The first half of the identity chapter, all of it descending from one distance-formula computation. Covers the even-odd identities and their proof by reflection across the x-axis, the difference identity for cosine proved from equal chords, the cofunction identities that fall out of it and explain the prefix co, the sum and difference identities for sine derived through the cofunction identities, and the tangent formula obtained by dividing. Includes finding exact values of non-special angles by decomposing them into special ones.
Subject: Trigonometry · 65 slides · symbolic lesson
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Title
Trigonometry · Chapter 10 — Foundations of Trigonometry
§10.4 Trigonometric Identities, pp. 770-776
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 770-776 — the pages these objectives are drawn from
Warm-up
A tempting shortcut is about to be ruled out, and it is worth ruling out by experiment rather than by assertion.
Discussion prompt
Is the cosine of a sum equal to the sum of the cosines? Test it with alpha equal to pi over three and beta equal to pi over six, where you know all the values involved.
Hint: Compute both sides separately and compare. Do not reason about it — just evaluate.
Answer:
\[ \cos\left(\tfrac{\pi}{3} + \tfrac{\pi}{6}\right) = \cos\left(\tfrac{\pi}{2}\right) = 0 \]
\[ \cos\left(\tfrac{\pi}{3}\right) + \cos\left(\tfrac{\pi}{6}\right) = \tfrac{1}{2} + \tfrac{\sqrt{3}}{2} \approx 1.366 \]
Zero against 1.366: emphatically not equal. Cosine does not distribute over addition, and neither does any other circular function. Whatever the cosine of a sum is, it is not the sum of the cosines — and this lesson exists to say what it actually is.
Concept
The whole of this section rests on a single geometric observation and one application of the distance formula: two central angles of equal measure cut off chords of equal length. Writing that equality out algebraically produces the difference identity for cosine, and every other identity in the section is derived from it.
\[ \cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta) \]
There is genuinely one theorem here and a great many consequences. Knowing which is which makes the section far less to remember than it first appears.
Figure (svg): Two chords on the unit circle of equal length, one joining the points at alpha and beta and the other joining the point at alpha minus beta to the point one comma zero, showing the congruent angles that make the chords equal
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 771-772
Section
Section 1
Concept
Replacing an angle by its negative reflects the corresponding point across the x-axis. That reflection keeps the first coordinate and negates the second, so the cosine is unchanged and the sine flips sign.
even and odd functions — A function is even when negating the input leaves the output unchanged, and odd when negating the input negates the output. Cosine and secant are even; sine, cosecant, tangent and cotangent are odd.
\[ \cos(-\theta) = \cos(\theta), \qquad \sin(-\theta) = -\sin(\theta) \]
The other four follow immediately, because each is built from these two. Secant inherits evenness from cosine; cosecant inherits oddness from sine; and tangent and cotangent are odd because each is a quotient with exactly one odd factor.
Figure (svg): An angle theta and its negative drawn on the same unit circle, with their two points shown as reflections of each other across the x-axis so the cosines agree and the sines are opposite
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 770-771
Picture it
The split is not even-handed, and the reason is that only two of the six read the horizontal coordinate.
Figure (svg): Two columns separating the even circular functions from the odd ones
Cosine and secant are the only even circular functions. If you can remember that, every other even-odd question answers itself.
Worked example
Use the even-odd identities to move the minus sign out, then evaluate a familiar angle.
\[ \text{Find } \cos\left(-\frac{\pi}{6}\right) \text{ and } \sin\left(-\frac{\pi}{6}\right) \text{ and } \tan\left(-\frac{\pi}{6}\right). \]
Apply the even identity to the cosine
Why: Cosine is even, so the minus sign simply disappears.
\[ \cos(-\frac{\pi}{6}) = \cos(\frac{\pi}{6}) \]
Evaluate the cosine
Why: The cosine of pi over six is root three over two.
\[ = \sqrt{3} / 2 \]
Apply the odd identity to the sine
Why: Sine is odd, so the minus sign comes out front.
\[ \sin(-\frac{\pi}{6}) = -\sin(\frac{\pi}{6}) = -\frac{1}{2} \]
Apply the odd identity to the tangent
Why: Tangent is odd too.
\[ \tan(-\frac{\pi}{6}) = -\sqrt{3} / 3 \]
Figure (svg): The solution to Worked example evaluate at a negative angle shown as a ladder of expressions, one row per legal move
\[ \cos\left(-\tfrac{\pi}{6}\right) = \tfrac{\sqrt{3}}{2}, \quad \sin\left(-\tfrac{\pi}{6}\right) = -\tfrac{1}{2}, \quad \tan\left(-\tfrac{\pi}{6}\right) = -\tfrac{\sqrt{3}}{3} \]
Verify: check by the quadrant instead
Why: Negative pi over six is in quadrant four, where the cosine is positive and the sine is negative, with reference angle pi over six. Both routes give the same three values, which is what the identities were asserting.
Sorting
Ask which coordinate the function reads, and what reflection across the x-axis does to it.
Sort into buckets
Sort each function.
