Sum and difference formulas — sin(A±B), cos(A±B), tan(A±B) — proofs: Worked examples

This page is the practice ground for the parent proofs. We already know the six identities; here we throw every kind of input at them so you never meet a case you haven't seen.

The scenario matrix

Cell	What makes it tricky	Example
C1 — nice split, both QI	angles chosen so both are "known" ( $30°, 45°, 60°$ )	Ex 1: $cos 15°$
C2 — one angle unknown, given ratios	you're handed $sin, cos$ not the angle	Ex 2: $sin (A + B)$ from triangles
C3 — a quadrant-II angle	one component is negative, sign bookkeeping	Ex 3: $A$ in QII
C4a — a quadrant-III angle	both $sin, cos$ negative	Ex 4: $B$ in QIII
C4b — a quadrant-IV angle	$cos > 0, sin < 0$	Ex 4b: $B$ in QIV
C5 — degenerate input $B = 0$ or $A = B$	formula must collapse to a known truth	Ex 5: $sin (A + 0)$ , $tan$ , double-angle
C6 — tangent denominator $\to 0$ (limiting)	$A + B = 90°$ , $tan$ undefined	Ex 6: $tan A tan B = 1$
C7 — real-world word problem	translate a story into $A \pm B$	Ex 7: two ramps
C8 — exam twist, hidden angle	prove an identity / phase shift	Ex 8: $cos (x + \frac{π}{3})$ expansion

Prerequisites you may want open: Unit Circle and Trig Definitions, Pythagorean Identity, Cofunction Identities.

Example 1 — Cell C1: nice split, both angles in QI

Worked example Find the exact value of

$cos 15°$ . Forecast: Guess — will the answer be bigger or smaller than $cos 30° \approx 0.866$ ? ( $15°$ is closer to $0°$ , so $cos$ should be bigger, near $0.97$ .)

Step 1. Write $15° = 45° - 30°$ . Why this step? $45°$ and $30°$ have exact known values; $15°$ does not. We split an ugly angle into two nice ones — the whole reason these formulas exist.

Step 2. Use $cos (A - B) = cos A cos B + sin A sin B$ with $A = 45°, B = 30°$ : $cos 15° = cos 45° cos 30° + sin 45° sin 30°.$ Why this step? We need $cos$ of a difference, and the difference form of cosine uses a plus (cosine of a difference keeps a $+$ inside).

Step 3. Substitute exact values, then multiply the fractions: $= \frac{2}{2} \cdot \frac{3}{2} + \frac{2}{2} \cdot \frac{1}{2} = \frac{6}{4} + \frac{2}{4} = \frac{6 + 2}{4} .$ Why this step? $\frac{2}{2} \cdot \frac{3}{2} = \frac{6}{4}$ because $23 = 6$ and $2 \cdot 2 = 4$ ; the two terms share the same denominator $4$ , so we just add the numerators.

Verify: $\frac{6 + 2}{4} \approx \frac{2.449 + 1.414}{4} = \frac{3.863}{4} \approx 0.966$ . Matches our forecast of "near $0.97$ ", and a calculator gives $cos 15° = 0.9659$ . ✓

Example 2 — Cell C2: angles given as ratios, both acute

Worked example

$A$ and $B$ are acute (QI). Given $sin A = \frac{3}{5}$ and $cos B = \frac{5}{13}$ , find $sin (A + B)$ . Forecast: Both angles are in QI, so $A + B$ could exceed $90°$ . Expect the answer to be positive but you can't guess the exact value yet.

Step 1. Find the missing components with the Pythagorean Identity. Since $A$ is acute, $cos A > 0$ : $cos A = 1 - (3/5)^{2} = 1 - 9/25 = \frac{4}{5}$ . Since $B$ is acute, $sin B > 0$ : $sin B = 1 - (5/13)^{2} = 1 - 25/169 = \frac{12}{13}$ . Why this step? The formula needs all four of $sin A, cos A, sin B, cos B$ . We only got two; Pythagoras recovers the rest, and the acute assumption fixes the signs to positive.

Step 2. Apply $sin (A + B) = sin A cos B + cos A sin B$ : $= \frac{3}{5} \cdot \frac{5}{13} + \frac{4}{5} \cdot \frac{12}{13} .$ Why this step? "Sine keeps company" — same signs, sin-cos plus cos-sin.

