Converting a trig sum to a single periodic function

The conversion formula we use in ODEs comes from the trig identity:

cos(A−B) = cos(A) cos(B) + sin(A) sin(B)

If we to determine R, δ so that:

C1 cos(ωt) + C2 sin(ωt) = R cos(ωt− δ)

we can say that:

R cos(ωt− δ) = R (cos(ωt) cos(δ) + sin(ωt) sin(δ))

= (R cos(δ)) cos(ωt) + (R sin(δ)) sin(ωt)

Therefore,C1 = R cos(δ) and C2 = R sin(δ)

From these equations, we can get formulas for R, δ in terms of C1, C2:

C21 + C2

2 = R2 cos2(δ) + R2 sin2(δ) = R2

so that R =√

C21 + C2

2

Furthermore,C2

C1=

R sin(δ)R cos(δ)

= tan(δ)

so that tan−1

(C2

C1

)= δ if −π

2< δ <

π

2(recall that the domain of the tangent

needs a restriction so that it is invertible).We need to be a bit careful in this computation. With C1 = R cos(δ) and

C2 = R sin(δ), we can visualize C1 and C2 as coordinates on the circle of radiusR, where C1 is the “x-”coordinate and C2 is the “y-”coordinate. If x > 0, thenwe can use the angle that the calculator provides:

If C1 > 0, then tan−1(C2/C1) = δ.If x < 0, then we should be in Quadrants II or III of the unit circle, but the

calculator will produce angles in Quadrants I and IV. Therefore:If C1 < 0, add π to the number provided by the calculator, δ = tan−1(C2/C1)+

π. Some calculators provide a four-quadrant inverse. If yours does this, the com-puter command will be something like: arctan(y,x) where you have to inputtwo numbers instead of the fraction.

Examples: Be sure to try these yourself!

1. Rewrite −12

cos(t) +√

32

sin(t) as R cos(ωt− δ).

In this case, R = 1 and δ = tan−1(−√

3) + π = −π3 + π = 2π

3

1

Therefore,

−12

cos(t) +√

32

sin(t) = cos(

t− 2π

3

)NOTE: If you don’t want to remember to add π in some cases, we canALWAYS make C1 > 0. For example, an alternative solution to thisproblem would be:

−

(12

cos(t)−√

32

sin(t)

)= −R cos(t− δ)

so that R = 1 and δ = −π3 :

−

(12

cos(t)−√

32

sin(t)

)= − cos

(t +

π

3

)

2. Solve y′′ + y′ + 2y = 0, y(0) = 12 , y′(0) = −1, and write the solution as

Aeαt cos(ωt− δ).

The solutions to the characteristic equation, r2 + r + 2 = 0 are

r = −12±√

72

i

so that the general solution is:

y(t) = e−t2

(C1 cos

(√7

2t

)+ C2 sin

(√7

2t

))

Solving for C1, C2, we get:

y(t) = e−t2

(12

cos

(√7

2t

)− 3

2√

7sin

(√7

2t

))

so that R =2√7

and δ = tan−1

(−3√

7

)≈ −0.848

Therefore,

y(t) =2√7e−

t2 cos

(√7

2+ 0.848

)

2