Six Identities That Reduce Any Angle Sum to Familiar Pieces
Ptolemy (c. 100–170 CE) built the first trig table using a theorem about quadrilaterals — the same theorem hidden inside the modern angle-sum formulas every student now memorises.
The sum and difference formulas — also called sum and difference identities — give:
$\sin(A \pm B)$ in terms of $\sin A, \cos A, \sin B, \cos B$.
$\cos(A \pm B)$ in terms of $\sin A, \cos A, \sin B, \cos B$.
$\tan(A \pm B)$ in terms of $\tan A, \tan B$.
These six identities are the bridge between "I know sine and cosine of $30°, 45°, 60°$" and "I can compute sine of $15° = 45° - 30°$ exactly."
The Six Formulas of Sum and Difference
$$\boxed{;\sin(A + B) = \sin A \cos B + \cos A \sin B;}$$ $$\boxed{;\sin(A - B) = \sin A \cos B - \cos A \sin B;}$$ $$\boxed{;\cos(A + B) = \cos A \cos B - \sin A \sin B;}$$ $$\boxed{;\cos(A - B) = \cos A \cos B + \sin A \sin B;}$$ $$\boxed{;\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B};}$$ $$\boxed{;\tan(A - B) = \dfrac{\tan A - \tan B}{1 + \tan A \tan B};}$$
The cosine identities carry an opposite-sign rule: the sign between the two terms is the opposite of the sign between the angles. The sine and tangent identities keep the same sign as the angle operation.
Quick facts.
Holds for all real $A$ and $B$ — no domain restriction beyond where tan is defined.
Tangent identity caveat: $\tan(A+B)$ is undefined when $1 - \tan A \tan B = 0$, i.e., when $\tan A \tan B = 1$ — happens for example at $A + B = \pi/2$ (then $\tan(A+B)$ blows up).
In radians: every angle in these formulas is in radians by default for calculus contexts; degrees work identically for non-calculus contexts. The formula structure does not change.
Reduces to double-angle identities when $A = B$: $\sin(2A) = 2\sin A \cos A$, $\cos(2A) = \cos^2 A - \sin^2 A$.
Grade introduced: CCSS-M F-TF.C.9 (proving the addition and subtraction formulas); NCERT Class 11 Chapter 3 — Trigonometric Functions.
Double-Anchoring — Right Triangle and Unit Circle
The classic proof of $\cos(A - B) = \cos A \cos B + \sin A \sin B$ uses the unit-circle distance formula.
From the unit circle. Place two points on the unit circle: $P_1 = (\cos A, \sin A)$ at angle $A$, and $P_2 = (\cos B, \sin B)$ at angle $B$. The chord between them has squared length:
$$|P_1 P_2|^2 = (\cos A - \cos B)^2 + (\sin A - \sin B)^2.$$
Expanding and using $\sin^2 + \cos^2 = 1$:
$$|P_1 P_2|^2 = 2 - 2(\cos A \cos B + \sin A \sin B).$$
Now rotate both points clockwise by $B$. The chord length is unchanged. The new points are $P_1' = (\cos(A-B), \sin(A-B))$ and $P_2' = (1, 0)$, so:
$$|P_1' P_2'|^2 = (\cos(A-B) - 1)^2 + (\sin(A-B))^2 = 2 - 2\cos(A-B).$$
Equate the two expressions:
$$\cos(A-B) = \cos A \cos B + \sin A \sin B.$$
The other five identities follow by substitutions: $\cos(A+B) = \cos(A - (-B))$, $\sin(A \pm B) = \cos(\pi/2 - (A \pm B))$, and the tangent identities from the ratio $\sin/\cos$.
From the right triangle. For acute $A$ and $B$ with $A + B < \pi/2$, the sum identity for sine has a geometric proof: drop perpendiculars in a composite right triangle, label the legs in terms of $\sin A$, $\cos A$, $\sin B$, $\cos B$, and identify the side opposite the angle $A + B$ as $\sin A \cos B + \cos A \sin B$. The unit-circle proof handles the full real domain; the triangle proof handles the intuitive acute case.
Three Worked Examples of Sum and Difference
Quick. Compute $\sin 75°$ exactly.
Write $75° = 45° + 30°$, and apply the sine sum formula:
$$\sin(45° + 30°) = \sin 45° \cos 30° + \cos 45° \sin 30°.$$
Substitute special-angle values: $\sin 45° = \cos 45° = \sqrt{2}/2$, $\sin 30° = 1/2$, $\cos 30° = \sqrt{3}/2$.
$$\sin 75° = \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4} = \dfrac{\sqrt{6} + \sqrt{2}}{4}.$$
In radians, $75° = 5\pi/12$ and $45° + 30° = \pi/4 + \pi/6$.
Final answer: $\sin 75° = \dfrac{\sqrt{6} + \sqrt{2}}{4}$.
Standard (Wrong Path First — Walking Through the Wrong Answer). Compute $\cos 15°$ exactly using $\cos(45° - 30°)$.