Step 3. Multiply each fraction then add: $= \frac{15}{65} + \frac{48}{65} = \frac{63}{65} .$ Why this step? $3 \cdot 5 = 15$ and $4 \cdot 12 = 48$ over the common denominator $5 \cdot 13 = 65$ ; equal denominators let us add the numerators directly ( $15 + 48 = 63$ ).

Verify: $\frac{63}{65} \approx 0.969 \leq 1$ — a legal sine value. ✓ (And $A \approx 36.87°$ , $B \approx 67.38°$ , so $A + B \approx 104.25°$ ; $sin 104.25° \approx 0.969$ . ✓)

Example 3 — Cell C3: one angle in quadrant II (a negative component)

Figure — Sum and difference formulas — sin(A±B), cos(A±B), tan(A±B) — proofs

Read the figure first. The blue arrow is the angle $A$ drawn on the unit circle; its tip lands in the upper-left region (quadrant II). The dashed red segment is the height $sin A = \frac{4}{5}$ (positive, above the axis), and the orange segment is the horizontal reach $cos A$ — it points left of the origin, so as a coordinate it is negative. That leftward orange arrow is the whole reason the next step carries a minus sign.

Step 1. Get $cos A$ . Using the picture: in QII the point sits left of the $y$ -axis, so its $x$ -coordinate (which is $cos A$ ) is negative. $cos A = - 1 - (4/5)^{2} = - 1 - 16/25 = - \frac{3}{5} .$ Why this step? Pythagoras gives the size $\frac{3}{5}$ ; the quadrant supplies the sign. Forgetting this minus is the #1 error here.

Step 2. Get $sin B$ . Since $B$ is acute, $sin B > 0$ , and the Pythagorean Identity gives $sin B = 1 - cos^{2} B = 1 - (12/13)^{2} = 1 - 144/169 = \frac{5}{13} .$ Why this step? The cosine formula needs $sin B$ , but we were only handed $cos B$ ; the Pythagorean identity recovers it, and the acute assumption tells us to take the positive root.

Step 3. Apply $cos (A + B) = cos A cos B - sin A sin B$ ("cosine is contrary" — minus in the middle): $= (- \frac{3}{5}) \frac{12}{13} - \frac{4}{5} \cdot \frac{5}{13} .$ Why this step? The sum form of cosine flips signs, and we carefully carry the negative $cos A$ .

Step 4. Multiply and add: $= - \frac{36}{65} - \frac{20}{65} = - \frac{56}{65} .$ Why this step? $(- 3) (12) = - 36$ and $(4) (5) = 20$ over $65$ ; both terms are subtracted, so we combine $- 36 - 20 = - 56$ .

Verify: $- \frac{56}{65} \approx - 0.862$ , in $[- 1, 1]$ and negative as forecast. ✓ ( $A \approx 126.87°$ , $B \approx 22.62°$ , sum $\approx 149.5°$ ; $cos 149.5° \approx - 0.862$ . ✓)

Example 4 — Cell C4a: an angle in quadrant III (two negatives)

Worked example

$A$ is acute with $tan A = \frac{3}{4}$ . $B$ is in quadrant III with $tan B = \frac{5}{12}$ . Find $tan (A - B)$ . Forecast: In QIII, tangent is actually positive (both $sin$ and $cos$ negative → ratio positive), so we can use the given $tan B$ directly. Guess the sign of $tan (A - B)$ ? Since $tan A > tan B$ , numerator is positive — likely positive.

Step 1. Note $tan A > 0$ , $tan B > 0$ (QIII keeps tangent positive because two minuses cancel in $sin / cos$ ). Why this step? Unlike sine or cosine, tangent doesn't need us to split into components — the formula uses tangents straight. But we must still confirm the sign, and QIII gives a positive tangent.

Step 2. Apply $tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}$ (difference form → plus below): $= \frac{\frac{3}{4} - \frac{5}{12}}{1 + \frac{3}{4} \cdot \frac{5}{12}} .$ Why this step? "Tangent's below is opposite": top has $-$ , bottom has $+$ .

