plot[{1-8 ((Cos[t])^2)+8 ((Cos[t])^4),Cos[t]},{t,-2 Pi,2 Pi}]

We ques that the linear combination equals Cos[4 t]. We verify it with a plot

Plot[{1-8 ((Cos[t])^2)+8 ((Cos[t])^4),Cos[t],Cos[4t]},{t,-2 Pi,2 Pi}]

Does Mathematica know this identity?

Simplify[1-8 ((Cos[t])^2)+8 ((Cos[t])^4)-Cos[4t]]

Yes!

Let us check the next linear combination:

{-1,18,-48,32}.{1,(((Cos[t]))^2),(((Cos[t]))^4),(((Cos[t]))^6)}

FullSimplify[{-1,18,-48,32}.{1,(((Cos[t]))^2),(((Cos[t]))^4),(((Cos[t]))^6)}-Cos[6t]]

Now we explore the four functions

{1, Cos[2 t],Cos[4 t],Cos[6 t]}

The coordinate vectors of these functions are verified below:

FullSimplify[{1,0,0,0}.{1,(((Cos[t]))^2),(((Cos[t]))^4),(((Cos[t]))^6)}-1]

FullSimplify[{-1,2,0,0}.{1,(((Cos[t]))^2),(((Cos[t]))^4),(((Cos[t]))^6)}-Cos[2t]]

FullSimplify[{1,-8,8,0}.{1,(((Cos[t]))^2),(((Cos[t]))^4),(((Cos[t]))^6)}-Cos[4t]]

FullSimplify[{-1,18,-48,32}.{1,(((Cos[t]))^2),(((Cos[t]))^4),(((Cos[t]))^6)}-Cos[6t]]

Next we use the coordinate vectors to prove that the functions

{1, Cos[2 t],Cos[4 t],Cos[6 t]}

are linearly independent:

RowReduce[{{1,0,0,0},{-1,2,0,0},{1,-8,8,0},{-1,18,-48,32}}]

Thus, the coordinate vectors are linearly independent. Therefore, (citation needed) the functions

{1, Cos[2 t],Cos[4 t],Cos[6 t]}

are linearly independent.

Cos[8 t]

FullSimplify[Cos[2*(4 t)]==(-1+2((Cos[4 t])^2))]

{1,-8,8,0}.{1,(((Cos[t]))^2),(((Cos[t]))^4),(((Cos[t]))^6)}

FullSimplify[{1,-8,8,0}.{1,(((Cos[t]))^2),(((Cos[t]))^4),(((Cos[t]))^6)}-Cos[4t]]

{1,-8,8,0}.{1,(((Cos[t]))^2),(((Cos[t]))^4),(((Cos[t]))^6)}

FullSimplify[Cos[2*(4 t)]==(-1+2(((1-8 ((Cos[t])^2)+8 ((Cos[t])^4)))^2))]

Expand[(-1+2(((1-8 ((Cos[t])^2)+8 ((Cos[t])^4)))^2))]