States the sum and difference identities for the sine, cosine and tangent, and puts them to work finding exact values for angles beyond the special ones. Explains the cosine's reversed sign pattern, derives the cofunction identities as a consequence, and uses the identities to verify and to simplify.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 7 — Trigonometric Identities and Equations
§7.2 Sum and Difference Identities, pp. 803-820
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 803-820 — the pages these objectives are drawn from
Warm-up
It looks as though it should, which is exactly the danger.
Discussion prompt
Is the sine of a sum equal to the sum of the sines? Test it at thirty degrees plus thirty degrees.
Hint: Compute both sides.
Answer:
The left side is the sine of sixty degrees, which is about 0.866. The right side is one half plus one half, which is 1.
They are not equal, so the sine does not distribute over a sum. Nor does any of the six functions.
The reason is that the sine is a function, not a multiplier. Distribution is a property of multiplication, and applying it to a function name is a category error — which is why this section needs genuine identities rather than a shortcut.
Concept
There is a formula for the sine or cosine of a sum, but it involves both functions of both angles rather than distributing.
\[ \cos(A\pm B)=\cos A\cos B\mp\sin A\sin B \]
Notice the structure: each term is a product of one function of A with one function of B. Nothing simpler is possible, and that is why the false distribution rule is so wrong — it produces one term where there must be two.
Figure (svg): Three cards giving the sum and difference identities for the cosine, the sine and the tangent, with their sign patterns marked
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 803-808
Section
Section 1
Concept
The cosine of a sum uses a minus sign, and the cosine of a difference uses a plus. This reversal is the single most-forgotten fact in the section.
A quick check on the sign: put B equal to A in the sum formula. The result should be the cosine of twice A, and with the minus sign it gives cosine squared minus sine squared, which is correct. With a plus sign it would give 1, which is obviously wrong.
Figure (svg): Three cards giving the sum and difference identities for the cosine, the sine and the tangent, with their sign patterns marked
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 803-809
Picture it
Note the marked sign patterns on each card.
Figure (svg): Three cards giving the sum and difference identities for the cosine, the sine and the tangent, with their sign patterns marked
The cosine opposes, the sine matches, and the tangent does both — matching on top and opposing on the bottom. Those three notes carry the whole section.
Worked example
Write it as a difference of special angles.
\[ \text{Find } \cos 15^\circ \text{ exactly.} \]
Write as a difference
Why: Both parts are special angles.
\[ 45 - 30 \]
Apply the difference formula
Why: Cosine times cosine plus sine times sine.
\[ \cos 45 \cos 30 + \sin 45 \sin 30 \]
Substitute the exact values
Why: From the special triangles.
\[ (r 2 / 2) (r 3 / 2) + (r 2 / 2) (\frac{1}{2}) \]
Combine over a common denominator
Why: Both have a four.
\[ \frac{r 6 + r 2}{4} \]
Figure (svg): A diagram showing fifteen degrees written as the difference of forty-five and thirty, both of which are special angles
\[ \cos 15^\circ=\frac{\sqrt{6}+\sqrt{2}}{4} \]
Verify: check numerically
Why: Root six is about 2.449 and root two about 1.414, so the sum over four is about 0.966. A calculator gives the cosine of fifteen degrees as 0.9659 — matching. And the value being just under 1 is right, since fifteen degrees is a small angle.
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 804-807
Prediction
You are expanding the cosine of a difference.
Predict first
Which sign appears between the two products?
Correct: A plus sign.
Why: The cosine's signs are opposite to the operation in the argument, so a difference inside gives a sum of products. This reversal applies only to the cosine; the sine's signs match the argument.
Worked example
Setting the two angles equal is a fast test.
\[ \text{Verify the sign in } \cos(A+B) \text{ by putting } B=A. \]
Substitute B equal to A
Why: Both angles the same.
\[ \cos(2 A) \]
Apply the sum formula with a minus
Why: Products of equal terms.
\[ \cos ^{2} A - \sin ^{2} A \]
Test at A equal to forty-five degrees
Why: Both squares are one half.
\[ 0 \]
Compare
Why: The cosine of ninety degrees is zero.
