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Given the simultaneous equations $$A\cos{(\sqrt{\lambda}\pi)} + B\sin{(\sqrt{\lambda}\pi)} = 0$$ $$A\cos{(2\sqrt{\lambda}\pi)}+B\sin{(2\sqrt{\lambda}\pi)} = 0$$ We want to show this has not trivial solutions (ie. solutions when $A\not=0$ and $B\not= 0$). In my notes I have that this gives non-trivial solutions when $$\sin{(2\sqrt{\lambda}\pi)}\cos{(\sqrt{\lambda}\pi)} - \cos{(2\sqrt{\lambda}\pi)}\sin{(\sqrt{\lambda}\pi)} = 0$$ but can't quite see why. Can someone explain, thanks.

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  • $\begingroup$ Double-angle formulas look promising... $\endgroup$ – abiessu Nov 14 '14 at 15:45
  • $\begingroup$ Double-angle formulas would be useful for transforming the last equation into something more informative, but it's not the best way of getting the equation in the first place. $\endgroup$ – Teepeemm Nov 14 '14 at 20:46
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If the system $$Au+Bv=0$$ $$Aw+Bz=0$$ has a non-trivial solution, then $$u=-\frac{Bv}A=\frac{Avw}{Az}=\frac{vw}z$$ Therefore, $$uz-vw=0$$

Remark: The case $z=0$ must be considered in a different but easy way.

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    $\begingroup$ Just in case OP does not know about determinants. $\endgroup$ – ajotatxe Nov 14 '14 at 15:55
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A system $$ Ax=0 $$ has a notrivial solution if and only if $\det A=0$.

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If you set up the corresponding augmented coefficient matrix, you'll see that the matrix has determinant $$\left(\sin{(2\sqrt{\lambda}\pi)}\cos{(\sqrt{\lambda}\pi)} - \cos{(2\sqrt{\lambda}\pi)}\sin{(\sqrt{\lambda}\pi)}\right)$$

We want the determinant to be zero in order to ensure there is a non-trivial solution.

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