Worked example
The real use of these identities is simplification, not evaluation.
\[ \text{Simplify } \frac{\sin(-\theta)\cos(-\theta)}{\tan(-\theta)}. \]
Apply the identities term by term
Why: Sine and tangent are odd; cosine is even.
\[ = (-\sin) (\cos) / (-\tan) \]
Cancel the two minus signs
Why: A negative over a negative is positive.
\[ = \sin \cos / \tan \]
Replace the tangent by its quotient form
Why: Dividing by sine over cosine is multiplying by cosine over sine.
\[ = \sin \cos(\cos / \sin) \]
Cancel the sine
Why: Valid wherever the tangent was defined and nonzero.
\[ = \cos ^{2} \theta \]
Figure (svg): The solution to Worked example simplify using even-odd shown as a ladder of expressions, one row per legal move
\[ \frac{\sin(-\theta)\cos(-\theta)}{\tan(-\theta)} = \cos^2(\theta) \]
Verify: test at pi over three
Why: The original is sine of negative pi over three, which is negative root three over two, times cosine of negative pi over three, which is one half, over tangent of negative pi over three, which is negative root three. That gives negative root three over four divided by negative root three, which is one quarter. And cosine squared of pi over three is one quarter. They agree.
Trap
\[ \cos\left(-\tfrac{\pi}{3}\right) = -\tfrac{1}{2} \]
Carry the minus sign from the angle into the answer
Why: The minus sign is visible in the input and it feels like it must go somewhere.
But cosine is even, so it goes nowhere at all. Negative pi over three is in quadrant four, where the cosine is positive.
\[ \cos\left(-\tfrac{\pi}{3}\right) = \cos\left(\tfrac{\pi}{3}\right) = \tfrac{1}{2} \]
Ask which function it is before deciding what the minus sign does
Why: For cosine and secant it disappears; for the other four it comes out front.
The check that never fails: draw the angle. Negative pi over three sweeps clockwise into quadrant four, where cosine is positive and sine is negative — which is exactly what even for cosine and odd for sine predict.
Fill the middle
Show that the tangent is odd, using the quotient identity and the even-odd identities for cosine and sine.
Fill in the blanks
\tan(-\theta) = \frac-sin thetacos theta = \frac-tan theta}___} = ___
Why: The sine in the numerator is odd, so it contributes a minus sign; the cosine in the denominator is even, so it contributes nothing. One minus sign survives, which makes the whole quotient odd. The same argument on cotangent gives the same conclusion, since it is the reciprocal.
Prediction
The cosine of some angle is 0.6.
Predict first
What is the cosine of the negative of that angle?
Correct: 0.6.
Why: Cosine is even, so negating the input leaves the output unchanged. Geometrically the two points are reflections across the x-axis, and reflection across a horizontal line does not move anything horizontally. Note that the sine would have flipped sign under the same operation, so the two functions genuinely behave differently and the answer would be different if the question had asked about sine.
Socratic
The book's proof begins by replacing theta with a coterminal angle between zero and two pi.
Discussion prompt
Why is that reduction needed at all, and why does it not weaken the result?
Hint: What does the picture with the two reflected points actually assume about the angle?
Answer:
The picture is drawn for an angle inside one revolution, so the argument as drawn only obviously applies to those. The reduction makes the picture legitimate for every angle.
It does not weaken the result because coterminal angles share their cosine and sine exactly — they share the point. So proving it for the reduced angle proves it for the original, and this pattern recurs throughout Section 10.4: reduce to a case the picture covers, then transfer back by coterminality.
Section
Section 2
Concept
Two central angles of equal measure subtend chords of equal length. Applying the distance formula to a pair of such chords and simplifying with the Pythagorean identity gives the difference formula for cosine, and the sum formula follows from it using the even-odd identities.
Theorem 10.13 — The sum and difference identities for cosine: the cosine of a sum is the product of the cosines minus the product of the sines, and the cosine of a difference is the product of the cosines plus the product of the sines.
\[ \cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta \]
Two features are worth noting because they are the two things people get wrong: the terms are matched — cosine with cosine, sine with sine — and the sign in the answer is the opposite of the sign in the bracket.
Figure (svg): Two chords on the unit circle of equal length, one joining the points at alpha and beta and the other joining the point at alpha minus beta to the point one comma zero, showing the congruent angles that make the chords equal
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 771-772
Picture it
Cosine is the odd one out in both respects, which is why it is worth learning first and separately.
Figure (svg): The sum and difference identities for cosine, sine and tangent stacked as three cards, each showing the combined plus-or-minus form
For sine the terms are mixed and the sign agrees with the bracket; for cosine the terms are matched and the sign is reversed. Those two sentences are the whole of what has to be remembered.
Worked example
Example 10.4.1, part 1. Fifteen degrees is not special, but it is the difference of two angles that are.
\[ \text{Find the exact value of } \cos(15^\circ). \]
Write 15 as a difference of special angles
Why: Forty-five minus thirty is fifteen, and both are special.
\[ 15 = 45 - 30 \]
Apply the difference identity
Why: Matched terms, and the sign flips to a plus.
\[ \cos 45 \cos 30 + \sin 45 \sin 30 \]
Substitute the four known values
Why: Root two over two, root three over two, root two over two, one half.
\[ (\sqrt{2} / 2) (\sqrt{3} / 2) + (\sqrt{2} / 2) (\frac{1}{2}) \]
Combine over a common denominator
Why: Root six over four plus root two over four.
\[ \frac{\sqrt{6} + \sqrt{2}}{4} \]
Figure (svg): The solution to Worked example the exact cosine of 15 degrees shown as a ladder of expressions, one row per legal move
\[ \cos(15^\circ) = \frac{\sqrt{6} + \sqrt{2}}{4} \]
Verify: check the decimal
Why: Root six is about 2.449 and root two about 1.414, so the answer is about 3.863 over 4, which is about 0.966. A calculator gives the cosine of 15 degrees as 0.9659. And the size is right: 15 degrees is a small angle, so its cosine should be close to 1.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 772-772
Elimination
Expand the cosine of alpha plus beta.
