# Linear independence of functions: $x_1(t) = 3$, $x_2(t)=3\sin^2t$, $x_3(t)=4\cos^2t$

I want to determine whether 3 functions are linearly independent:

\begin{align*} x_1(t) & = 3 \\ x_2(t) & = 3\sin^2(t) \\ x_3(t) & = 4\cos^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_2\sin^2(t) + 4c_3\cos^2(t) = 0 \end{align}

My first idea is to differentiate both sides and get:

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

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?

• How could you take advantage of the Pythagorean identity? – David Mitra Jun 26 '13 at 22:13
• Since $\cos^2 \varphi + \sin^2 \varphi \equiv 1$, you can directly see that $4x_2 + 3x_3$ is a constant. – Daniel Fischer Jun 26 '13 at 22:13
• I was thinking about that, but I how do I deal with the coefficients? $(\sqrt(3c_2)\sin(t))^2 + (\sqrt(4c_3)\sin(t))^2$ – CodeKingPlusPlus Jun 26 '13 at 22:15
• @CodeKingPlusPlus Linear independence is invariant under multiplication of individual vectors by nonzero constants. So you can just multiply your three vectors by respectively $\frac13,\frac13,\frac14$, and the problem becomes easy. – Marc van Leeuwen Dec 26 '14 at 6:24

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 $$x_1(t) - \left[(x_2(t) +\frac 34 x_3(t)\right] = 3 - (3 \sin^2 t + 3\cos^2 t)= 3 - 3\left(\underbrace{\sin^2(t) + \cos^2(t)}_{\large = 1}\right)=0$$

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 $x_1(t)-x_2(t)-\frac{3}{4}x_3(t)=3-3\cos^2{t}-3\sin^2{t}=0$, 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.