DWG NO. 5.2 — Lesson 2 of 5

Sum and Difference Formulas

Unit 5: Trigonometric Identities and Equations · ~25 min

Objective Apply the sum and difference formulas to find exact trig values for angles built from two special angles, and to expand expressions like sin(x + π).

The unit circle from 4.2 only gives exact values at special angles — multiples of 30°, 45°, and 60°. But 75° isn't special on its own; it's special because it's 45° + 30°, a sum of two angles you already know. Sum and difference formulas let you break an unfamiliar angle into a combination of familiar ones. One thing they are not: sin(A + B) does not equal sinA + sinB — the formulas below are the real relationship, and it's worth memorizing that the naive version is wrong before it becomes a habit.

The formulas

sin(A ± B) = sinA cosB ± cosA sinB
cos(A ± B) = cosA cosB ∓ sinA sinB (note the sign flips)
tan(A ± B) = \(\frac{tanA \pm tanB}{1 ∓ tanA tanB}\)

The cosine formula's sign is easy to mix up: cos(A + B) uses minus on the right, and cos(A − B) uses plus. The tangent version comes from dividing the sine formula by the cosine formula and simplifying — it's built from the other two, not a separate fact to memorize independently.

Angle B is added on top of angle A; the sum formulas rebuild the resulting ray's coordinates from the two starting angles

Worked Example 1 · Exact value with the difference formula

ProblemFind the exact value of cos75° by writing 75° = 45° + 30°.

1cos(45° + 30°) = cos45°cos30° − sin45°sin30°.

2Substitute exact values: (\(\frac{\sqrt{2}}{2}\))(\(\frac{\sqrt{3}}{2}\)) − (\(\frac{\sqrt{2}}{2}\))(½) = \(\frac{\sqrt{6}}{4}\) − \(\frac{\sqrt{2}}{4.}\)

cos75° = \(\frac{\sqrt{6} - \sqrt{2}}{4}\)

Worked Example 2 · Sum formula from given ratios

ProblemGiven sinA = \(\frac{3}{5}\) with A in Quadrant I, and cosB = \(\frac{-12}{13}\) with B in Quadrant II, find sin(A + B).

1Find the missing ratios: cosA = \(\frac{4}{5}\) (positive in QI); sinB = \(\frac{5}{13}\) (positive in QII).

2sin(A+B) = sinA cosB + cosA sinB = (\(\frac{3}{5}\))(\(\frac{-12}{13}\)) + (\(\frac{4}{5}\))(\(\frac{5}{13}\)).

3= \(\frac{-36}{65}\) + \(\frac{20}{65}\) = \(\frac{-16}{65.}\)

sin(A + B) = \(\frac{-16}{65}\)

Worked Example 3 · Expanding an expression

ProblemSimplify sin(x + π) using the sum formula.

1sin(x + π) = sinx cosπ + cosx sinπ.

2cosπ = −1 and sinπ = 0, so this becomes sinx(−1) + cosx(0).

sin(x + π) = −sinx

Guided practice

1.Find the exact value of cos15° using 15° = 45° − 30°.
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cos45°cos30° + sin45°sin30° = \(\frac{\sqrt{2}}{2} \frac{\sqrt{3}}{2}\) + \(\frac{\sqrt{2}}{2} \frac{1}{2}\) = \(\frac{\sqrt{6} + \sqrt{2}}{4}\).
2.Find the exact value of sin105° using 105° = 60° + 45°.
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sin60°cos45° + cos60°sin45° = \(\frac{\sqrt{3}}{2} \frac{\sqrt{2}}{2}\) + \(\frac{1}{2} \frac{\sqrt{2}}{2}\) = \(\frac{\sqrt{6} + \sqrt{2}}{4}\).
3.Given cosA = \(\frac{4}{5}\) (A in QI) and sinB = \(\frac{-3}{5}\) (B in QIV), find cos(A − B).
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sinA = \(\frac{3}{5,}\) cosB = \(\frac{4}{5}\) (both positive ratios needed here given the quadrants). cos(A−B) = cosAcosB + sinAsinB = (\(\frac{4}{5}\))(\(\frac{4}{5}\)) + (\(\frac{3}{5}\))(\(\frac{-3}{5}\)) = \(\frac{16}{25}\) − \(\frac{9}{25}\) = \(\frac{7}{25}\).
4.Simplify cos(x − \(\frac{\pi}{2}\)).
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cosx cos(\(\frac{\pi}{2}\)) + sinx sin(\(\frac{\pi}{2}\)) = cosx(0) + sinx(1) = sinx.
5.Explain where the "1 ∓ tanA tanB" denominator in the tangent sum/difference formula comes from.
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tan(A+B) = sin\(\frac{A+B}{cos}\)(A+B). Writing out both sum formulas and dividing every term by cosA cosB turns the sine terms into tanA and tanB, and turns the cosine terms into 1 and tanA tanB — producing the compact ratio form.

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