Eliminate the wrong options
Which is the correct expansion?
Survives elimination: B
Why: For cosine the terms are matched — cosine times cosine, sine times sine — and the sign in the answer is the opposite of the sign in the bracket. So a sum inside gives a minus outside. Both features are what distinguish the cosine formula from the sine formula, and both are what the distractors get wrong.
Worked example
Example 10.4.1, part 2. This one identity is the seed of the whole cofunction family.
\[ \text{Verify that } \cos\left(\frac{\pi}{2} - \theta\right) = \sin(\theta). \]
Apply the difference identity with alpha equal to pi over two
Why: Matched terms, with the sign flipping to a plus.
\[ \cos(\frac{\pi}{2}) \cos(\theta) + \sin(\frac{\pi}{2}) \sin(\theta) \]
Substitute the two quadrantal values
Why: The cosine of pi over two is 0 and the sine is 1.
\[ (0) \cos(\theta) + (1) \sin(\theta) \]
Simplify
Why: The first term vanishes entirely.
\[ = \sin(\theta) \]
Note what has been proved
Why: This holds for every angle, not just acute ones, which the triangle picture could never have shown.
Figure (svg): The solution to Worked example verify a cofunction identity shown as a ladder of expressions, one row per legal move
\[ \cos\left(\frac{\pi}{2} - \theta\right) = \sin(\theta) \]
Verify: test at an obtuse angle
Why: Take theta equal to two pi over three. The left side is the cosine of pi over two minus two pi over three, which is the cosine of negative pi over six, which is root three over two. The right side is the sine of two pi over three, also root three over two. It holds outside the range any triangle could reach.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 773-773
Error analysis
A student expands the cosine of a difference.
Annotate
On: \( \cos\left(\frac{\pi}{2} - \theta\right) = \cos\left(\frac{\pi}{2}\right) - \cos(\theta) = -\cos(\theta) \)
Substituting a single convenient value is the fastest way to test any proposed identity. One counterexample settles it, and choosing a quadrantal angle usually makes the arithmetic trivial.
Faded example
Find the exact cosine of 75 degrees.
Fill in the blanks
\cos(75^\circ) = \cos(45^\circ + 30^\circ) = \cos45\cos30 - \sin45\sin30 = \frac(sqrt6 - sqrt2)/4}___ ___ \frac___}___ = ___
Why: Seventy-five degrees is 45 plus 30, so the sum formula applies and the sign in the answer is a minus. The two products are root six over four and root two over four, giving root six minus root two over four, about 0.259. That is small, as it should be for an angle close to 90 degrees where the cosine approaches zero. Compare with the cosine of 15 degrees, which had the same two terms added instead.
Prediction
You are about to expand the cosine of alpha minus beta.
Predict first
What sign will appear between the two terms?
Correct: Plus, opposite the bracket.
Why: The cosine formulas reverse the sign: a difference inside produces a sum outside, and a sum inside produces a difference. This is the single most-missed detail in the section, and the mnemonic is that cosine is contrary. Sine, by contrast, keeps the sign the bracket had, which is why the two formulas must be kept clearly apart.
Explain it to yourself
The proof used the fact that equal angles subtend equal chords.
Discussion prompt
Explain why that geometric fact is enough to produce an algebraic identity. What turns a statement about lengths into a statement about cosines?
Hint: How do you compute the length of a chord when you know the endpoints' coordinates?
Answer:
The endpoints of both chords are points on the unit circle, so their coordinates are cosines and sines by definition. The distance formula turns each chord's length into an expression in those coordinates.
Setting the two expressions equal therefore sets two trigonometric expressions equal, and the Pythagorean identity collapses most of the terms. Geometry supplies the equality; the distance formula supplies the algebra. That combination is how almost every classical identity was originally found, and it is why the Pythagorean identity keeps appearing in the middle of the proofs.
Section
Section 3
Concept
The identity just verified says the cosine of an angle is the sine of its complement. Applying it twice, and then to the other four functions, gives a complete family: each function of an angle equals its cofunction of the complement.
cofunction identities — Each circular function of an angle equals its cofunction of the complementary angle. Cosine pairs with sine, cotangent with tangent, and cosecant with secant.
\[ \cos\left(\tfrac{\pi}{2} - \theta\right) = \sin\theta, \quad \tan\left(\tfrac{\pi}{2} - \theta\right) = \cot\theta, \quad \sec\left(\tfrac{\pi}{2} - \theta\right) = \csc\theta \]
This finally explains the naming. Cosine is short for complement's sine, cotangent for complement's tangent, cosecant for complement's secant — and it is why the reciprocal pairings looked crossed back in Lesson 10.3a.
Figure (svg): A right triangle with its two acute angles marked as theta and its complement, showing that the side opposite one is adjacent to the other so the cofunction identities hold
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 773-773
Picture it
For an acute angle the identity needs no algebra at all. The two acute angles simply trade roles.