Figure (svg): The solution to Worked example check the sign shown as a ladder of expressions, one row per legal move
\[ \cos 2A=\cos^2A-\sin^2A \]
Verify: test the wrong sign too
Why: With a plus sign the formula would give cosine squared plus sine squared, which is 1 for every angle — so the cosine of any doubled angle would be 1, which is plainly false. The test distinguishes the two signs decisively in one line.
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 807-809
Trap
\[ \cos(A+B)=\cos A+\cos B \]
Treat the cosine as a factor
Why: The function name is distributed across the sum as though it were a coefficient.
At sixty and thirty degrees this claims one half plus root three over two, about 1.37 — but the cosine of ninety degrees is zero.
A function name is not a multiplier. Distribution is a property of multiplication over addition, and the cosine is neither.
The correct formula produces two products, each mixing one function of each angle, which is why nothing simpler exists.
Test any suspected shortcut numerically before using it. One substitution of easy angles settles the question in seconds, and this particular error survives only because it is never tested.
Sorting
Compare the sign in the argument to the sign in the expansion.
Sort into buckets
Sort each function.
Faded example
Fill in the products and the sign.
Fill in the blanks
\cos(A+B)=\cos A\cos B - \sin A\sin B
Why: The two products each pair one function of each angle, and the minus sign is the cosine's reversal. Writing cosine cosine first and sine sine second keeps the pattern consistent across all the formulas.
Explain it to yourself
Setting the two angles equal tests the sign in one line.
Discussion prompt
Explain why that test works.
Hint: What is the cosine of twice a forty-five degree angle?
Answer:
Putting B equal to A turns the formula into an expression for the cosine of twice A, which can be checked at a convenient angle.
At forty-five degrees the double is ninety degrees, whose cosine is zero. The minus version gives cosine squared minus sine squared, which is one half minus one half — zero, correct.
The plus version would give 1 for every angle, which is absurd. So the test is decisive rather than suggestive, and it takes one line. A formula you can re-derive or re-check is safer than one you only remember.
Section
Section 2
Concept
The sine of a sum expands with a plus and the sine of a difference with a minus, which is the intuitive behaviour — and it is exactly why the cosine's reversal is so easy to forget.
The tangent formula is worth deriving once from the sine over the cosine, dividing numerator and denominator by the product of the cosines. That derivation explains where the 1 in the denominator comes from and why the sign there is reversed.
Figure (svg): Three cards giving the sum and difference identities for the cosine, the sine and the tangent, with their sign patterns marked
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 809-815
Picture it
Three cards, three different patterns.
Figure (svg): Three cards giving the sum and difference identities for the cosine, the sine and the tangent, with their sign patterns marked
The sine is the easy one and the cosine the trap. The tangent combines both behaviours, which makes it the one worth deriving rather than recalling.
Worked example
A sum of special angles this time.
\[ \text{Find } \sin 75^\circ \text{ exactly.} \]
Write as a sum
Why: Two special angles.
\[ 45 + 30 \]
Apply the sum formula
Why: Sine cosine plus cosine sine.
\[ \sin 45 \cos 30 + \cos 45 \sin 30 \]
Substitute exact values
Why: From the special triangles.
\[ (r 2 / 2) (r 3 / 2) + (r 2 / 2) (\frac{1}{2}) \]
Combine
Why: A common denominator of four.
\[ \frac{r 6 + r 2}{4} \]
Figure (svg): A right triangle with its two acute angles marked as complementary, showing that one angle's sine is the other's cosine
\[ \sin 75^\circ=\frac{\sqrt{6}+\sqrt{2}}{4} \]
Verify: compare with the cosine of fifteen
Why: This is the same value found earlier for the cosine of fifteen degrees, and it should be — seventy-five and fifteen are complementary, so one's sine equals the other's cosine. That agreement is a genuine check on both computations.
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 810-812
Faded example
The sine's signs match the argument.
Fill in the blanks
\sin(A-B)=\sin A\cos B - \cos A\sin B
Why: A difference inside gives a difference of the two products, since the sine's signs match. Only the cosine reverses, which is why keeping the two straight is worth one deliberate check per problem.
