Prove the addition and subtraction formulas for sine, cosine and tangent, and use them to find exact values and solve problems

Learning goal: Prove the addition and subtraction formulas for sine, cosine and tangent, and use them to solve problems: exact values at angles such as π/12 and 5π/12, values of sin(α + β) from the values at α and β, and rewriting combined waves.

Scope note. F.TF.9 is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full, as extension content for students on that path.

The formulas. For all real α and β:
cos(α − β) = cos α cos β + sin α sin β; cos(α + β) = cos α cos β − sin α sin β;
sin(α + β) = sin α cos β + cos α sin β; sin(α − β) = sin α cos β − cos α sin β;
tan(α + β) = (tan α + tan β)/(1 − tan α tan β) and tan(α − β) = (tan α − tan β)/(1 + tan α tan β), whenever every tangent involved is defined and the denominator is not zero. Notice what the formulas are not: sin(α + β) is not sin α + sin β. At α = β = π/4, sin(π/2) = 1 but sin(π/4) + sin(π/4) = √2 ≈ 1.414. Sine is not a linear function, and the formulas say precisely how it fails to be one.

Proof of the cosine difference formula. Everything rests on this one formula, proved from the unit circle. Let A = P(α) = (cos α, sin α) and B = P(β) = (cos β, sin β). The distance formula gives AB² = (cos α − cos β)² + (sin α − sin β)². Expand: cos²α − 2 cos α cos β + cos²β + sin²α − 2 sin α sin β + sin²β. By the Pythagorean identity, cos²α + sin²α = 1 and cos²β + sin²β = 1, so AB² = 2 − 2(cos α cos β + sin α sin β). Now rotate the whole picture clockwise about the origin through the angle β. Rotation preserves distances, A moves to C = P(α − β) = (cos(α − β), sin(α − β)) and B moves to D = P(0) = (1, 0). So CD = AB. Compute CD² by the same distance formula: (cos(α − β) − 1)² + (sin(α − β) − 0)² = cos²(α − β) − 2 cos(α − β) + 1 + sin²(α − β) = 2 − 2 cos(α − β), again using the Pythagorean identity. Setting AB² = CD²: 2 − 2(cos α cos β + sin α sin β) = 2 − 2 cos(α − β), and dividing by −2 gives cos(α − β) = cos α cos β + sin α sin β. The argument is valid for all real α and β, because P(α), P(β) and P(α − β) exist for all of them and rotation preserves distance regardless of where the points are.

The other five follow. Replace β by −β in the difference formula and use cos(−β) = cos β and sin(−β) = −sin β (cosine is even, sine is odd): cos(α + β) = cos α cos β − sin α sin β. For sine, first note the cofunction identity cos(π/2 − x) = sin x, which is the difference formula with α = π/2: cos(π/2 − x) = cos(π/2) cos x + sin(π/2) sin x = 0 + sin x. Replacing x by π/2 − x in it gives sin(π/2 − x) = cos x as well. Then sin(α + β) = cos(π/2 − (α + β)) = cos((π/2 − α) − β) = cos(π/2 − α) cos β + sin(π/2 − α) sin β = sin α cos β + cos α sin β. Replacing β by −β gives the sine difference formula. For tangent, divide sin(α + β) by cos(α + β) and then divide numerator and denominator by cos α cos β: (sin α cos β + cos α sin β)/(cos α cos β − sin α sin β) = (tan α + tan β)/(1 − tan α tan β). The subtraction version follows from the oddness of tangent.

A fully worked model. Find sin 75° exactly, that is sin(5π/12). Write 5π/12 = π/4 + π/6. Then sin(π/4 + π/6) = sin(π/4) cos(π/6) + cos(π/4) sin(π/6) = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6 + √2)/4. Check with a calculator: (2.4495 + 1.4142)/4 = 0.9659, and sin(1.3090) ≈ 0.9659 ✓. In the same way cos(5π/12) = cos(π/4) cos(π/6) − sin(π/4) sin(π/6) = √6/4 − √2/4 = (√6 − √2)/4 ≈ 0.2588, and tan(5π/12) = sin/cos = (√6 + √2)/(√6 − √2); multiplying top and bottom by (√6 + √2) gives (6 + 2√12 + 2)/(6 − 2) = (8 + 4√3)/4 = 2 + √3 ≈ 3.732. Because π/12 = π/2 − 5π/12, the cofunction identities give cos(π/12) = (√6 + √2)/4, sin(π/12) = (√6 − √2)/4 and tan(π/12) = 1/(2 + √3) = 2 − √3 ≈ 0.2679.

Using values without knowing the angles. Suppose sin α = 3/5 with α in quadrant I and cos β = 5/13 with β in quadrant I. By the Pythagorean identity and the quadrant, cos α = 4/5 and sin β = 12/13. Then sin(α + β) = (3/5)(5/13) + (4/5)(12/13) = 15/65 + 48/65 = 63/65, and cos(α + β) = (4/5)(5/13) − (3/5)(12/13) = 20/65 − 36/65 = −16/65. So α + β lies in quadrant II: sine positive, cosine negative. Check numerically: α = arcsin(0.6) ≈ 0.6435, β = arccos(5/13) ≈ 1.1760, α + β ≈ 1.8195, and sin(1.8195) ≈ 0.9692 = 63/65 ✓. The formulas produce exact values for an angle nobody ever named.