Step 3. Simplify top and bottom separately. Top: rewrite $\frac{3}{4} = \frac{9}{12}$ so both fractions share denominator $12$ , giving $\frac{9}{12} - \frac{5}{12} = \frac{4}{12} = \frac{1}{3}$ . Bottom: $\frac{3}{4} \cdot \frac{5}{12} = \frac{15}{48} = \frac{5}{16}$ , so $1 + \frac{5}{16} = \frac{21}{16}$ . $tan (A - B) = \frac{1/3}{21/16} = \frac{1}{3} \cdot \frac{16}{21} = \frac{16}{63} .$ Why this step? Dividing by a fraction means multiplying by its reciprocal ( $\div \frac{21}{16} = \times \frac{16}{21}$ ), and $3 \cdot 21 = 63$ gives the denominator.

Verify: $\frac{16}{63} \approx 0.254 > 0$ as forecast. ✓

Example 4b — Cell C4b: an angle in quadrant IV ( $cos > 0, sin < 0$ )

Worked example

$A$ is acute with $cos A = \frac{8}{17}$ . $B$ is in quadrant IV with $cos B = \frac{3}{5}$ . Find $sin (A + B)$ . Forecast: In QIV, sine is negative ( $cos > 0$ but $sin < 0$ ), so $sin B < 0$ . That negative piece may drag $sin (A + B)$ down — the answer could even be negative. Hold that guess.

Step 1. Complete $A$ (acute, QI): $sin A = 1 - (8/17)^{2} = 1 - 64/289 = \frac{15}{17}$ (positive). Why this step? Pythagoras recovers $sin A$ ; the acute assumption forces the positive root.

Step 2. Complete $B$ in QIV: the magnitude is $1 - (3/5)^{2} = \frac{4}{5}$ , but in quadrant IV the height is below the axis, so $sin B = - \frac{4}{5} .$ Why this step? This is the QIV twist: $cos B > 0$ but $sin B < 0$ . Pythagoras gives the size $\frac{4}{5}$ ; the quadrant flips the sign to negative — exactly the subcase C3 and C4a did not cover.

Step 3. Apply $sin (A + B) = sin A cos B + cos A sin B$ : $= \frac{15}{17} \cdot \frac{3}{5} + \frac{8}{17} \cdot (- \frac{4}{5}) .$ Why this step? "Sine keeps company" (same signs), and we faithfully carry the negative $sin B$ .

Step 4. Multiply and add: $= \frac{45}{85} - \frac{32}{85} = \frac{13}{85} .$ Why this step? $15 \cdot 3 = 45$ and $8 \cdot 4 = 32$ over $17 \cdot 5 = 85$ ; the second term is negative, so $45 - 32 = 13$ .

Verify: $\frac{13}{85} \approx 0.153$ , inside $[- 1, 1]$ . The negative $sin B$ pulled the sum down toward zero (small positive) rather than negative — a fine outcome consistent with our "could be small/negative" forecast. ✓

Example 5 — Cell C5: degenerate inputs (formula must collapse)

Worked example Show the formulas survive the degenerate cases (a)

$B = 0$ , (b) $B = A$ , and (c) the tangent degeneracies. Forecast: With $B = 0$ nothing changes, so $sin (A + 0)$ must equal $sin A$ . With $B = A$ we should recover a double-angle formula. And $tan (A - A)$ should be $tan 0 = 0$ .

Step 1 (a). Put $B = 0$ into $sin (A + B) = sin A cos B + cos A sin B$ : $sin (A + 0) = sin A cos 0 + cos A sin 0 = sin A \cdot 1 + cos A \cdot 0 = sin A .$ Why this step? A degenerate input is a free correctness test: adding nothing must give the original. Using $cos 0 = 1, sin 0 = 0$ the second term dies. ✓

Step 2 (b). Put $B = A$ into $cos (A + B) = cos A cos B - sin A sin B$ : $cos (A + A) = cos A cos A - sin A sin A = cos^{2} A - sin^{2} A = cos 2 A .$ Why this step? Setting the two angles equal is the bridge to double-angle identities — the "same-angle" edge case.