Worked example
Watch the denominator's sign.
\[ \text{Find } \tan 15^\circ \text{ exactly.} \]
Write as a difference
Why: Two special angles.
\[ 45 - 30 \]
Apply the difference formula
Why: Minus on top, plus on the bottom.
\[ \frac{\tan 45 - \tan 30}{1 + \tan 45 \tan 30} \]
Substitute
Why: Tangent of 45 is 1, of 30 is one over root three.
\[ \frac{1 - 1 / r 3}{1 + 1 / r 3} \]
Simplify
Why: Multiply through by root three and rationalise.
\[ 2 - r 3 \]
Figure (svg): The solution to Worked example use the tangent formula shown as a ladder of expressions, one row per legal move
\[ \tan 15^\circ=2-\sqrt{3} \]
Verify: check numerically
Why: Root three is about 1.732, so the answer is about 0.268. A calculator gives the tangent of fifteen degrees as 0.2679 — matching. And a value well under 1 is right, since fifteen degrees is well under the forty-five degrees where the tangent equals 1.
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 812-815
Error analysis
A student expands a tangent of a sum.
Annotate
On: \( \tan(A+B)=\frac{\tan A+\tan B}{1+\tan A\tan B} \)
The tangent formula behaves like the sine on top and like the cosine on the bottom, which is unsurprising once it is derived from their quotient. Deriving it once removes the need to recall two separate sign rules.
Prediction
You are expanding the tangent of a difference.
Predict first
What sign appears in the denominator?
Correct: A plus sign.
Why: The tangent's denominator sign is always opposite to the numerator's, so a difference inside gives a minus on top and a plus on the bottom. This mirrors the cosine's reversal, which makes sense since the denominator comes from the cosine formula.
Matching
The first product tells you which formula is in use.
Match the pairs
Why: The cosine formulas begin with cosine times cosine and the sine formulas with sine times cosine. Once the first term is right, the sign rule — opposite for the cosine, matching for the sine — settles the rest.
Step zero
You are asked for the exact sine of a non-special angle.
Discussion prompt
What do you look for first?
Hint: What makes the formula usable?
Answer:
Look for a way to write the angle as a sum or difference of two special angles — thirty, forty-five, sixty, ninety, and their multiples.
Fifteen is forty-five minus thirty. Seventy-five is forty-five plus thirty. A hundred and five is sixty plus forty-five. If no such decomposition exists, the identity cannot help and the value is not exactly expressible this way.
Only then choose sum or difference and write down the formula. Finding the decomposition is the whole creative step; everything after it is substitution and arithmetic.
Section
Section 3
Concept
Before this section only the special angles had exact values. Now every sum and difference of them does, which roughly triples the table.
Fifteen degrees is a twelfth of a half turn, which is why every multiple of it becomes accessible: each is a sum of thirties and forty-fives. Angles like ten or twenty degrees remain out of reach, and no elementary formula gives their exact values.
Figure (svg): A diagram showing fifteen degrees written as the difference of forty-five and thirty, both of which are special angles
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 806-813
Picture it
Two known angles combine into a new one.
Figure (svg): A diagram showing fifteen degrees written as the difference of forty-five and thirty, both of which are special angles
The decomposition is the only choice to be made. There are usually two ways to write a target angle, and either works — checking that both give the same answer is a free verification.
Worked example
Several decompositions exist, and all agree.
\[ \text{Find } \cos 105^\circ \text{ exactly.} \]
Choose a decomposition
Why: Sixty plus forty-five.
\[ 60 + 45 \]
Apply the sum formula
Why: The cosine's sign reverses.
\[ \cos 60 \cos 45 - \sin 60 \sin 45 \]
Substitute
Why: From the special triangles.
\[ (\frac{1}{2}) (r 2 / 2) - (r 3 / 2) (r 2 / 2) \]
Combine
Why: A common denominator of four.
\[ \frac{r 2 - r 6}{4} \]
Figure (svg): A diagram showing fifteen degrees written as the difference of forty-five and thirty, both of which are special angles
\[ \cos 105^\circ=\frac{\sqrt{2}-\sqrt{6}}{4} \]
Verify: check the sign is plausible
Why: A hundred and five degrees is in the second quadrant, so its cosine must be negative — and root two is smaller than root six, so the answer is negative. The numerical value is about negative 0.259, matching a calculator.
