May 15th, 2018, 01:25 AM  #1 
Member Joined: Apr 2017 From: India Posts: 56 Thanks: 0  Linear dependence
Why the functions cos(3x) and cos(3x+pi/2) are linearly dependent functions?

May 15th, 2018, 02:50 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
They aren't! Cos(a+ b)= cos(a)cos(b) sin(a)sin(b) so cos(3x+ pi/2)= cos(3x)cos(pi/2) sin(3x)sin(pi/2)= sin(3x) which is NOT a multiple of cos(3x). 
May 15th, 2018, 10:55 AM  #3 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry 
Perhaps you mean algebraically dependent? In that case, it follows from the standard trig identity $\sin^2\left(x\right) + \cos^2\left(x\right) \equiv 1$. To fill in the details, let $f(X,Y)$ be the polynomial $X^2 + Y^2  1$. Then, as Country Boy shows, $\cos\left(3x + \frac{\pi}{2}\right) = \sin\left(3x\right)$, so $\begin{align*} f\left(\cos\left(3x\right), \cos\left(3x + \frac{\pi}{2}\right) \right) &= f(\cos\left(3x\right), \sin\left(3x\right)) \\ &= (\cos\left(3x\right))^2 + (\sin\left(3x\right))^2  1 \\ &= \cos^2\left(3x\right) + \sin^2\left(3x\right)  1 \\ &= 1  1 \\ &= 0 \end{align*}$ 
May 15th, 2018, 12:52 PM  #4 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
By definition, f(x) and g(x) are linearly independent over an interval if no constants a and b exist st af(x) + bg(x) =0 on the interval. If you're familiar with Wronskians, if W(f,g) $\displaystyle \neq$ 0 for some x on the interval, the functions are LI on the interval. W(f,g)=1. 
May 15th, 2018, 02:59 PM  #5 
Member Joined: Apr 2017 From: India Posts: 56 Thanks: 0 
I really appreciate the response.Then how we will judge for the linear independency of the functions e^x and xe^(x)? The wronskian comes out to be e^(2x) 
May 16th, 2018, 01:04 PM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
Which is never 0! From the definition, let for all x. Taking x= 0, we have . Then, taklng x= 1, so B= 0. The two functions are linearly independent. 

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