Step 3 (c). Now stress-test the tangent formula on its degeneracies. Put $B = 0$ into $tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B}$ : $tan (A + 0) = \frac{t a n A + t a n 0}{1 - t a n A t a n 0} = \frac{t a n A + 0}{1 - 0} = tan A .$ Put $B = A$ into the difference form $tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}$ : $tan (A - A) = \frac{t a n A - t a n A}{1 + t a n ^{2} A} = \frac{0}{1 + t a n ^{2} A} = 0.$ Why this step? The tangent formula has a denominator, so its degenerate tests are extra important — we confirm it neither blows up nor misbehaves. With $B = 0$ the numerator loses a term and the denominator becomes $1$ , returning $tan A$ ; with $B = A$ the numerator becomes $0$ (numerator wins over any nonzero denominator), giving $tan 0 = 0$ exactly as it should. ✓

Verify (numeric). Take $A = 50°$ : $sin (50° + 0) = sin 50° = 0.766$ . ✓ Take $A = 30°$ : $cos (2 \cdot 30°) = cos 60° = 0.5$ while $cos^{2} 30° - sin^{2} 30° = 0.75 - 0.25 = 0.5$ . ✓ Take $A = 40°$ : $tan (40° + 0) = tan 40° \approx 0.839$ ✓ and $tan (40° - 40°) = tan 0° = 0$ . ✓

Example 6 — Cell C6: tangent's limiting case (denominator → 0)

Worked example

$A$ and $B$ are acute with $tan A = \frac{1}{2}$ and $tan B = 2$ . Evaluate $tan (A + B)$ , and explain the result geometrically. Forecast: Notice $tan A tan B = \frac{1}{2} \cdot 2 = 1$ . The denominator $1 - tan A tan B$ becomes $0$ . Division by zero → the value is undefined / infinite. Guess: $A + B = 90°$ .

Step 1. Write $tan (A + B) = \frac{tan A + tan B}{1 - tan A tan B} = \frac{\frac{1}{2} + 2}{1 - 1} = \frac{5/2}{0}$ . Why this step? The formula is faithfully reporting a limiting situation: the output angle is one whose tangent doesn't exist.

Step 2. Interpret. $tan θ = \frac{sin θ}{cos θ}$ is undefined exactly when $cos θ = 0$ , i.e. $θ = 90°$ . So the denominator hitting $0$ is the formula screaming " $A + B = 90°$ ". Why this step? A zero denominator is never a "mistake" here — it's the algebra of a right angle. Whenever $tan A tan B = 1$ , the two acute angles are complementary (they add to $90°$ ).

Verify: $A = arctan (0.5) \approx 26.565°$ , $B = arctan 2 \approx 63.435°$ ; sum $= 90.000°$ . ✓ Equivalently $tan A = \frac{1}{2}$ and $cot B = \frac{1}{t a n B} = \frac{1}{2}$ , so $tan A = cot B$ , which is exactly the cofunction condition for complementary angles.

Example 7 — Cell C7: real-world word problem

Read the figure first. The gray arrow is the horizontal (the sea). The blue arrow is the line of sight after turning up by the first angle $A$ (small blue arc at the origin). The orange arrow is the line of sight after the extra rise $B$ (green arc stacked on top of the blue one); it makes the total angle $A + B$ with the horizontal. The picture shows why we add the angles: turning up by $A$ and then by $B$ lands the sight-line at elevation $A + B$ — precisely the input to our cosine formula.

Step 1. From $tan A = \frac{7}{24}$ build the right triangle: opposite $7$ , adjacent $24$ , hypotenuse $7^{2} + 2 4^{2} = 625 = 25$ . So $sin A = \frac{7}{25}, cos A = \frac{24}{25}$ . Why this step? The story hands us a tangent (a ratio of two sides), but the cosine sum formula needs $sin A$ and $cos A$ separately. Reading $tan A$ as "opposite $7$ over adjacent $24$ " fixes a $7$ - $24$ - $25$ right triangle; its hypotenuse $25$ then gives each of $sin A$ and $cos A$ directly, both positive because $A$ is acute.

Step 2. From $sin B = \frac{3}{5}$ (B acute) get $cos B = \frac{4}{5}$ via the Pythagorean Identity. Why this step? $cos B = 1 - (3/5)^{2} = 1 - 9/25 = \frac{4}{5}$ , positive because $B$ is acute; the formula needs $cos B$ which the story didn't hand us.

