An Intuitive Derivation of
D sin(Bx) + F cos(Bx) = A sin(Bx + C)

By Debra Woods

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Statement of Problem

The purpose of this lesson is to investigate the sum of sine and cosine functions each with the same period, and to derive the relationship between this sum and a single sine function.

Examples

First, we will look at the function  f(x) = sin x + cos x.

Clear[f,x];
f[x_] = Sin[x] + Cos[x];

Plot[f[x],{x,-2Pi,2Pi},AspectRatio->Automatic,
					PlotStyle->{Green}];

It is interesting to note that this function looks a lot like a sine function with an amplitude change and a phase shift.  Let's investigate another possibility by giving sine and cosine different amplitudes, but keeping the coefficient of the x term the same for both functions.

Clear[g,h,x];
g[x_] = 2Sin[3x] + 3Cos[3x];
Plot[g[x],{x,-2Pi,2Pi},AspectRatio->Automatic,
						PlotStyle->{Purple}];

Note that the new function still looks like a single sine function.  Now let's see what happens if we change the coefficient of x.  Recall, that this will change the periods of each function in our sum.

h[x_] = Sin[2x] + Cos[3x];
Plot[h[x],{x,-2Pi,2Pi},AspectRatio->Automatic,
						PlotStyle->{Red}];

This function is clearly not a sine function.  So, it is reasonable to conjecture that the sum of a sine function and a cosine function with the same period will yield a sine function with the same period, but different amplitude.

Try different values of D, F, and B to make yourself feel more comfortable with the conjecture.

Derivation

Let's look into the algebra involved and see if we can produce a single sine function from D sin(Bx) + F cos(Bx).  That is,
													D sin(Bx) + F cos(Bx) = A sin(Bx + C)
													
Recall that A sin(Bx + C) = A[sin(Bx)cos(C) + cos(Bx)sin(C)].  Now, it is clear that  D = Acos(C) and F = Asin(C).  Thus, 
						D¤ + F¤ = A¤(cos¤(C) + sin¤(C)) 
						D¤ + F¤ = A¤   and
						F/D = tan C, which leads to tan—∕(F/D) = C.

Demonstration of derivation correctness

Let's apply this to the first function above, f(x) = sin x + cos x.
In this case D = 1 and E = 1, so A = é2  and C = í¢Ÿ .

Now let's superimpose the plots.  The wider plot is the original function, and the narrow plot is the new form of the function.

Clear[t];
t[x_] = Sqrt[2] Sin[x + Pi/4];
Plot[{f[x],t[x]},{x,-2Pi,2Pi},AspectRatio->Automatic,
				PlotStyle->{{Thickness[0.02],Red},
						{{Thickness[0.04],Black}}}];

We have now derived a new trigonometric identity.  This formula will prove its usefulness in applied subjects, such as harmonic motion.