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So, it is stated that if $C_1g(t)+C_2f(t)=0$ then $g(t)=\sin{t}$ and $f(t)=\cos{t}$ are linearly dependent, i.e., $C_1g(t)+C_2f(t)=0 \iff C_1=C_2=0$. It is proved simply by assuming two values of $t$, first $t=0$ and then $t=\frac{\pi}{2}$. In the first case $C_2=0$ and in the second case $C_1$ is $0$.

I don't get how is that a prove, if you can find other values of $t$, where you can express one function by the the other. Like if $t=\frac{\pi}{4}$, then if $C_1=1$ and $C_2=-1$ $\Rightarrow C_1g(t)+C_2f(t)=0$. What am I missing here?

Valentin
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user
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1 Answers1

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First of all $g(t) = \sin{t}$ and $f(t) = \cos{t}$ are linearly independent, and no assumption about their linear combination is going to change that fact.

I, therefore, assume that what is in fact being asked is to show that any vanishing linear combination of them ($C_1g(t)+C_2f(t)=0$) is necessarily trivial ($C_1 = C_2 = 0$).

This can indeed be proved by choosing particular values of $t=0,\frac{\pi}{2}$, because $C_1$, $C_2$ are constants, so if they can be shown to be 0 somewhere, they are 0 everywhere.

If you set $t=\frac{\pi}{4}$ then all you can conclude is that $C_1+C_2=0$, which is of course also satisfied by the trivial solution. You cannot, at the same time, pick arbitrary values for the constants themselves.

Valentin
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