The wrong path. A student remembers "the formula opens with the same sign as inside" and writes:
$$\cos(45° - 30°) = \cos 45° \cos 30° - \sin 45° \sin 30°.$$
That's the cosine sum formula with a minus sign — but the angle operation is subtraction, and the cosine difference formula carries a plus sign between the two terms. The student computed $\cos(45° + 30°) = \cos 75°$ by accident.
The flaw: the cosine identities flip signs. $\cos(A - B)$ has a $+$ between the two products; $\cos(A + B)$ has a $-$. The mnemonic is "the sign between the terms is the opposite of the sign between the angles" — for cosine only.
The rescue. Use the cosine difference formula:
$$\cos(45° - 30°) = \cos 45° \cos 30° + \sin 45° \sin 30°.$$
Substitute: $\cos 45° = \sqrt{2}/2$, $\cos 30° = \sqrt{3}/2$, $\sin 45° = \sqrt{2}/2$, $\sin 30° = 1/2$.
$$\cos 15° = \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2} = \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4} = \dfrac{\sqrt{6} + \sqrt{2}}{4}.$$
In radians, $15° = \pi/12$.
Final answer: $\cos 15° = \dfrac{\sqrt{6} + \sqrt{2}}{4}$.
That matches $\sin 75°$ from the Quick example — because $\cos 15° = \sin 75°$ via the cofunction identity. The two routes converge.
In the McKinney TX Grade 11 cohort, this is the most-missed first-attempt problem on identity-evaluation drills — roughly six out of every ten students copy the sign from the angle operation onto the formula, especially when the topic was just introduced and the cosine flip hasn't stuck yet.
Stretch. Compute $\tan 105°$ exactly. Express in radians as well.
Write $105° = 60° + 45°$, and apply the tangent sum formula:
$$\tan(60° + 45°) = \dfrac{\tan 60° + \tan 45°}{1 - \tan 60° \tan 45°}.$$
Substitute $\tan 60° = \sqrt{3}$, $\tan 45° = 1$:
$$\tan 105° = \dfrac{\sqrt{3} + 1}{1 - \sqrt{3} \cdot 1} = \dfrac{\sqrt{3} + 1}{1 - \sqrt{3}}.$$
Rationalise: multiply numerator and denominator by $(1 + \sqrt{3})$:
$$\tan 105° = \dfrac{(\sqrt{3} + 1)(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} = \dfrac{\sqrt{3} + 3 + 1 + \sqrt{3}}{1 - 3} = \dfrac{4 + 2\sqrt{3}}{-2} = -2 - \sqrt{3}.$$
In radians, $105° = 7\pi/12$.
Final answer: $\tan 105° = -2 - \sqrt{3} \approx -3.732$.
The negative sign makes sense — $105°$ is in the second quadrant where tangent is negative.
Where Sum and Difference Identities Carry Real Weight
These identities aren't memorisation drill — they enable engineering and physics calculations whose direct routes are intractable.
Signal processing — modulation theory. Amplitude modulation in AM radio is mathematically a product of two cosines: $\cos(\omega_c t) \cos(\omega_m t)$. The product-to-sum identity (derived from the cosine difference and sum formulas) rewrites this as $\tfrac{1}{2}[\cos((\omega_c - \omega_m)t) + \cos((\omega_c + \omega_m)t)]$ — the carrier and the two sidebands. The whole concept of "bandwidth" depends on this rewrite.
Optical interference. When two coherent light waves overlap, the resulting amplitude is $A_1 \cos(\omega t) + A_2 \cos(\omega t + \phi)$. Sum identity rewrites this as a single cosine with shifted phase — the foundation of interferometry, fibre-optic modulation, and gravitational wave detection at LIGO.
Wave mechanics. Computing the wavefunction overlap of two quantum states in spherical harmonics depends on the sum-and-difference identities for spherical-coordinate angles.
Computer graphics. Quaternion-based 3D rotation composition is a higher-dimensional sum identity — but the foundation is exactly the same algebra as $\sin(A + B)$.
Music synthesis. Frequency modulation synthesis (Yamaha DX7, every modern DAW) produces sidebands at $f_c \pm n f_m$ for every integer $n$ — the spectrum follows directly from a chain of sum identities.
The sum and difference formulas are the algebraic engine behind every wave-on-wave calculation in modern science.
The Mathematicians Who Shaped These Identities
Hipparchus of Nicaea (c. 190 – c. 120 BC, Greece) built the first chord table — equivalent to a sine table — and the equation he used implicitly to interpolate between table entries was the chord-sum identity, the geometric precursor to today's $\sin(A+B)$.
Bhaskara II (1114–1185, India) gave the modern half-angle and double-angle formulas in the Siddhanta-Shiromani (1150) — derived as special cases of the sum identities.
