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 Argument

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 ${\displaystyle \Delta OBA}$ 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 Argument

The algebraic argument is essentially an algebraic translation of the insights from the geometric argument. We're also in the special case that ${\displaystyle \lambda _{1}=0}$and ${\displaystyle \angle OAB=90^{\circ }}$ . 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 ${\displaystyle \tan(y)={\frac {b}{a}}}$ .

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

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

${\displaystyle {\begin{aligned}f(x)&=a\sin(x)+b\cos(x)\\&={\sqrt {a^{2}+b^{2}}}\left({\frac {a}{\sqrt {a^{2}+b^{2}}}}\sin(x)+{\frac {b}{\sqrt {a^{2}+b^{2}}}}\cos(x)\right)\\&={\sqrt {a^{2}+b^{2}}}{\Big (}\sin(x)\cos(y)+\cos(x)\sin(y){\Big )}\\&={\sqrt {a^{2}+b^{2}}}\sin(x+y)\\\end{aligned}}}$ ,

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