Step 3. Apply $cos (A + B) = cos A cos B - sin A sin B$ : $= \frac{24}{25} \cdot \frac{4}{5} - \frac{7}{25} \cdot \frac{3}{5} = \frac{96}{125} - \frac{21}{125} = \frac{75}{125} = \frac{3}{5} .$ Why this step? We combine the two elevation turns into one; cosine's "contrary" minus sign mixes them. Numerically $24 \cdot 4 = 96$ and $7 \cdot 3 = 21$ over $125$ , and $96 - 21 = 75$ , which reduces $\frac{75}{125} = \frac{3}{5}$ (divide top and bottom by $25$ ).

Verify: $\frac{3}{5} = 0.6$ . Since $cos A = \frac{24}{25} = 0.96$ and the total angle is larger, $0.6 < 0.96$ — smaller cosine, exactly as forecast. ✓ Also $A \approx 16.26°$ , $B \approx 36.87°$ , sum $\approx 53.13°$ , $cos 53.13° = 0.6$ . ✓

Example 8 — Cell C8: exam twist (hidden angle, expand a general expression)

Worked example Expand

$cos (x + \frac{π}{3})$ in terms of $sin x$ and $cos x$ , then use it to find $cos (x + \frac{π}{3})$ when $x = \frac{π}{6}$ . Forecast: $\frac{π}{6} + \frac{π}{3} = \frac{π}{2} = 90°$ , and $cos 90° = 0$ . So the final number should be $0$ — a good self-check.

Step 1. Apply $cos (A + B) = cos A cos B - sin A sin B$ with $A = x, B = \frac{π}{3}$ : $cos (x + \frac{π}{3}) = cos x cos \frac{π}{3} - sin x sin \frac{π}{3} .$ Why this step? Here $B$ is a fixed known angle but $x$ is a variable — the exam twist. Nothing changes; we still just plug into the sum formula.

Step 2. Use $cos \frac{π}{3} = \frac{1}{2}, sin \frac{π}{3} = \frac{3}{2}$ : $= \frac{1}{2} cos x - \frac{3}{2} sin x .$ Why this step? This is the finished expanded form — a linear combination of $cos x$ and $sin x$ , which is exactly what "expand in terms of $sin x$ and $cos x$ " asks for.

Step 3. Substitute $x = \frac{π}{6}$ , so $cos \frac{π}{6} = \frac{3}{2}, sin \frac{π}{6} = \frac{1}{2}$ : $= \frac{1}{2} \cdot \frac{3}{2} - \frac{3}{2} \cdot \frac{1}{2} = \frac{3}{4} - \frac{3}{4} = 0.$ Why this step? Both products equal $\frac{3}{4}$ , so they cancel exactly — the identical terms subtract to zero.

Verify: Matches the forecast $cos \frac{π}{2} = 0$ . ✓

Recall

Recall Which cell forces you to insert a minus sign by hand?

C3/C4a/C4b — a quadrant-II/III/IV angle. Pythagoras gives the magnitude; the quadrant gives the sign of the missing component. ::: Quadrant fixes the sign.

Recall In quadrant IV, which of

$sin$ and $cos$ is negative? $sin < 0$ (below the axis), $cos > 0$ (right of the axis) — see Ex 4b. ::: Only sine is negative in QIV.

Recall What does a zero denominator in

$tan (A + B)$ mean? $tan A tan B = 1 \Rightarrow A + B = 90°$ , where tangent is undefined — the angles are complementary. ::: $A + B = 90°$ .

Recall How do you sanity-check a

$sin$ or $cos$ answer instantly? It must lie in $[- 1, 1]$ ; if it's outside, you slipped a sign or a component. ::: Range check $[- 1, 1]$ .

Connections

Parent proofs (Hinglish) — where these identities come from.
Unit Circle and Trig Definitions — quadrant signs of $sin, cos$ .
Pythagorean Identity — recovers the missing component in Ex 2, 3, 4b, 7.
Double Angle Formulas — the $B = A$ collapse in Ex 5.
Cofunction Identities — the complementary-angle insight in Ex 6.