Figure (svg): A right triangle with its two acute angles marked as theta and its complement, showing that the side opposite one is adjacent to the other so the cofunction identities hold
The algebraic proof matters because it extends the result to every angle, including obtuse and negative ones where no triangle exists. The triangle explains why it is true; the identity says how far it reaches.
Worked example
Apply the cosine version to the complement and watch it fold back on itself.
\[ \text{Show that } \sin\left(\frac{\pi}{2} - \theta\right) = \cos(\theta). \]
Start from the identity already proved
Why: The cosine of the complement is the sine.
\[ \cos(\frac{\pi}{2} - x) = \sin(x) \]
Apply it with x equal to pi over two minus theta
Why: Substituting into the known identity.
\[ \sin(\frac{\pi}{2} - \theta) = \cos(\frac{\pi}{2} - (\frac{\pi}{2} - \theta)) \]
Simplify the inner expression
Why: The two halves of pi cancel and the double negative on theta resolves.
\[ \frac{\pi}{2} - \frac{\pi}{2} + \theta = \theta \]
Read off the result
Why: The complement of the complement is the angle itself.
\[ = \cos(\theta) \]
Figure (svg): The solution to Worked example derive the sine cofunction identity shown as a ladder of expressions, one row per legal move
\[ \sin\left(\frac{\pi}{2} - \theta\right) = \cos(\theta) \]
Verify: test at pi over six
Why: The left side is the sine of pi over two minus pi over six, which is the sine of pi over three, root three over two. The right side is the cosine of pi over six, also root three over two.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 773-773
Matching
In each case the two angles are complementary.
Match the pairs
Why: In each pair the two angles sum to 90 degrees and the function changes to its cofunction. Sine becomes cosine, tangent becomes cotangent, secant becomes cosecant, and each of those swaps runs in both directions. Checking one numerically: the secant of 80 degrees is about 5.759 and the cosecant of 10 degrees is also about 5.759.
Worked example
Recognising a complement inside an expression is the whole skill here.
\[ \text{Simplify } \frac{\sin(40^\circ)}{\cos(50^\circ)} \text{ and } \tan(20^\circ)\tan(70^\circ). \]
Notice the first pair are complementary
Why: Forty and fifty add to ninety.
\[ 50 = 90 - 40 \]
Rewrite the denominator as a cofunction
Why: The cosine of the complement of 40 is the sine of 40.
\[ \cos 50 = \sin 40 \]
Conclude for the first
Why: A quantity divided by itself is 1.
\[ = 1 \]
Handle the second the same way
Why: Twenty and seventy are complementary, so the tangent of 70 is the cotangent of 20, which is the reciprocal of the tangent of 20.
\[ \tan 20 \cot 20 = 1 \]
Figure (svg): The solution to Worked example simplify using a cofunction shown as a ladder of expressions, one row per legal move
\[ \frac{\sin(40^\circ)}{\cos(50^\circ)} = 1 \qquad \tan(20^\circ)\tan(70^\circ) = 1 \]
Verify: check numerically
Why: The sine of 40 degrees and the cosine of 50 degrees are both about 0.6428, so their quotient is 1. The tangent of 20 degrees is about 0.3640 and the tangent of 70 degrees about 2.7475, and their product is 1.0000. Both check.
Trap
\[ \sin(40^\circ) = \cos(140^\circ) \]
Pair 40 with 140 because they are related through 180
Why: Supplementary pairs do appear elsewhere in trigonometry, so the habit of looking for a sum of 180 is real.
But cofunction identities are about the complement, which sums to 90, not the supplement, which sums to 180. The cosine of 140 degrees is negative, while the sine of 40 degrees is positive, so the two cannot be equal.
\[ \sin(40^\circ) = \cos(50^\circ) \]
Pair the angle with its complement, summing to 90 degrees
Why: Cofunction means complement's function, and the complement is what is left of a right angle.
The supplement relationship is real but it is a different one: the sine of an angle equals the sine of its supplement, with no change of function. That is why the sine of 40 equals the sine of 140, and it is what makes the Law of Sines ambiguous in Lesson 11.2. Two genuine relationships, easily confused, and worth keeping straight by asking whether the total is 90 or 180.
Faded example
Simplify the expression cosine of 18 degrees over sine of 72 degrees.
Fill in the blanks
72^\circ = 90^\circ - 18 deg \;\Longrightarrow\; \sin(72^\circ) = \cos(1) \;\Longrightarrow\; \frac______ = ___
Why: Eighteen and seventy-two are complementary, so the sine of 72 equals the cosine of 18 by the cofunction identity. The expression is then a quantity over itself, which is 1. Recognising a complementary pair inside an expression is what turns a calculator problem into a one-line answer, and pairs summing to 90 are worth spotting on sight.
Two truths and a lie
Rule out the statements that are true. The survivor is the false one.
Eliminate the wrong options
One of these statements about cofunction identities is wrong.
Survives elimination: B
Why: Statement B is false: the cofunction identities are about the complement, summing to a right angle, not the supplement summing to a straight angle. There is a genuine supplement relationship — the sine of an angle equals the sine of its supplement — but it involves no change of function, which is exactly what distinguishes it from a cofunction identity.
Edge cases
The cofunction identity for tangent says the tangent of the complement is the cotangent.
Discussion prompt
What happens when theta is zero? Check both sides and say what the identity is claiming there.