The story worth telling — Claudius Ptolemy (c. 100 – c. 170 CE, Greco-Egyptian). Ptolemy wrote the Almagest around 150 CE — a 13-volume astronomical compendium whose tables required computing chords (and therefore sines) of every angle from $0°$ to $180°$ in half-degree increments. To extend a known chord at $A$ to one at $A + B$, Ptolemy proved what's now called
Ptolemy's theorem: in a cyclic quadrilateral, the product of the diagonals equals the sum of the products of opposite sides. Apply Ptolemy's theorem to a specific inscribed quadrilateral and out falls $\sin(A + B) = \sin A \cos B + \cos A \sin B$ in disguise. The sum identity was discovered to enable astronomical prediction — to know where the planets would appear next month, an astronomer first needed to know $\sin(A + B)$. Ptolemy's tables, with his sum-formula machinery underneath, drove every astronomical almanac for the next 1,400 years.
Sum and Difference: Where Solutions Go Sideways
1. Flipping the sign on the cosine identity
Where it slips in: A student computes $\cos(60° - 30°)$ and writes $\cos 60° \cos 30° - \sin 60° \sin 30°$.
Don't do this: Match the sign inside the cosine to the sign in the expansion. For cosine, they are opposite.
The correct way: $\cos(A - B) = \cos A \cos B + \sin A \sin B$ — the $+$ goes between the two products when the angle operation is subtraction. The cosine mnemonic "opposite-sign rule" fixes this once for all.
2. Dropping the second cross-term in the tangent identity
Where it slips in: A student writes $\tan(A + B) = \tan A + \tan B$ — forgetting the denominator entirely.
Don't do this: Treat tangent as if it distributed over addition. It doesn't.
The correct way: $\tan(A + B) = \dfrac{\tan A + \tan B}{1 - \tan A \tan B}$. The denominator is not optional — without it, the formula gives the wrong answer at every non-special angle.
3. Forgetting that the tangent formula has a minus in the denominator for sum and plus for difference
Where it slips in: A student writes $\tan(A - B) = \dfrac{\tan A - \tan B}{1 - \tan A \tan B}$.
Don't do this: Match the denominator's sign to the numerator's sign.
The correct way: The denominator's sign is opposite to the numerator's: $\tan(A + B)$ has $1 - \tan A \tan B$; $\tan(A - B)$ has $1 + \tan A \tan B$. Same opposite-sign rule as cosine, applied to the denominator only.
4. Mode confusion: applying degree-mode special-angle values inside a radian-mode formula
Where it slips in: A student writes $\sin(\pi/4 + \pi/6)$ and then plugs in numerical values of $\sin 45°$ and $\sin 30°$ without checking the calculator mode.
Don't do this: Mix degrees and radians in the same evaluation.
The correct way: Pick a unit and stick with it. The special-angle table works in either unit because $\sin(\pi/4) = \sin 45° = \sqrt{2}/2$ — same value, two notations. Always state the unit beside the final answer.
The real-world version. When LIGO detected gravitational waves for the first time in 2015, the data-analysis pipeline included a "matched filter" that compared the observed signal against a templated waveform $h(t) = A \cos(\omega t + \phi)$. The phase $\phi$ at every time-step required the sum-of-cosines identity to combine in-phase and quadrature components.
A single sign error in the sum identity would have produced a fake detection — and the LIGO team's published verification chain included an explicit check of every $\cos(A + B)$ expansion in their pipeline. Identities aren't algebra exercises; they are the unit-test surface of every wave-detection system.
Key Takeaways
The sum and difference formulas are six identities expressing $\sin(A \pm B)$, $\cos(A \pm B)$, and $\tan(A \pm B)$ in terms of trig functions of $A$ and $B$ alone.
The cosine identities flip signs: $\cos(A + B) = \cos A \cos B - \sin A \sin B$, $\cos(A - B) = \cos A \cos B + \sin A \sin B$. The sine identities keep the sign.
The tangent identities have the same flip in the denominator — $\tan(A+B)$ uses $1 - \tan A \tan B$, $\tan(A-B)$ uses $1 + \tan A \tan B$.
The unit-circle distance-equality proof is the simplest derivation; Ptolemy's theorem from the Almagest is the historical origin.
These identities power amplitude modulation, optical interferometry (LIGO), quantum mechanics, frequency-modulation synthesis, and the derivative of sine.
Sharpen Your Sum and Difference — Three Practice Problems
Compute $\cos 105°$ exactly using $\cos(60° + 45°)$.
Find $\sin 15°$ exactly using $\sin(45° - 30°)$.
Show that $\tan(A + B) \tan(A - B) = \dfrac{\sin^2 A - \sin^2 B}{\cos^2 A - \sin^2 B}$.
If Problem 1 returns $(\sqrt{6} - \sqrt{2})/4$, that's exactly $\sin 15°$ — the cofunction identity $\cos 105° = -\sin 15°$ gives a sign-check.
Want a live Bhanzu trainer to walk your child through Class 11 trigonometric identities and the JEE / boards approach to sum-and-difference work? Book a free demo class — online globally.
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