Linear Independence of Cosine Function

Question

Is it true that the the set $\{\cos(x+y), \cos(x+y), \cos x, \cos y\}$ is linearly independent? I claim that this is LI (Linearly independent). I start with the following. Let $a$, $b$, $c$ and $d$ be real number where $$ a\cos(x+y) + b\cos(x-y) + c\cos x +d\cos y = 0$$ for all $(x,y) \in R^2$ Then, for $ (x,y) = (π/2,π/2),(2π,2π),(π/4, 3π/4), (0,π/2)$. Then, finding $a,b,c$ and $d$ using elimination then i obtain $a = b = c = d = 0$. Is my solution correct?

Answer 1

I did not check your computations, but your approach is totally correct.

Answer 2

I assume you're considering the vector space of (continuous) functions $\mathbb{R}^2\to\mathbb{R}$.

Your approach is sensible and likely correct.

Here's another way to solve the problem, using differentiation: suppose $$ f(x,y)=a\cos(x+y)+b\cos(x-y)+c\cos x+d\cos y $$ is identically zero. Then, evaluating at $(0,0)$, $a+b+c+d=0$; considering partial derivatives \begin{align} \frac{\partial f}{\partial x}\Big|_{(\pi/2,0)} &=(-a\sin(x+y)-b\sin(x-y)-c\sin x)\Big|_{(\pi/2,0)}=-a-b-c=0 \\[6px] \frac{\partial f}{\partial x}\Big|_{(0,\pi/2)} &=(-a\sin(x+y)-b\sin(x-y)-c\sin x)\Big|_{(0,\pi/2)}=-a+b=0 \\[6px] \frac{\partial f}{\partial y}\Big|_{(0,\pi/2)} &=(-a\sin(x+y)+b\sin(x-y)-c\sin y)\Big|_{(0,\pi/2)}=-a+b-c=0 \end{align} From the first and third equations we get $b=0$ and $a+c=0$; the second equation now yields $a=0$ and so also $c=0$.