Combining waves. The formulas also run backwards. Take sin x + √3 cos x. Compare with 2 sin(x + π/3) = 2(sin x cos(π/3) + cos x sin(π/3)) = 2(sin x × 1/2 + cos x × √3/2) = sin x + √3 cos x. So sin x + √3 cos x = 2 sin(x + π/3): the sum of two waves of the same period is a single wave with amplitude 2 and a phase shift. Check at x = 0.7: sin 0.7 + √3 cos 0.7 ≈ 0.6442 + 1.3247 = 1.9690 and 2 sin(0.7 + 1.0472) = 2 sin(1.7472) ≈ 1.9690 ✓. This is how a modeling problem with two sinusoidal contributions of the same frequency is reduced to one with an amplitude and a shift.

The misconception. Beyond "sine distributes," students often mix the signs: cosine of a sum has a minus in the middle, and sine of a sum has a plus. A quick check settles it: cos(π/2) = cos(π/4 + π/4) must be 0, and cos²(π/4) − sin²(π/4) = 1/2 − 1/2 = 0 while cos²(π/4) + sin²(π/4) = 1 would be wrong. Test any remembered formula on a case whose answer you already know.

Worked, a cosine value: Find cos(7π/12) exactly. 7π/12 = π/3 + π/4, so cos(7π/12) = cos(π/3) cos(π/4) − sin(π/3) sin(π/4) = (1/2)(√2/2) − (√3/2)(√2/2) = √2/4 − √6/4 = (√2 − √6)/4 ≈ (1.4142 − 2.4495)/4 ≈ −0.2588. The angle 7π/12 = 105° is in quadrant II, so a negative cosine is expected. Calculator: cos(1.8326) ≈ −0.2588 ✓.

Worked, tangent of a sum from given tangents: tan α = 3/4 and tan β = 12/5, both angles in quadrant I. tan(α + β) = (3/4 + 12/5)/(1 − (3/4)(12/5)) = (15/20 + 48/20)/(1 − 36/20) = (63/20)/(−16/20) = −63/16. The negative tangent shows α + β is in quadrant II (both angles are acute, so their sum is less than π). This agrees with the earlier result sin(α + β) = 63/65 and cos(α + β) = −16/65 for the same angles, since (63/65)/(−16/65) = −63/16.

Worked, a difference of two given angles: With sin α = 3/5, cos α = 4/5, cos β = 5/13 and sin β = 12/13 as before, sin(α − β) = sin α cos β − cos α sin β = 15/65 − 48/65 = −33/65 and cos(α − β) = cos α cos β + sin α sin β = 20/65 + 36/65 = 56/65. Check the identity: (−33/65)² + (56/65)² = (1089 + 3136)/4225 = 4225/4225 = 1 ✓. Since α ≈ 0.6435 < β ≈ 1.1760, α − β is negative and small, in quadrant IV, matching the signs.

Worked, a double-angle consequence: Set β = α in the addition formulas: sin(2α) = 2 sin α cos α and cos(2α) = cos²α − sin²α. With sin α = 3/5 and cos α = 4/5, sin(2α) = 2 × 3/5 × 4/5 = 24/25 = 0.96 and cos(2α) = 16/25 − 9/25 = 7/25 = 0.28. Calculator check: 2α ≈ 1.2870, sin(1.2870) ≈ 0.96 and cos(1.2870) ≈ 0.28 ✓. A problem that asks for the height of a projectile or the length of a shadow at double a known angle is solved this way without ever computing the angle.

Representation task: Draw the unit circle with A = P(5π/12) ≈ (0.259, 0.966), B = P(π/6) ≈ (0.866, 0.5), C = P(π/4) ≈ (0.707, 0.707) and D = (1, 0). Draw chords AB and CD, measure both, and record that they are equal (each is √(2 − √2) ≈ 0.765 units) because C and D are A and B rotated clockwise by π/6. Beside the drawing write the two distance-formula computations AB² = 2 − 2(cos α cos β + sin α sin β) and CD² = 2 − 2 cos(α − β), and the conclusion cos(π/4) = cos(5π/12) cos(π/6) + sin(5π/12) sin(π/6); check it numerically to three decimal places.

Independent performance: Write a complete proof of cos(α − β) = cos α cos β + sin α sin β with a labeled diagram, then derive the other five formulas from it, naming the symmetry or cofunction fact used at each step. Then solve three problems: the exact value of one of sin, cos or tan at π/12, 5π/12, 7π/12 or 11π/12, solved with a subtraction formula that you show explicitly; the value of sin(α + β) or cos(α − β) for two angles given by one trigonometric value and a quadrant each, with the quadrant of α + β or α − β identified; and the rewriting of a sum a sin x + b cos x that you choose (with a² + b² a perfect square) in the form R sin(x + φ). Success criteria: every derivation step is justified; every exact value is simplified with no radical in a denominator and matches a calculator to three decimals; the quadrant of the combined angle is stated from the signs; and the rewritten wave is checked at one numerical x.