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Manipulating sin/cos functions

by HighCommander4
Tags: functions, manipulating, sin or cos
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HighCommander4
#1
Sep11-07, 10:59 PM
P: 7
I'm not sure if this is considered pre- or post-calculus (I am doing it for a calculus course, but I doubt it involves actual calculus) but I'll go ahead and post here.

1. The problem statement, all variables and given/known data

I am required to re-write the function sin x + 2 cos (x - pi/6) in the form r sin (x + theta).

2. Relevant equations

cos (x) = sin (x + pi/2)
perhaps equations for sin (a + b) and cos (a + b)?

3. The attempt at a solution

So I tried to get everything to be sine:
sin x + 2 cos (x - pi/6) = sin x + 2 sin (x - pi/6 + pi/2)
= sin x + 2 sin (x + pi/3)

I now tried to expand the second term using the equation sin(a+b) = sin(a)cos(b) + cos(a)sin(b):
= sin x + 2 [ sin (x) cos(pi/3) + cos (x) sin (pi/3)]
= sin x + 2 [sin(x)(1/2) + cos(x)(sqrt(3)/2)]
= sin x + sin x + sqrt(3) cos x
= 2 sin x + sqrt(3) cos x

But I don't know where to go from here. Trying to convert this new cosine to sine using cos (x) = sin (x + pi/2) just brings you back to the same thing after simplifying.

So where do I go from here to get it to be in the form r sin (x + theta)?
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Dick
#2
Sep11-07, 11:24 PM
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Use the addition formulas to write both sides completely in terms of sin(x) and cos(x). Then equate the coefficients of sin(x) and cos(x) on both sides. This will allow you to determine r and theta. Two equations, two unknowns. The principle involved is that sin(x) and cos(x) are linearly independent functions, but you don't need to know that.
HighCommander4
#3
Sep12-07, 03:16 PM
P: 7
Thank you for your reply.

I'm guessing you meant to set sin(x) + 2cos(x - pi/6) equal to r sin(x + theta). I tried that and followed your instructions and this is what I got:

sin(x) + 2cos(x - pi/6) = r sin(x + theta)
sin(x) + 2[cos(x)cos(pi/6) + sin(x)sin(pi/6)] = r[sin(x)cos(theta) + cos(x)sin(theta)]
sin(x) + 2[cos(x)(sqrt(3)/2) + sin(x)(1/2)] = r sin(x)cos(theta) + r cos(x)sin(theta)
sin(x) + sqrt(3)cos(x) + sin(x) = r sin(x)cos(theta) + r cos(x)sin(theta)
2sin(x) + sqrt(3)cos(x) = (r cos(theta))sin(x) + (r sin(theta))cos(x)

So now I have everything in terms of sin(x) and cos(x). If I'm understanding correctly, you're telling me to set their coefficients equal to each other, that is,
2 = r cos(theta)
sqrt(3) = r sin(theta)

How do I know I am allowed to do that? More specifically, how do I know that other values of r and theta (different from the ones I get after solving this system of equations) don't exist that also satisfy the equation? I would somehow have to prove that other solutions don't exist before I go on. I would really appreciate if you showed me how I can do that.

Anyways, let's assume I proved that no other solutions exist, I now proceed to solve this system of equations:

r = 2 / cos(theta)
sqrt(3) = (2 / cos(theta))sin(theta)
sqrt(3) = 2 tan(theta)
sqrt(3) / 2 = tan(theta)
theta = arctan(sqrt(3) / 2)
r = 2 / cos(arctan(sqrt(3) / 2))

My equation in the form r sin(x + theta) would then be
(2 / cos(arctan(sqrt(3) / 2))) sin(x + arctan(sqrt(3) / 2))
Am I doing something wrong that I am getting results that are this ugly, or is there perhaps a way of simplifying them?

Dick
#4
Sep12-07, 03:42 PM
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Manipulating sin/cos functions

Starting from:
2 = r cos(theta)
sqrt(3) = r sin(theta)
there is much easier way to proceed. Square both sides of the equations and add them, you should easily get r. Now divide the two equations by each other. You should easily get theta (as you already did).
The proof that if Asin(x)+Bcos(x)=Csin(x)+Dcos(x), then A=C and B=D involves the notion of linear independence of functions. Have you done that?
HighCommander4
#5
Sep12-07, 03:50 PM
P: 7
Thank you for your continued help. Squaring and adding does indeed make it much easier. I now get r = sqrt(7), but for theta I still get arcsin(sqrt(3/7)). Is that as good as it gets?

I've done linear independence of vectors in R2 and R3, but not of functions like sine and cosine. If this solution requires knowledge of higher concepts than we have learned, perhaps there is an alternate solution relying on simpler concepts?
Dick
#6
Sep12-07, 03:58 PM
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Yep, good as it gets. Here's a simpler proof. Showing if Asin(x)+Bcos(x)=Csin(x)+Dcos(x), then A=C and B=D is equivalent to showing if Asin(x)+Bcos(x)=0 then A=B=0, right? Integrate (Asin(x)+Bcos(x))^2 from 0 to 2pi. I'll save you some time and tell you what you get. It's pi*(A^2+B^2). If that is 0, what do you conclude about A and B?
HighCommander4
#7
Sep12-07, 07:00 PM
P: 7
I follow your logic, but I haven't done integration yet either. I'm basically just starting my first-year calculus course and I'm expected to solve this problem using what I already learned in high school. Is there a way of solving it using the more basic techniques taught in high school?
Dick
#8
Sep12-07, 09:31 PM
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Ok, I'm still with you. Even simpler. Asin(x)+Bcos(x)=0. Put x=0 and tell me what you conclude about B. Put x=pi/2 and tell me what you conclude about A. Sorry I didn't think about how simple it could be earlier. Try the same thing for Asin(x)+Bcos(x)=Csin(x)+Dcos(x) if you don't believe me yet.
HighCommander4
#9
Sep12-07, 09:40 PM
P: 7
Working with Asin(x)+Bcos(x)=Csin(x)+Dcos(x),
x = 0 yields B = D
x = pi/2 yields A = C
But how does that prove that A = C and B = D in general?
Dick
#10
Sep12-07, 09:47 PM
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Your equation is supposed to hold for all x. If it doesn't hold for x=0 and x=pi/2, what's the point in writing it down?
HighCommander4
#11
Sep12-07, 09:51 PM
P: 7
What I mean to say is, how do I know that A = C and B = D for x = pi/4 or x = 1.23456 or any other value of x?
Dick
#12
Sep12-07, 09:56 PM
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A,B,C and D are not functions of x. They are constants. If you solve for the relation between A,B,C and D by plugging in specific values of x, how can they not hold for all x? You've shown A=C and B=D. So A*sin(x)+B*cos(x)=C*sin(x)+D*cos(x) becomes Asin(x)+Bcos(x)=Asin(x)+Bcos(x). Looks to me like that would hold for all x.
HighCommander4
#13
Sep12-07, 10:18 PM
P: 7
I get it now. A, B, C and D are constants so they don't depend on the value of x.

So I can finally present a well-formed solution to this problem. A big thank you for all your help!
Dick
#14
Sep12-07, 10:19 PM
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You are talcum.


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