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

1. Nov 17, 2012

### essedbl

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.

2. Nov 17, 2012

### SteamKing

Staff Emeritus
Try applying the angle addition formulas for sin and cos first until you obtain sin(x) and cos(x).

3. Nov 18, 2012

### Mentallic

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.

4. Nov 18, 2012

### Ray Vickson

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