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 808-811
Sorting
It must be a sum or difference of special angles.
Sort into buckets
Sort each angle.
Worked example
The identities work with any two angles, not just special ones.
\[ \text{Given } \sin A=\tfrac{3}{5} \text{ in QI and } \cos B=\tfrac{5}{13} \text{ in QI, find } \sin(A+B). \]
Find the missing ratios
Why: By the Pythagorean identity.
\[ \cos A = \frac{4}{5}, \sin B = \frac{12}{13} \]
Apply the sum formula
Why: Signs match for the sine.
\[ \sin A \cos B + \cos A \sin B \]
Substitute
Why: All four values known.
\[ (\frac{3}{5}) (\frac{5}{13}) + (\frac{4}{5}) (\frac{12}{13}) \]
Combine
Why: A common denominator of 65.
\[ \frac{15}{65} + \frac{48}{65} \]
Figure (svg): The solution to Worked example use given values rather than special angles shown as a ladder of expressions, one row per legal move
\[ \sin(A+B)=\frac{63}{65} \]
Verify: check the value is legal
Why: The result is about 0.969, which is between negative one and one as any sine must be. And both angles being in the first quadrant with fairly large sines makes a sum close to a right angle plausible, so a sine near 1 is expected.
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 811-815
Trap
\[ \text{given } \sin A \text{ and } \cos B, \text{ apply the formula directly} \]
Substitute only the two given values
Why: The formula's other two slots are left empty or filled with the same numbers.
The expansion needs four values and only two were supplied.
The formula needs both functions of both angles — four values in total. Two are given and two must be found.
Find them with the Pythagorean identity, using the stated quadrant to fix each sign. Skipping the quadrant gives the right size and possibly the wrong sign.
Listing all four values before substituting makes the omission impossible to miss, and it is the standard first step in every problem of this type.
Faded example
A hundred and sixty-five degrees.
Fill in the blanks
165^\circ = 120^\circ + 45^\circ, \text2 ___
Why: A hundred and twenty degrees is a second-quadrant special angle with reference angle sixty, and forty-five is a first-quadrant one. Their sum reaches a hundred and sixty-five, so the sum formula gives its exact value.
Prediction
You are computing the cosine of a hundred and five degrees.
Predict first
What sign should the answer have?
Correct: Negative, since the angle is in the second quadrant.
Why: The cosine is negative for any angle between ninety and a hundred and eighty degrees, regardless of how the angle was decomposed. Predicting the sign before computing catches sign errors in the expansion immediately.
Explain it
The table of exact values just got bigger.
Discussion prompt
Explain to a classmate which angles are now exactly computable and which are not.
Hint: What is the smallest new angle?
Answer:
Every multiple of fifteen degrees is now reachable, because fifteen is forty-five minus thirty and everything else is built by adding thirties, forty-fives and sixties.
Angles that are not multiples of fifteen — ten, twenty, thirty-five — remain out of reach. No combination of the special angles produces them.
So the extension is real but bounded. A good explanation adds why: the identities only combine angles you already know, so the reachable set is whatever the special angles generate by addition, and that set is exactly the multiples of fifteen.
Section
Section 4
Concept
Substituting a right angle into the difference formula produces the cofunction identities: each function of an angle equals its co-function of the complement.
In a right triangle the two acute angles are complementary, and one angle's opposite side is the other's adjacent side. So the ratio one calls a sine the other calls a cosine, which is the geometric content of these identities.
Figure (svg): A right triangle with its two acute angles marked as complementary, showing that one angle's sine is the other's cosine
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 815-818
Picture it
The two acute angles of a right triangle share the same sides.
Figure (svg): A right triangle with its two acute angles marked as complementary, showing that one angle's sine is the other's cosine
The naming was never arbitrary. Every co-function is its partner evaluated at the complementary angle, which is where the prefix comes from.
