Jtdylktuy

I want to determine whether 3 functions are linearly independent:

begin{align*}
x_1(t) & = 3 \
x_2(t) & = 3sin^2(t) \
x_3(t) & = 4cos^2(t)
end{align*}

Definition of Linear Independence: $c_1x_1 + c_2x_2 + c_3x_3 = 0 implies c_1=c_2=c_3=0$ (only the trivial solution)

So we have:
begin{align}
3c_1 + 3c_2sin^2(t) + 4c_3cos^2(t) = 0
end{align}

My first idea is to differentiate both sides and get:

$6c_2sin(t)cos(t) - 8c_3cos(t)sin(t) = 0$

Then we can factor to get:

$sin(t)cos(t)(6c_2 - 8c_3) = 0$

So $c_3= dfrac{6}{8}c_2$ gives the equation equals zero. Thus all $c$ are not $0$ and thus $x_1, x_2, x_3$ are linearly dependent.

Is this correct? Or is there a cleaner way to do this?

Yes, indeed, your answer is fine. And it would have been a particularly fine determining the linear (in)dependence of a system of equations that doesn't readily admit of another observation about the relationship between $cos^2 t$ and $sin^2 t$ $(dagger)$. Indeed, you're one step away from working with the Wronskian, which is a useful tool to prove linear independence.

$(dagger)$ Now, to the observation previously noted: You could have also used the fact that $$x_1(t) - left[(x_2(t) +frac 34 x_3(t)right] = 3 - (3 sin^2 t + 3cos^2 t)= 3 - 3left(underbrace{sin^2(t) + cos^2(t)}_{large = 1}right)=0$$

and saved yourself a little bit of work: you can read off the nonzero coefficients $c_i$ to demonstrate their existence: $c_1 = 1, c_2 = -1, c_3 = -frac 34$, or you could simply express $x_1$ as a linear combination of $x_2, x_3$, to conclude the linear dependence of the vectors. (But don't count on just any random set of vectors turning out so nicely!)

It is much more easier to use known identity $sin^2{t}+cos^2{t}=1$. We have $x_1(t)-x_2(t)-frac{3}{4}x_3(t)=3-3cos^2{t}-3sin^2{t}=0$, so functions are linearly dependent.

This is a pretty straightforward question, firstly, let me remind you the definition of linearly dependent functions. It says,

A set of functions $mathrm{f_1(x), f_2(x), ... , f_n(x)}$ are called linearly dependent if

$mathrm{c_1f_1(x)+ c_2f_2(x)+ ... + c_n f_n(x) = 0, where c_1, c_2, ... , c_n are arbitrary constants}$ holds true for atleast two non-zero c's.

Coming to the question, then

$mathrm{f_1(x) = 3, f_2(x) = 3sin^2x, f_3(x) = 4cos^2x}$

Consider $mathrm{c_1, c_2, c_3}$ as arbitrary constants,

$mathrm{3c_1 + 3c_2sin^2x + 4c_3cos^2x}$

Easily, if $mathrm{c_1 = frac{-1}{3}, c_2 = frac{1}{3}, c_3 = frac{1}{4}}$, then

$mathrm{3c_1 + 3c_2sin^2x + 4c_3cos^2x}$ = $0$

As all the arbitrary constants are non-zero, the functions are definitely LD.