How do you use linear combinations to solve trigonometric equations?

1 Answer
Apr 27, 2018

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) ) #