Worked example
Put a right angle into the difference formula.
\[ \text{Show } \cos\bigl(\tfrac{\pi}{2}-\theta\bigr)=\sin\theta. \]
Apply the cosine difference formula
Why: With A a right angle.
\[ \cos(\frac{\pi}{2}) \cos + \sin(\frac{\pi}{2}) \sin \]
Substitute the known values
Why: Cosine of a right angle is zero, sine is one.
\[ 0 \times \cos + 1 \times \sin \]
Simplify
Why: The first term vanishes.
State the result
Why: The cofunction identity.
Figure (svg): A right triangle with its two acute angles marked as complementary, showing that one angle's sine is the other's cosine
\[ \cos\bigl(\tfrac{\pi}{2}-\theta\bigr)=\sin\theta \]
Verify: test at a convenient angle
Why: At thirty degrees the complement is sixty, and the cosine of sixty is one half — which is the sine of thirty. The identity holds there, and the derivation shows it holds for every angle since nothing about theta was assumed.
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 816-817
Prediction
The sine of twenty degrees.
Predict first
Which of these equals it?
Correct: The cosine of seventy degrees.
Why: Twenty and seventy are complementary, so the sine of one equals the cosine of the other. The cofunction identity always pairs an angle with what is left after subtracting it from a right angle.
Worked example
It converts between the two halves of the table.
\[ \text{Simplify } \frac{\sin 40^\circ}{\cos 50^\circ}. \]
Check the angles
Why: They add to ninety.
Apply the cofunction identity
Why: The cosine of fifty is the sine of forty.
\[ \cos 50 = \sin 40 \]
Substitute
Why: The two are equal.
\[ \sin 40 / \sin 40 \]
Simplify
Why: Anything nonzero over itself.
\[ 1 \]
Figure (svg): The solution to Worked example use a cofunction identity shown as a ladder of expressions, one row per legal move
\[ \frac{\sin 40^\circ}{\cos 50^\circ}=1 \]
Verify: check numerically
Why: Both the sine of forty and the cosine of fifty are about 0.643, so the quotient is 1. Spotting that two angles are complementary is often the whole trick in a problem that otherwise looks like it needs a calculator.
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 817-818
Error analysis
A student simplifies a quotient of two trigonometric values.
Annotate
On: \( \frac{\sin 40^\circ}{\cos 40^\circ}=1 \)
The cofunction rule has a precondition, and it is easy to apply it on autopilot when two trigonometric functions appear together. Adding the angles takes a second and confirms the identity applies.
Faded example
The tangent's partner.
Fill in the blanks
\tan\theta=\cot\bigl(\tfrac22}-\theta\bigr), \text___\theta \text___ \tfrac______}
Why: The complement is what remains after subtracting from a right angle, which is pi over two in radians. Every function pairs with its co-function under this substitution, which is where the prefix comes from.
Sorting
The two angles must be complementary.
Sort into buckets
Sort each pair.
Explain it to yourself
Cosine, cotangent, cosecant — all with the same prefix.
Discussion prompt
Explain what the co- prefix means and why.
Hint: What does the identity say?
Answer:
Co- is short for complementary. Each co-function is its partner evaluated at the complementary angle.
So the cosine of an angle is the sine of its complement, and the same relation holds for the cotangent and cosecant with their partners.
The right triangle makes it visible: the two acute angles are complementary, and one's opposite side is the other's adjacent side. The naming records a genuine relationship rather than being an arbitrary label, which is why every co-function pairs with exactly one ordinary function.
Section
Section 5
Concept
Verifications involving sums and differences follow one pattern: expand the compound argument, simplify the resulting products, and recognise what is left.
When one of the two angles is a multiple of a right angle, the expansion collapses immediately because one of the two products has a zero factor. That collapse is exactly how the cofunction and reduction identities are derived.
Figure (svg): Three cards giving the sum and difference identities for the cosine, the sine and the tangent, with their sign patterns marked
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 813-820
Picture it
Every verification here begins by using one of these three.
Figure (svg): Three cards giving the sum and difference identities for the cosine, the sine and the tangent, with their sign patterns marked
Expansion is almost always the first move, because it converts an unfamiliar compound argument into products of things already known.
