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


 * $$f(x)=A\sin(x+\phi)$$

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

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 $$\Delta OBA$$ 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 $$\lambda_1+\angle BOA$$.

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 $$\angle OAB=90^\circ$$. 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 $$\lambda_1+\angle BOA$$ 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 $$\cos(y)=\frac{a}{\sqrt{a^2+b^2}}$$.

Check that $$\sin^2(x)+\cos^2(x)=1$$ as expected.


 * $$\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)\\ &=A\sin(x+\phi) \end{aligned}$$ , which is (drum roll) a sine wave of amplitude $$A=\sqrt{a^2+b^2}$$ and phase $$\phi=y$$.

Check each step in the formula.
 * What trig formulae did we use?

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 $$\triangle OBC$$ is not a right triangle.
 * What additional trig formulas did you need?