Hint: Evaluate each side separately before comparing.
Answer:
\[ \tan\left(\tfrac{\pi}{2} - 0\right) = \tan\left(\tfrac{\pi}{2}\right) \quad\text{undefined} \]
\[ \cot(0) = \frac{\cos 0}{\sin 0} = \frac{1}{0} \quad\text{undefined} \]
Both sides are undefined, so the identity is not violated — it simply has nothing to say at this angle, which is why the theorem is stated for all applicable angles.
This is the right way to read every restriction in the chapter. An identity with a restriction is not a weaker claim about a smaller set of angles so much as a claim that goes silent exactly where one of its terms ceases to exist. Notice too that the two sides fail together, which is not an accident: they are the same quantity.
Section
Section 4
Concept
The sine formulas are derived from the cosine ones by routing through a cofunction identity. The result differs from the cosine version in both of the ways that matter: the terms are mixed rather than matched, and the sign agrees with the bracket rather than reversing it.
\[ \sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta \]
Comparing the two families side by side is the fastest way to keep them apart, because each is memorable exactly by how it differs from the other.
Figure (svg): The sum and difference identities for cosine, sine and tangent stacked as three cards, each showing the combined plus-or-minus form
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 773-774
Picture it
The only creative step in these problems is deciding which two special angles to combine.
Figure (svg): A table showing how non-special angles are written as sums or differences of special ones, with fifteen degrees as forty-five minus thirty and nineteen pi over twelve as four pi over three plus pi over four
Read the denominator. Twelfths come from thirds and quarters, because one third plus one quarter is seven twelfths and one third minus one quarter is one twelfth.
Worked example
Example 10.4.2, part 1. The denominator of 12 tells you which angles to combine.
\[ \text{Find the exact value of } \sin\left(\frac{19\pi}{12}\right). \]
Decompose the angle
Why: A denominator of 12 suggests combining thirds and quarters; four pi over three plus pi over four is nineteen pi over twelve.
\[ 19 \pi / 12 = 4 \pi / 3 + \frac{\pi}{4} \]
Apply the sum identity for sine
Why: Mixed terms, and the sign stays a plus.
\[ \sin(4 \pi / 3) \cos(\frac{\pi}{4}) + \cos(4 \pi / 3) \sin(\frac{\pi}{4}) \]
Substitute the four values
Why: Four pi over three is in quadrant three with reference angle pi over three.
\[ (-\sqrt{3} / 2) (\sqrt{2} / 2) + (-\frac{1}{2}) (\sqrt{2} / 2) \]
Combine over a common denominator
Why: Negative root six over four minus root two over four.
\[ \frac{-\sqrt{6} - \sqrt{2}}{4} \]
Figure (svg): The solution to Worked example the exact sine of nineteen pi over twelve shown as a ladder of expressions, one row per legal move
\[ \sin\left(\frac{19\pi}{12}\right) = \frac{-\sqrt{6} - \sqrt{2}}{4} \]
Verify: check the sign and size
Why: Nineteen pi over twelve is nineteen twenty-fourths of a revolution, which is in quadrant four where the sine is negative — and the answer is negative. Numerically it is about negative 0.966, and the angle is close to three pi over two where the sine is negative one, so a value near negative one is right.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 774-774
Comparison
Fill the blanks from memory. Each formula is memorable by how it differs from its neighbour.
Comparison matrix
| Cosine | Sine | |
|---|---|---|
| Terms are | matched | mixed |
| Sign in the answer | opposite the bracket | same as the bracket |
| Sum expansion | cos cos - sin sin | sin cos + cos sin |
| Difference expansion | cos cos + sin sin | sin cos - cos sin |
Two questions distinguish all four formulas: are the terms matched or mixed, and does the sign keep or reverse. Cosine reverses and matches; sine keeps and mixes.
Worked example
Example 10.4.2, part 2. Neither angle is named; both must be reconstructed first.
\[ \alpha \text{ in QII with } \sin\alpha = \tfrac{5}{13}; \; \beta \text{ in QIII with } \tan\beta = 2. \text{ Find } \sin(\alpha - \beta). \]
Find the cosine of alpha
Why: Use the Pythagorean identity; quadrant two makes it negative.
\[ \cos \alpha = -\frac{12}{13} \]
Find the secant of beta from its tangent
Why: One plus tangent squared is secant squared, so secant squared is 5; quadrant three makes the secant negative.
\[ \sec \beta = -\sqrt{5}, \cos \beta = -\sqrt{5} / 5 \]
Find the sine of beta
Why: The tangent times the cosine gives the sine.
\[ \sin \beta = 2(-\sqrt{5} / 5) = -2 \sqrt{5} / 5 \]
Apply the difference identity for sine
Why: Mixed terms, sign stays a minus.
\[ \sin a \cos b - \cos a \sin b \]
Figure (svg): The solution to Worked example angles given by values rather than measures shown as a ladder of expressions, one row per legal move
\[ \sin(\alpha - \beta) = \left(\tfrac{5}{13}\right)\left(-\tfrac{\sqrt{5}}{5}\right) - \left(-\tfrac{12}{13}\right)\left(-\tfrac{2\sqrt{5}}{5}\right) = -\frac{29\sqrt{5}}{65} \]
Verify: check the size is legitimate
Why: Twenty-nine root five over sixty-five is about 0.998, which is just inside the interval from negative one to one — so it is a possible sine. A value even slightly outside would have signalled an arithmetic error, and this one is close enough to the boundary to be worth checking.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 774-774
Error analysis
A student expands the sine of a sum.
