Trigonometry/Simplifying a sin(x) + b cos(x)

Consider the function

\displaystyle f(x) = a \sin x  + b \cos x

We shall show that this is a sinusoidal wave, and find its amplitude and phase.

To make things a little simpler, we shall assume that a and b are both positive numbers. This isn't necessary, and after studying this section you may like to think what would happen if either of a or b is zero or negative.

Geometric ArgumentEdit

to-do: add diagram.

We'll first use a geometric argument that actually shows a more general result, that:

\displaystyle g(\theta) = a_1 \sin( \theta + \lambda_1) + a_2 \sin( \theta + \lambda_2)

is a sinusoidal wave. Since we can set \displaystyle \lambda_1=0^\circ, \lambda_2=90^\circ the result we are trying for with \displaystyle f follows as a special case.

We use the 'unit circle' definition of sine. \displaystyle a_1 \sin( \theta + \lambda_1) is the y coordinate of a line of length \displaystyle a_1 at angle \displaystyle \theta + \lambda_1 to the x axis, from O the origin, to a point A.

If we now draw a line \displaystyle \overline{AB} of length \displaystyle a_2 at angle \displaystyle \theta + \lambda_2 (where that angle is measure relative to a line parallel to the x axis), its y coordinate is the sum of the two sines.

However, there is another way to look at the y coordinate of point \displaystyle B. The line \displaystyle \overline{OB} does not change in length as we change \displaystyle \theta, because the lengths of \displaystyle \overline{OA} and \displaystyle \overline{AB} and the angle between them do not change. All that happens is that the triangle \displaystyle \Delta OBC rotates about O. In particular \displaystyle \overline{OB} rotates about O.

This then brings us back to a 'unit circle' like definition of a sinusoidal function. The amplitude is the length of \displaystyle \overline{OB} and the phase is \displaystyle \lambda_1 + \angle BOA.

Algebraic ArgumentEdit

The algebraic argument is essentially an algebraic translation of the insights from the geometric argument. We're also in the special case that \displaystyle \lambda_1 = 0and \displaystyle \angle OAB=90^\circ. The x's and y's in use in this section are now no longer coordinates. The 'y' is going to play the role of \displaystyle \lambda_1 + \angle BOA and the 'x' plays the role of \displaystyle \theta.

We define the angle y by \tan(y) = \frac{b}{a}.

By considering a right-angled triangle with the short sides of length a and b, you should be able to see that

\sin(y) = \frac{b}{\sqrt{a^2+b^2}} and \cos(y) = \frac{a}{\sqrt{a^2+b^2}}.
Check this

Check that \displaystyle \sin^2 x + \cos^2 x = 1 as expected.

f(x) = a \sin(x) + b \cos(x) = \sqrt{a^2+b^2} [\frac{a}{\sqrt{a^2+b^2}} \sin(x) + \frac{b}{\sqrt{a^2+b^2}} \cos(x)] = \sqrt{a^2+b^2}[\sin(x)\cos(y) + \cos(x)\sin(y)] = \sqrt{a^2+b^2}\sin(x+y),

which is (drum roll) a sine wave of amplitude \sqrt{a^2+b^2} and phase y.

Check this

Check each step in the formula.

  • What trig formulas did we use?
The more general case

Can you do the full algebraic version for the more general case:

\displaystyle g(\theta) = a_1 \sin( \theta + \lambda_1) + a_2 \sin( \theta + \lambda_2)

using the geometric argument as a hint? It is quite a bit harder because \displaystyle \Delta OBC is not a right triangle.

  • What additional trig formulas did you need?

Last modified on 16 February 2011, at 10:55