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Can someone help me on this one I been having problems with the independent functions lately? Thanks

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  • $\begingroup$ Were are you stuck ? Assume you have a linear combination of those functions that is zero everywhere, and show that the coefficients of this linear combination are all zero. You may first try to plug some values of $t$... $\endgroup$
    – Joel Cohen
    Oct 24, 2015 at 15:46

3 Answers 3

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They are linearly independant if the only solution to the equation $a\sin(t)+b\cos(t)+ct\sin(t)+dt\cos(t)=0$ has only one solution $a=b=c=d=0$.

But you have only one equation and 4 variables. Since the equation should be true for all values of $t$, one way of overcoming this difficulty is replacing $t$ by values which simplify the calculations.

For example, if you put $t=0$ in the equation, you get, $a\cdot 0+b\cdot 1+c\cdot 0+d\cdot 0=0%$. Thus, $b=0$ is the only value for $b$ and you wind up with one equation with 3 variables: $a\sin(t)+ct\sin(t)+dt\cos(t)=0$ Suitable choices of $t$ will let you find $a$, $c$, $d$, which, hopefully will all be 0.

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  • $\begingroup$ Thank you so much for that, I was doing the same setting that you provide $\endgroup$
    – elarqui
    Oct 24, 2015 at 15:58
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Hint:

Start fro a linear relation between them: $$\lambda \sin t+\mu \cos t+\nu t\sin t +\xi t\sin t =0$$ and give $t$ $3$ well-chosen values. For instance, setting $t=0$ yields at once $\mu=0$.

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If $a\sin t+b t \sin t +c\cos t +d t \cos t=0$ for all $t$,with $a,b,c,d$ not all $0$ then $$\tan t=-\frac {c+d t}{a+b t}$$ whenever $\cos t \ne 0$ and $a+b t\ne 0$. But for any $r$, there are infinitely many $t$ for which $\tan t=r$and but at most one $t$ for which $(c+d t)/(a+b t)=r$..... If we are restricted to $t\in [u,v]$ with $u<v$ we can use the method given in the other answers :Take various values of $t$ and get a triplet of linear equations in $a,b,c,d$ that are inconsistent.

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