Annotate
On: \( \sin(\alpha + \beta) = \sin\alpha\sin\beta + \cos\alpha\cos\beta \)
Matched terms belong to cosine and mixed terms belong to sine. A quick way to remember it: the cosine formula never mixes the two functions within a term, and the sine formula always does.
Faded example
Find the exact sine of 75 degrees.
Fill in the blanks
\sin(75^\circ) = \sin(45^\circ + 30^\circ) = \sin45\cos30 + \cos45\sin30 = \frac(sqrt6 + sqrt2)/4}___ ___ \frac___}___ = ___
Why: For sine the sign in the answer matches the sign in the bracket, so a sum gives a plus. The result, root six plus root two over four, is about 0.966. Compare with the cosine of 75 degrees, which came out as the same two terms subtracted — and note that this value equals the cosine of 15 degrees, exactly as the cofunction identity requires.
Ranking
Derive the sum formula for sine from the cosine formulas.
Put in order
Why: The derivation is a round trip through the cofunction identity. Converting the sine into a cosine of a complement makes the cosine difference formula applicable; regrouping is what puts it into the shape that formula expects; expanding produces terms that are themselves cofunctions; and converting those back gives the mixed-term sine formula. Every step is an identity already proved, which is why nothing new had to be assumed.
Missing information
A problem says: alpha has sine three fifths and beta has cosine five thirteenths. Find the sine of alpha plus beta.
Discussion prompt
Work out what you can and say precisely what is missing. How many possible answers are there as the problem stands?
Hint: The sum formula needs four values. How many are determined?
Answer:
The formula needs the sine and cosine of both angles. The sine of alpha and the cosine of beta are given, but the cosine of alpha and the sine of beta are each determined only up to sign by the Pythagorean identity: the cosine of alpha is plus or minus four fifths, and the sine of beta is plus or minus twelve thirteenths.
What is missing is the quadrant of each angle, or equivalently the sign of each missing value. With two independent sign choices there are four possible answers, and the problem as stated does not choose between them.
This is why every textbook problem of this shape names both quadrants. It is not decoration — without it the question genuinely has four answers, and any single one of them would be an incomplete response.
Section
Section 5
Concept
The tangent formula is not proved from scratch. It is the quotient of the sine and cosine formulas, with numerator and denominator divided through by the product of the two cosines so that only tangents remain.
\[ \tan(\alpha \pm \beta) = \frac{\tan\alpha \pm \tan\beta}{1 \mp \tan\alpha\tan\beta} \]
The signs follow the same pattern as before: the numerator agrees with the bracket, and the denominator reverses it. Since the denominator's sign comes from the cosine formula, that reversal is inherited rather than arbitrary.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 775-775
Picture it
This is the complete statement of Theorem 10.16, which the rest of the chapter uses constantly.
Figure (svg): The sum and difference identities for cosine, sine and tangent stacked as three cards, each showing the combined plus-or-minus form
The top sign throughout goes with a sum and the bottom sign with a difference. Reading the tangent line carefully: the numerator takes the bracket's sign, the denominator takes the opposite.
Worked example
Example 10.4.2, part 3. Two divisions and the tangents appear.
\[ \text{Derive a formula for } \tan(\alpha + \beta) \text{ in terms of } \tan\alpha \text{ and } \tan\beta. \]
Write the tangent as a quotient and expand both parts
Why: Use the sine and cosine sum formulas.
\[ \frac{\sin a \cos b + \cos a \sin b}{\cos a \cos b - \sin a \sin b} \]
Divide numerator and denominator by the product of the cosines
Why: Multiplying by a form of 1, chosen to turn every term into a tangent.
Simplify the numerator
Why: Each term loses one cosine and becomes a tangent.
\[ \tan a + \tan b \]
Simplify the denominator
Why: The first term becomes 1 and the second becomes a product of tangents.
\[ 1 - \tan a \tan b \]
Figure (svg): The solution to Worked example derive the tangent sum formula shown as a ladder of expressions, one row per legal move
\[ \tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} \]
Verify: test at two known angles
Why: Take alpha and beta both equal to pi over four, where each tangent is 1. The formula gives 1 plus 1 over 1 minus 1, which is 2 over 0 — undefined. And the tangent of pi over two, the actual sum, is indeed undefined. The formula's own denominator vanishes exactly where it should.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 775-775
Fill the middle
Find the exact tangent of 15 degrees.
Fill in the blanks
\tan(45^\circ - 30^\circ) = \frac+/3}2 - sqrt3} \sqrt___/3} = \frac___}___} \sqrt___} = ___
Why: For a difference the numerator takes a minus and the denominator takes a plus. Clearing the inner fractions by multiplying through by 3 and then rationalising gives 2 minus root three, about 0.268. That is small, as it should be for a shallow angle, and it is the reciprocal of the tangent of 75 degrees — which the cofunction identity requires, since 15 and 75 are complementary.
