View source for Trigonometry/Simplifying a sin(x) + b cos(x)

Consider the function 
:<math>f(x)=a\sin(x)+b\cos(x)</math>

We shall show that this is a sinusoidal wave

:<math>f(x)=A\sin(x+\phi)</math>

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

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:
:<math>g(\theta)=a\sin(\theta+\lambda_1)+b\sin(\theta+\lambda_2)</math>
is a sinusoidal wave.

By setting <math>\lambda_1=0^\circ\ ,\ \lambda_2=90^\circ</math> , it will follow that <math>f(\theta)=a\sin(\theta)+b\cos(\theta)</math> is sinusoidal.

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

We now draw a line <math>\overline{AB}</math> of length <math>b</math> at angle <math>\theta+\lambda_2</math> (where that angle is measured relative to a line parallel to the x axis). The y-coordinate of <math>B</math> is the y-coordinate of <math>A</math> plus the vertical displacement from <math>A</math> to <math>B</math>. In other words its y-coordinate is <math>g(\theta)</math>.

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

Hence, the y-coordinate of <math>B</math> is a sinusoidal function (we can see this from the 'unit circle' definition mentioned earlier). The amplitude is the length of <math>\overline{OB}</math> and the phase is <math>\lambda_1+\angle BOA</math> .

==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 <math>\lambda_1=0</math>and <math>\angle OAB=90^\circ</math> . 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 <math>\lambda_1+\angle BOA</math> and the 'x' plays the role of <math>\theta</math> .

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

By considering a right-angled triangle with the short sides of length a and b, you should be able to see that
:<math>\sin(y)=\frac{b}{\sqrt{a^2+b^2}}</math> and <math>\cos(y)=\frac{a}{\sqrt{a^2+b^2}}</math> .

{{ExerciseRobox|title=Check this}}
Check that <math>\sin^2(x)+\cos^2(x)=1</math> as expected.
{{Robox/Close}}

:<math>\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}</math> ,
which is (drum roll) a sine wave of amplitude <math>A=\sqrt{a^2+b^2}</math> and phase <math>\phi=y</math>.
{{ExerciseRobox|title=Check this}}
Check each step in the formula.
*What trig formulae did we use?
{{Robox/Close}}

{{ExerciseRobox|title=The more general case}}
Can you do the full algebraic version for the more general case:
:<math>g(\theta)=a_1\sin(\theta+\lambda_1)+a_2\sin(\theta+\lambda_2)</math>
using the geometric argument as a hint?
It is quite a bit harder because <math>\triangle OBC</math> is not a right triangle.
*What additional trig formulas did you need?
{{Robox/Close}}

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