Worked example
One angle is a straight angle, so the expansion collapses.
\[ \text{Verify } \sin(\pi-\theta)=\sin\theta. \]
Expand the left side
Why: The sine difference formula.
\[ \sin(\pi) \cos - \cos(\pi) \sin \]
Substitute the known values
Why: Sine of pi is zero, cosine is negative one.
\[ 0 - (-1) \sin \]
Simplify
Why: Two negatives.
Compare
Why: It matches the right side.
Figure (svg): The solution to Worked example verify a shift identity shown as a ladder of expressions, one row per legal move
\[ \sin(\pi-\theta)=\sin\theta \]
Verify: test at a convenient angle
Why: At pi over six the left side is the sine of five pi over six, which is one half, and the right side is the sine of pi over six, also one half. The identity is the reflection rule used in §6.3 to find second solutions, now derived rather than asserted.
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 814-816
Prediction
You expand the sine of a sum and the sine of a difference and add them.
Predict first
Which terms cancel?
Correct: The cosA sinB terms, since they have opposite signs.
Why: The sine's signs match the argument, so the sum expansion has a plus on the second product and the difference expansion has a minus. Adding annihilates them, leaving twice the first product — which is the sum-to-product formula in embryo.
Worked example
Expanding both compound arguments makes terms cancel.
\[ \text{Verify } \sin(A+B)+\sin(A-B)=2\sin A\cos B. \]
Expand the first term
Why: Signs match for the sine.
\[ \sin A \cos B + \cos A \sin B \]
Expand the second term
Why: A difference gives a minus.
\[ \sin A \cos B - \cos A \sin B \]
Add them
Why: The second products cancel.
\[ 2 \sin A \cos B \]
Compare
Why: It matches the right side.
Figure (svg): Three cards giving the sum and difference identities for the cosine, the sine and the tangent, with their sign patterns marked
\[ \sin(A+B)+\sin(A-B)=2\sin A\cos B \]
Verify: test at convenient angles
Why: At A sixty and B thirty degrees the left side is the sine of ninety plus the sine of thirty, which is 1 plus one half, or 1.5. The right side is twice root three over two times root three over two, which is 1.5 — matching. This identity is the seed of the sum-to-product formulas in §7.4.
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 816-820
Trap
\[ \sin(A+B)+\sin(A-B) \;\to\; \text{expand only the first} \]
Expand one term and leave the other
Why: The second compound argument is left as it stands.
No cancellation appears, and the verification stalls.
Expand every compound argument. The cancellation depends on both expansions being present so that the cross terms can meet.
Here the two second products differ only in sign, so adding kills them. Leaving one unexpanded hides that entirely.
Symmetric expressions usually need symmetric treatment. When two similar terms appear, expanding both is almost always the productive move.
Faded example
The cosine of a straight angle minus theta.
Fill in the blanks
\cos(\pi-\theta)=\cos\pi\cos\theta+\sin\pi\sin\theta=(-1)\cos\theta+0=-\cos\theta
Why: The cosine of a straight angle is negative one and its sine is zero, so the second product vanishes and the first picks up a sign. This derives the reflection rule used constantly when finding second solutions to equations.
Sorting
It collapses when one angle is a quadrantal angle.
Sort into buckets
Sort each expression.
Explain it
Some expansions reduce to one term immediately.
Discussion prompt
Explain to a classmate when and why that happens.
Hint: What is special about the quadrantal angles?
Answer:
It happens when one of the two angles is a quadrantal angle — a multiple of a right angle — because at those angles one of the sine and cosine is zero.
A zero factor kills one of the two products entirely, so what was a two-term expansion becomes a single term, often with a sign attached from the other factor being negative one.
That is how the cofunction identities and every reflection rule are derived. They are not separate facts to learn — each is one substitution into the sum or difference formula, which is worth pointing out because it cuts the list of things to remember by half.
Comparison
Fill the blanks from memory. Getting these three right is most of the section.
Comparison matrix
| cosine | sine | tangent | |
|---|---|---|---|
| first term | cosA cosB | sinA cosB | tanA over the sum |
| sign relative to the argument | opposite | matches | matches on top |
| denominator | none | none | 1 with the opposite sign |
| memory hook | the odd one out | the intuitive one | both behaviours at once |
The second row is the whole difficulty. The cosine opposes and the sine matches, and mixing them up is the error that shows up most often in this chapter's work.