Worked example
Once derived, it is a straight substitution.
\[ \text{Find the exact value of } \tan(75^\circ). \]
Decompose the angle
Why: Forty-five plus thirty is seventy-five.
\[ 75 = 45 + 30 \]
Substitute the two tangents
Why: The tangent of 45 is 1 and the tangent of 30 is root three over three.
\[ \frac{1 + \sqrt{3} / 3}{1 - \sqrt{3} / 3} \]
Clear the inner fractions
Why: Multiply top and bottom by 3.
\[ \frac{3 + \sqrt{3}}{3 - \sqrt{3}} \]
Rationalise the denominator
Why: Multiply top and bottom by 3 plus root three.
\[ = 2 + \sqrt{3} \]
Figure (svg): The solution to Worked example use the tangent formula shown as a ladder of expressions, one row per legal move
\[ \tan(75^\circ) = \frac{3 + \sqrt{3}}{3 - \sqrt{3}} = 2 + \sqrt{3} \approx 3.732 \]
Verify: cross-check against the sine and cosine
Why: The sine of 75 degrees is root six plus root two over four and the cosine is root six minus root two over four. Their quotient is root six plus root two over root six minus root two, which rationalises to 2 plus root three. The two routes agree.
Trap
\[ \tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 - \tan\alpha\tan\beta} \]
Use the same sign in both numerator and denominator
Why: Having correctly used a minus in the numerator for a difference, the same sign gets carried below the line.
The denominator came from the cosine formula, which reverses signs. For a difference the cosine formula gives a plus, so the denominator must have a plus.
\[ \tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta} \]
Let the numerator follow the bracket and the denominator oppose it
Why: The numerator inherits from the sine formula and the denominator from the cosine formula, and those two behave oppositely.
A free check: put alpha equal to beta in the difference formula. The numerator becomes zero, so the whole expression should be zero — and the tangent of zero is indeed zero. That works with either sign in the denominator, so try instead beta equal to zero: the formula must collapse to the tangent of alpha, and it does, since the denominator becomes 1 either way. The genuinely discriminating test is the one in the worked example, where the denominator has to vanish at 45 plus 45.
Prediction
You apply the tangent sum formula with alpha equal to 60 degrees and beta equal to 30 degrees.
Predict first
What will happen?
Correct: The denominator becomes zero.
Why: The tangent of 60 degrees is root three and the tangent of 30 is root three over three, and their product is exactly 1, so the denominator one minus that product is zero. That is not a failure of the formula but a correct report: the sum is 90 degrees, and the tangent of 90 degrees is undefined. The formula's denominator vanishes precisely when the sum is an odd multiple of pi over two, which is exactly where the answer should not exist.
Sorting
Reading the problem for what is given and what is wanted decides it.
Sort into buckets
Sort each task by the identity family it needs.
Counterexample
A student proposes: the tangent sum formula works for any two angles whatsoever.
Discussion prompt
Find two angles for which it fails, and say precisely what goes wrong. Is the failure a defect of the formula?
Hint: Try making one of the angles pi over two.
Answer:
Take alpha equal to pi over two and beta equal to pi over four. The formula requires the tangent of alpha, which does not exist, so the right-hand side cannot even be written down.
\[ \tan\left(\tfrac{\pi}{2} + \tfrac{\pi}{4}\right) = \tan\left(\tfrac{3\pi}{4}\right) = -1 \]
Meanwhile the left-hand side is perfectly well defined and equals negative one. So the formula fails to compute something that genuinely exists.
It is not a defect so much as a limitation of the method: the derivation divided by the product of the cosines, and that division is invalid when a cosine is zero. When the tangent formula cannot be used, going back to the sine and cosine formulas always can be — and that is the reliable fallback.
Comparison
Fill the blanks from memory. Six formulas, but really one theorem and five consequences.
Comparison matrix
| Identity | Statement | Where it came from |
|---|---|---|
| Even-odd | cos(-t) = cos t, sin(-t) = -sin t | reflection across the x-axis |
| Cosine difference | cos a cos b + sin a sin b | equal chords plus the distance formula |
| Cosine sum | cos a cos b - sin a sin b | the difference formula plus even-odd |
| Cofunction | cos of the complement is sin | the difference formula with a equal to pi over 2 |
| Sine sum | sin a cos b + cos a sin b | cofunction, then the cosine difference formula |
| Tangent sum | (tan a + tan b) over (1 - tan a tan b) | sine over cosine, divided by cos a cos b |
Read the right-hand column downward and the whole section is a single chain: one geometric fact, then five derivations. Nothing after the second row was proved from scratch.
Pattern
Whether the question asks for an exact value, a simplification or a verification, the same five moves cover it.
Step four is where most of the work in an exam question actually lies. The identity itself is a substitution; finding the four values to substitute is the part that takes time.
Check
Even and odd. Ask which function it is.
Check your understanding
If the sine of theta is 0.4, what is the sine of negative theta?
Answer: B
Why: Sine is odd, so negating the input negates the output. Geometrically, negating the angle reflects the point across the x-axis, and that reflection negates the vertical coordinate, which is exactly what the sine reads.
Check
A sum formula. Watch which terms pair with which.
Check your understanding
Which is the correct expansion of the sine of alpha minus beta?
Answer: A
Why: The sine formulas use mixed terms — sine times cosine, then cosine times sine — and the sign in the answer matches the sign in the bracket. A difference inside therefore gives a minus outside.
Check
A cofunction. Look for a complementary pair.
Check your understanding
The expression tangent of 27 degrees times cotangent of 27 degrees is being compared with tangent of 27 degrees times tangent of 63 degrees. What are their values?
