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

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To convert sin(6x) + cos(6x) into the form k sin(x + φ), it's important to recognize that the periods of the functions differ, making direct transformation impossible for all x. The expression sin(6x) cannot simply be represented as 6 sin(x) due to this periodicity issue. Instead, you can express it as k sin(6x + φ), which maintains the correct periodicity. The coefficients of sin(6x) and cos(6x) must match on both sides of the equation to derive the values of k and φ. Therefore, the approach should focus on the relationship between sin(6x) and cos(6x) rather than simplifying to sin(x).
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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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