Pattern
Five steps, and the first is the only one requiring judgement.
Step 5 catches almost every error in one line, since the quadrant fixes the sign independently of how the expansion went.
OpenStax Algebra and Trigonometry 2e, §9.2 Sum and Difference Identities §9.2
Check
The cosine's sign.
Check your understanding
What does the cosine of a sum expand to?
Answer: A
Why: The cosine's expansion carries the opposite sign to the argument, so a sum inside gives a difference of products. Setting the two angles equal confirms it: the result is cosine squared minus sine squared, which is correct for a doubled angle.
Check
Which angles are now reachable.
Check your understanding
Which angle can be given an exact value using these identities?
Answer: A
Why: Fifteen degrees is forty-five minus thirty, both special angles, so the difference formula applies. The reachable angles are exactly the multiples of fifteen degrees, since those are what the special angles generate by addition and subtraction.
Check
Cofunctions.
Check your understanding
The sine of thirty-five degrees equals which of these?
Answer: A
Why: Thirty-five and fifty-five are complementary, and the cofunction identity says a function of an angle equals its co-function of the complement. The check is simply that the two angles sum to ninety.
Real world
Combining two waves of the same frequency is a sum-of-angles problem.
Discussion prompt
Two speakers emit the same tone but one signal is delayed. Why do these identities describe what is heard?
Hint: A delay is a phase shift.
Answer:
A delay shifts the phase, so the second wave is the sine of the angle plus a constant. Adding the two waves means adding a sine and a shifted sine.
The sum formula expands the shifted one into two products, and collecting terms shows the total is a single wave of the same frequency with a new amplitude and phase.
Which is why two identical tones never produce a new pitch — only a louder or quieter version of the same one, depending on the delay. The identities predict exactly how loud, and the same computation governs interference in optics, antenna arrays and noise-cancelling headphones.
Commit first
State your confidence along with your answer.
Predict first
Why does the cosine of a sum use a minus sign?
Correct: It is verifiable by setting the two angles equal and testing.
Why: Putting B equal to A gives an expression for the cosine of a doubled angle, and testing at forty-five degrees shows the minus version gives zero while the plus version gives 1 for every angle. The sign is a derivable fact, not an arbitrary one, and the test takes a single line.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
A classmate writes that the sine of a sum is the sum of the sines. How do you convince them otherwise?
Hint: Pick two easy angles.
Answer:
Test it. At thirty degrees plus thirty degrees the left side is the sine of sixty, about 0.87, and the right side is one half plus one half, which is 1. They differ.
Then say why: the sine is a function, not a coefficient. Distribution is a property of multiplication over addition, and there is no multiplication here.
And offer the replacement: the real formula gives two products, each mixing one function of each angle. A good explanation ends by recommending the numerical test as a habit — any suspected shortcut can be checked in ten seconds, and this one only survives because it never is.
Exit ticket
One honest answer, so the next lesson can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The first is where nearly all the errors live, since the cosine's reversal is easy to lose. The second is the most concrete payoff and the easiest to check against a calculator.
Connect it up
One page, drawn from memory, is worth more than rereading the section.
Draw it
Write the three sum and difference formulas, marking on each whether its signs match or oppose the argument. Beside them, show fifteen degrees as forty-five minus thirty and carry out the cosine computation to its exact value. Underneath, derive one cofunction identity by substituting a right angle into the difference formula.
If your sign annotations are right and your cofunction derivation is a substitution rather than a recalled fact, the section's two hardest points are both on the page.
Recap
Five things, and the second is the one to check every time.
| if you remember one thing | it should be this |
|---|---|
| about the cosine | its signs are opposite to the argument's |
| about the sine | its signs match, which is why the cosine's reversal surprises |
| about exact values | multiples of fifteen degrees are now all reachable |
| about cofunctions | co- means complementary, and it is one substitution away |
Section 7.3 sets the two angles equal, turning these formulas into the double-angle identities — and then runs them backwards to get the half-angle formulas, which reach angles even smaller than fifteen degrees.
OpenStax, Precalculus, §7.2 Sum and Difference Identities §7.2, pp. 803-820 — everything on these slides traces back here
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