Answer: A
Why: The first is a function times its own reciprocal, which is 1. The second is 1 as well, because 27 and 63 are complementary, so the tangent of 63 equals the cotangent of 27 by the cofunction identity — making the second expression identical to the first.
Real world
Two speakers play a pure tone of the same frequency but one is slightly delayed, so at a listener's ear the two signals are the sine of omega t and the sine of omega t plus phi, where phi is a fixed phase shift.
Discussion prompt
Expand the second signal and explain, in terms of the sum identity, why adding two shifted copies of the same tone gives another tone of the same frequency rather than something new. What determines how loud the result is?
Hint: Expand, then collect the terms in sine of omega t and cosine of omega t.
Answer:
\[ \sin(\omega t) + \sin(\omega t + \phi) = \sin(\omega t)\big(1 + \cos\phi\big) + \cos(\omega t)\sin\phi \]
The result is a constant times the sine plus a constant times the cosine, both at the original frequency, so no new frequency has appeared. Any such combination can itself be rewritten as a single sinusoid of that same frequency, which Lesson 11.1 does explicitly.
The loudness is governed by the two constants. When phi is zero the second term vanishes and the first coefficient is 2, so the signals reinforce and the sound doubles in amplitude. When phi is pi, the first coefficient is zero and the second is zero too, so the signals cancel completely — which is exactly the principle behind noise-cancelling headphones, and it is the sum identity that predicts it.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
Is the sine of alpha plus beta ever equal to the sine of alpha plus the sine of beta?
Correct: Sometimes, for instance when both angles are zero.
\[ \alpha = \beta = 0: \quad \sin(0) = 0 = \sin(0) + \sin(0) \quad\checkmark \]
\[ \alpha = \beta = \tfrac{\pi}{2}: \quad \sin(\pi) = 0 \ne 2 = \sin\left(\tfrac{\pi}{2}\right) + \sin\left(\tfrac{\pi}{2}\right) \]
Why: The distributive form is not an identity, since it fails for almost all pairs of angles. But it is not never true either: at alpha and beta both zero, both sides are zero, and there are other coincidental solutions such as beta equal to zero with alpha anything at all. The distinction is worth making precisely: an identity must hold for every value, so a formula that holds only sometimes is an equation to be solved rather than an identity to be used. Calling it never true would be as inaccurate as calling it always true.
Explain it
They have been handed six formulas to memorise and are drowning in them.
Discussion prompt
In no more than five sentences, tell them which one to actually learn and how the others follow. Then give them the two questions that get the signs right every time.
Hint: Which of the six was proved from geometry, and which were derived from that one?
Answer:
A usable answer: only one of them was proved from scratch — the cosine of a difference, which comes from two equal chords and the distance formula. The cosine sum formula is that one with a negated angle; the cofunction identities are that one with pi over two put in; the sine formulas come from the cofunction identities; and the tangent formula is sine over cosine divided through. So there is one thing to learn and a chain to follow.
For the signs, ask two questions. Are the terms matched or mixed? Cosine matches, sine mixes. Does the sign keep or flip? Cosine flips, sine keeps. Those two answers determine all four formulas, and the tangent formula just does both at once — numerator keeps, denominator flips.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The cosine sign is fixed by the single sentence that cosine is contrary: the answer's sign opposes the bracket's. The decomposition is fixed by reading the denominator, since twelfths come from thirds and quarters. Reconstruction is fixed by writing down all four values you will need before touching the identity, so none is forgotten mid-substitution. Even and odd is fixed by remembering that only cosine and secant are even, because only they read the horizontal coordinate. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of the page draw the unit circle with an angle and its negative marked, and write the six even-odd identities beside it, grouped into the two even and the four odd. In the middle, write the cosine difference identity in a box and draw arrows from that box to four other boxes: the cosine sum identity, the cofunction identities, the sine identities, and the tangent identities, labelling each arrow with the one move that gets you there. To the right, draw a right triangle with both acute angles labelled and write three cofunction identities you can read directly off it. In the bottom left, work out the exact value of the cosine of 105 degrees completely, showing the decomposition you chose and why. In the bottom right, write the two questions that determine the signs in all four sum and difference formulas, and beside each write which family answers it which way. Finally, circle the one identity on your page that was proved rather than derived.
The circled identity should be the cosine of a difference. Everything else on the page descends from it, and knowing that is worth more than knowing any of the individual formulas.
Recap
Five things, and the second one is the one the next two lessons are built on.
| If the question says | Your first move is |
|---|---|
| Simplify an expression with negative angles | Move every minus out with the even-odd identities |
| Find the exact value at 15, 75 or 105 degrees | Decompose into 45 and 30 |
| Find the exact value at a twelfth of pi | Decompose into thirds and quarters |
| Given sin alpha and cos beta with quadrants | Reconstruct all four values before substituting |
| Two angles sum to 90 | Look for a cofunction identity |
The next lesson does one thing with these formulas: puts the two angles equal. Setting beta equal to alpha in the sum identities produces the double angle formulas, and running those backwards produces the power reduction and half angle formulas — three more families, all from substitution rather than new proof.
Stitz & Zeager, College Trigonometry, Ch. 10 Foundations of Trigonometry — §10.4 Trigonometric Identities §10.4, pp. 770-776 — everything on these slides traces back here
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