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December 14th, 2017, 05:18 PM   #1
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Linear dependence question

Let $\displaystyle f_1(x) = \sin x,~ f_2(x) = \cos(x + \frac{\pi}{6}), and~ f_3(x) = \sin(x - \frac{\pi}{4}) $ for $\displaystyle 0 <= x <= 2\pi$. Show that
{ $\displaystyle f_1, f_2, f_3$ } is linearly dependent by finding constants c and k such that $\displaystyle cf_1 - 2f_2 - kf_3 = 0 $.

what is c and k?

Last edited by skipjack; February 21st, 2018 at 07:09 PM.
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December 14th, 2017, 06:53 PM   #2
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Here is a hint:

http://www2.clarku.edu/~djoyce/trig/sumformulas.jpg
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December 14th, 2017, 07:00 PM   #3
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February 21st, 2018, 11:45 AM   #4
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Here's how I would do this, using the trig identities, cos(a+ b)= cos(a)cos(b)- sin(a)sin(b) and sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b):
$\displaystyle \cos(x+ \pi/6)= \cos(x)\cos(\pi/6)- \sin(x)\sin(\pi/6)= \frac{\sqrt{3}}{2}\cos(x)- \frac{1}{2}\sin(x)$

$\displaystyle \sin(x- \pi/4)= \sin(x)\cos(\pi/4)+ \cos(x)\sin(\pi/4)= \frac{\sqrt{2}}{2} \sin(x)+ \frac{\sqrt{2}}{2} \cos(x)$.

So $\displaystyle A \sin(x)+ B \cos(x+ \pi/6)+ C \sin(x- \pi/4)= A \sin(x)+ \frac{\sqrt{3}}{2}B\cos(x)- \frac{1}{2}B \sin(x)+ \frac{\sqrt{2}}{2}\sin(x)C \sin(x)+ \frac{\sqrt{2}}{2}C \cos(x)=0$

That is, $\displaystyle (A+ \frac{\sqrt{3}}{2}B+ \frac{\sqrt{2}}{2}C) \sin(x)+ (\frac{\sqrt{3}}{2}+ \frac{\sqrt{2}}{2}C) \cos(x)= 0$

sin(x) and cos(x) are independent so we must have $\displaystyle A+ \frac{\sqrt{3}}{2}B+ \frac{\sqrt{2}}{2}C= 0$ and $\displaystyle \frac{\sqrt{3}}{2}+ \frac{\sqrt{2}}{2}C= 0$, two equations in three unknowns.

We can solve those two equations for two unknowns, A and B, say, in terms of the third, C.

Last edited by skipjack; February 21st, 2018 at 05:10 PM.
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February 21st, 2018, 08:34 PM   #5
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You made some slips.
The equations should be $\displaystyle A - \frac12B + \frac{\sqrt2}{2}C = 0$ and $\displaystyle \frac{\sqrt3}{2}B - \frac{\sqrt2}{2}C = 0$.
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