Question

How do you use linear combinations to solve trigonometric equations?

Answer 1

Substitute

`a cos x + b sin x = sqrt{a^2 + b^2} cos(x -text{arctan2}( b //, a) )`

where arctan2 is the two parameter, four quadrant inverse tangent.

Explanation:

I've been answering all these old math questions. It's hard to know if anyone reads the answers.

The linear combination of a cosine and a sine of the same angle corresponds to a scaling and a phase shift. Let's explain how that works.

The linear combination of a cosine and a sine of the same angle is an expression of the form:

`a cos x + b sin x`

That looks very much like the sum angle formula for sine or the difference angle formula for cosine:

`sin(alpha + beta ) = sin alpha cos beta + cos alpha sin beta`

`cos(alpha - beta) = cos alpha cos beta + sin alpha sin beta`

Indeed, we can take our linear combination and transform it into a scaled version of either of these. Pro Tip: When given a choice, prefer cosine to sine.

Let's multiply by a scale factor `r` and set `alpha = x, beta=theta.`

`r cos(x - theta) = r cos theta cos x + r sin theta sin x`

Matching to our linear combination leads us to want to solve for `r` and `theta`:

`a = r cos theta`

`b = r sin theta`

We've seen this before. It's how we turn turn the polar coordinates `P(r, theta)` to rectangular coordinates, here `(a,b).` So our task is to turn `(a,b)` to polar coordinates.

Let's remind ourselves how to do that, perhaps in a bit more detail than usually seen.

Squaring and adding we get

`a^2 + b^2 = (r cos theta)^2 + (r sin theta)^2 = r^2(cos^2 theta + sin^2 theta) = r^2`

`r = sqrt{a^2 + b^2}`

`sin theta = b/r`

`cos theta = a/r`

`tan theta = b/a`

`theta = text{arctan2}( b //, a)`

Usually this is written as the regular arctangent, but that's not really right. The regular arctangent only covers two quadrants, and doesn't work on the y axis. This is the two parameter, four quadrant inverse tangent, which returns a valid theta for all real input pairs. The `//,` is deliberate, reminding us there are two separate parameters, and which is which.

So now we're assured `a = r cos theta` and `b = r sin theta` so

`a cos x + b sin x = r cos theta cos x + r sin theta sin x = r cos(x - theta)`

`a cos x + b sin x = sqrt{a^2 + b^2} cos(x -text{arctan2}( b //, a) )`