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December 14th, 2017, 04:18 PM  #1 
Senior Member Joined: Jan 2017 From: Toronto Posts: 178 Thanks: 2  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 06:09 PM. 
December 14th, 2017, 05:53 PM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 383 Thanks: 207 Math Focus: Dynamical systems, analytic function theory, numerics  
December 14th, 2017, 06:00 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 1,980 Thanks: 1027  
February 21st, 2018, 10:45 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,197 Thanks: 872 
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 04:10 PM. 
February 21st, 2018, 07:34 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,059 Thanks: 1619 
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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