How do I go from the form sin(6x) to the form A sin(x)?

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Homework Help Overview

The discussion revolves around the problem of transforming the expression sin(6x) + cos(6x) into the form k sin(x + φ). Participants are exploring the relationship between the functions and the implications of their periodicities.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to apply angle addition formulas and question the validity of directly relating sin(6x) to A sin(x). There is uncertainty about the coefficients and the implications of periodicity in the transformation.

Discussion Status

Some participants have offered guidance on using angle addition formulas, while others have raised concerns about the feasibility of the transformation due to differing periods of the functions involved. Multiple interpretations of the problem are being explored.

Contextual Notes

Participants note that the transformation from sin(6x) + cos(6x) to k sin(x + φ) may not hold for all x due to the differing periods of the functions, which is a significant constraint in the discussion.

essedbl
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I am working on a homework problem: sin(6x) + cos(6x). I have to represent it in the form:
k sin(x + \phi)

I have the formulas to change A sin(x) + B cos(x) int k sin(x + \phi), however, I don't know how to go from sin(6x) to A sin(x). Is it just 6 sin(x)?

I don't think it is, because if so, k will be equal to √(62+62), which is not a simple whole number, so I suspect my approach is wrong.
 
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Try applying the angle addition formulas for sin and cos first until you obtain sin(x) and cos(x).
 
essedbl said:
I am working on a homework problem: sin(6x) + cos(6x). I have to represent it in the form:
k sin(x + \phi)

I have the formulas to change A sin(x) + B cos(x) int k sin(x + \phi), however, I don't know how to go from sin(6x) to A sin(x). Is it just 6 sin(x)?

I don't think it is, because if so, k will be equal to √(62+62), which is not a simple whole number, so I suspect my approach is wrong.

If you let

\sin(6x)+\cos(6x)=k\cdot\sin(6x+\phi)

with some constants k and \phi, then upon expanding the RHS we get

\sin(6x)+\cos(6x)=k\left(\sin(6x)\cos(\phi)+\cos(6x)\sin(\phi)\right)

\sin(6x)+\cos(6x)=k\cdot\cos(\phi)\sin(6x)+k\cdot\sin(\phi)\cos(6x)

Now notice that for each side to be equal, the coefficient of \sin(6x) on the LHS must equal the coefficient of \sin(6x) on the RHS, and similarly for \cos(6x)

This will give you 2 equations in 2 unknowns, so you should be able to find the value of each.
 
essedbl said:
I am working on a homework problem: sin(6x) + cos(6x). I have to represent it in the form:
k sin(x + \phi)

I have the formulas to change A sin(x) + B cos(x) int k sin(x + \phi), however, I don't know how to go from sin(6x) to A sin(x). Is it just 6 sin(x)?

I don't think it is, because if so, k will be equal to √(62+62), which is not a simple whole number, so I suspect my approach is wrong.

You can't go from sin(6x) + cos(6x) to ksin(x+##\phi##), at least not for all x. The formula involving 6x has period π/3, while the other one has period 2π, no matter what values of k and ##\phi## you pick. And no, sin(6x) is not 6sin(x), and for exactly the same reason. Note, however, that you *can* go to ksin(6x+##\phi##).

RGV
 

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