Auxiliary-Angle Method and Products to Sums

The Auxiliary-Angle Form: Converting a sin(x) + b cos(x)

Any expression of the form a sin(x) + b cos(x), where a and b are real constants (not both zero), can be rewritten as a single sinusoidal function with amplitude R and a phase shift. This is called the auxiliary-angle method (also called the harmonic form). The two most useful forms are:

• R sin(x + α)
• R cos(x − α)

where R > 0 and α is the auxiliary angle, typically chosen so that 0 < α < π/2 (keeping α in the first quadrant simplifies most problems).

Derivation of R sin(x + α):

Expand R sin(x + α) using the compound-angle identity:

R sin(x + α) = R [sin x cos α + cos x sin α] = (R cos α) sin x + (R sin α) cos x

Matching coefficients with a sin x + b cos x gives the system:

• R cos α = a
• R sin α = b

Squaring and adding: R² cos²α + R² sin²α = a² + b², so R = √(a² + b²).

Dividing: R sin α / R cos α = b/a, so tan α = b/a (with α chosen to match the signs of a and b).

Derivation of R cos(x − α):

Expand R cos(x − α) = R [cos x cos α + sin x sin α] = (R cos α) cos x + (R sin α) sin x.

Matching: R sin α = a, R cos α = b, giving the same R = √(a² + b²) but now tan α = a/b.

The choice of form (sin or cos) usually depends on the question. Both give identical R; only the formula for α differs.

Step-by-Step Method and a Fully Worked Example

Follow this systematic procedure every time:

1. Identify a and b from a sin x + b cos x.
2. Compute R = √(a² + b²) — this is always positive.
3. Determine α from tan α = b/a (for R sin form) or tan α = a/b (for R cos form). Check the quadrant using the signs of both a and b.
4. Write the final auxiliary form and verify by expanding back.

Worked Example: Express f(x) = 3 sin x + 4 cos x in the form R sin(x + α), giving α correct to 2 decimal places.

Step 1: a = 3, b = 4.

Step 2: R = √(3² + 4²) = √(9 + 16) = √25 = 5. (Arithmetic check: 9 + 16 = 25 ✓, √25 = 5 ✓.)

Step 3: tan α = b/a = 4/3. Since a = 3 > 0 and b = 4 > 0, both R cos α = 3 and R sin α = 4 are positive, so α is in the first quadrant.

α = arctan(4/3) ≈ 0.9273 rad ≈ 0.93 rad (2 d.p.) [equivalently ≈ 53.13°].

Step 4: f(x) = 5 sin(x + 0.93).

Verification: 5 sin(x + 0.93) = 5[sin x cos(0.93) + cos x sin(0.93)]. Now cos(0.93) ≈ 0.6006 and sin(0.93) ≈ 0.7994 (check: 0.6006² + 0.7994² ≈ 0.3607 + 0.6390 ≈ 0.9997 ≈ 1 ✓). So 5 × 0.6006 ≈ 3.003 ≈ 3 and 5 × 0.7994 ≈ 3.997 ≈ 4 ✓. The coefficients recover correctly.

Second example — negative coefficients: Express g(x) = −√3 sin x + cos x in the form R sin(x + α).

a = −√3, b = 1. R = √(3 + 1) = 2. tan α = b/a = 1/(−√3). But now R cos α = −√3 < 0 and R sin α = 1 > 0, so α is in the second quadrant. Reference angle = arctan(1/√3) = π/6 = 30°, so α = π − π/6 = 5π/6. Therefore g(x) = 2 sin(x + 5π/6). Verify: 2[sin x cos(5π/6) + cos x sin(5π/6)] = 2[sin x(−√3/2) + cos x(1/2)] = −√3 sin x + cos x ✓.