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

Consider the function

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

We shall show that this is a sinusoidal wave

$f(x)=A\sin (x+\varphi )$

and find that the amplitude is $A=\sqrt{a^2+b^2}$ and the phase $\varphi =\arctan \frac{b}{a}$

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:

$g(\theta )=a\sin (\theta +{\lambda }_1)+b\sin (\theta +{\lambda }_2)$

is a sinusoidal wave.

By setting ${\lambda }_1=0^{\circ },{\lambda }_2=90^{\circ }$ , it will follow that $f(\theta )=a\sin (\theta )+b\cos (\theta )$ is sinusoidal.

We use the 'unit circle' definition of sine: $a\sin (\theta +{\lambda }_1)$ is the y coordinate of a line of length $a$ at angle $\theta +{\lambda }_1$ to the x axis, from O the origin, to a point A.

We now draw a line $\overline{AB}$ of length $b$ at angle $\theta +{\lambda }_2$ (where that angle is measured relative to a line parallel to the x axis). The y-coordinate of $B$ is the y-coordinate of $A$ plus the vertical displacement from $A$ to $B$. In other words its y-coordinate is $g(\theta )$.

However, there is another way to look at the y coordinate of point $B$ . The line $\overline{OB}$ does not change in length as we change $\theta$ - all that happens is that the triangle $\mathrm{\Delta }OBA$ rotates about O. In particular, $\overline{OB}$ rotates about O.

Hence, the y-coordinate of $B$ is a sinusoidal function (we can see this from the 'unit circle' definition mentioned earlier). The amplitude is the length of $\overline{OB}$ and the phase is ${\lambda }_1+\mathrm{\angle }BOA$ .

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 ${\lambda }_1=0$and $\mathrm{\angle }OAB=90^{\circ }$ . 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 ${\lambda }_1+\mathrm{\angle }BOA$ and the 'x' plays the role of $\theta$ .

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

Template:ExerciseRobox Check that ${\sin }^2(x)+{\cos }^2(x)=1$ as expected. Template:Robox/Close

$\begin{array}{cc}f(x)& =a\sin (x)+b\cos (x)\\ & =\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\sin (x)+\frac{b}{\sqrt{a^2+b^2}}\cos (x))\\ & =\sqrt{a^2+b^2}(\sin (x)\cos (y)+\cos (x)\sin (y))\\ & =\sqrt{a^2+b^2}\sin (x+y)\\ & =A\sin (x+\varphi )\end{array}$ ,

which is (drum roll) a sine wave of amplitude $A=\sqrt{a^2+b^2}$ and phase $\varphi =y$. Template:ExerciseRobox Check each step in the formula.

• What trig formulae did we use?

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

$g(\theta )=a_1\sin (\theta +{\lambda }_1)+a_2\sin (\theta +{\lambda }_2)$

using the geometric argument as a hint? It is quite a bit harder because $\mathrm{△}OBC$ is not a right triangle.

• What additional trig formulas did you need?

Template:Trig/NAV