The six sum and difference formulas for sine, cosine and tangent, finding exact values by splitting a special angle, evaluating with given function values and quadrant signs, simplifying expressions and deriving reduction identities, and solving trigonometric equations and models.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 14 — Trigonometric Graphs, Identities, and Equations
Apply Sum and Difference Formulas
Objectives
Five outcomes. Six formulas, and the exact values they unlock.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 949-953 — the lesson these objectives are drawn from
Warm-up
You know the exact values of the sine and cosine of 30, 45 and 60 degrees.
Discussion prompt
Is the sine of 75 degrees equal to the sine of 30 degrees plus the sine of 45 degrees? Check with a calculator before deciding.
Hint: Compute all three.
Answer:
\[ \sin 75\degree \approx 0.966; \quad \sin 30\degree+\sin 45\degree \approx 0.5+0.707 = 1.207 \]
They are nowhere near equal, so the sine does not distribute over a sum. Something more careful is needed.
\[ \sin(a+b) = \sin a\cos b+\cos a\sin b \]
That formula, and five like it, are what this lesson supplies — and they turn 75 degrees into an exact value rather than a decimal.
Concept
The sine, cosine and tangent of a sum or difference can be written using the sines, cosines and tangents of the two separate angles. That converts an unfamiliar angle into familiar ones.
sum and difference formulas — Six identities giving the sine, cosine and tangent of a sum or difference of two angles in terms of the functions of the separate angles.
\[ \sin(a\pm b) = \sin a\cos b \pm \cos a\sin b \]
Sine and tangent keep the sign that appears in the angle; cosine reverses it. That one asymmetry is the source of most errors here.
Figure (svg): Two columns comparing how sine and tangent handle the sign against how cosine does
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 949-949
Section
Section 1
Concept
Each of sine, cosine and tangent has a sum formula and a difference formula. Sine and tangent carry the same sign as the angle; cosine carries the opposite one.
\[ \cos(a+b) = \cos a\cos b-\sin a\sin b \]
None of the three functions distributes over addition, so the sine of a sum is never the sum of the sines. The formulas exist precisely because that shortcut fails.
Figure (svg): The six sum and difference formulas arranged in two columns
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 949-949 — Sum and Difference Formulas
Picture it
The sum formulas beside the difference formulas.
Figure (svg): The six sum and difference formulas arranged in two columns
Reading down the columns, only the cosine row swaps its sign between the two sides. Sine and tangent match the angle's sign.
Worked example
Applying the key concept.
\[ \text{Write the expansions of } \sin(a+b), \; \cos(a+b), \; \sin(a-b) \text{ and } \cos(a-b). \]
Sine of a sum
Why: The same sign as the angle.
\[ \sin a \cos b + \cos a \sin b \]
Cosine of a sum
Why: The opposite sign.
\[ \cos a \cos b - \sin a \sin b \]
Sine of a difference
Why: Again matching the angle.
\[ \sin a \cos b - \cos a \sin b \]
Cosine of a difference
Why: Again reversed.
\[ \cos a \cos b + \sin a \sin b \]
Figure (svg): The six sum and difference formulas arranged in two columns
\[ \sin: \pm; \qquad \cos: \mp \]
Verify: test the cosine formula at a known angle
Why: Take a and b both equal to 60 degrees. The formula gives cos 60 cos 60 minus sin 60 sin 60, which is one quarter minus three quarters, or negative one half — and cos 120 degrees is indeed negative one half. Had the sign been positive the result would have been 1, which is plainly wrong for an obtuse angle.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 949-949
Sorting
Sine and tangent match; cosine reverses.
Sort into buckets
Sort each expansion by the sign in the middle.
The two cosine entries sit in the opposite buckets from where the angle's sign would suggest, which is exactly the pattern to memorise.
Worked example
Why the formulas are needed at all.
\[ \text{Compare } \cos(60\degree+60\degree) \text{ with } \cos 60\degree+\cos 60\degree. \]
Compute the left side
Why: The cosine of 120 degrees.
\[ -\frac{1}{2} \]
Compute the right side
Why: One half plus one half.
\[ 1 \]
Compare
Why: Negative one half is not 1.
State the general rule
Why: None of the three functions distributes.
Figure (svg): The solution to Worked example show the shortcut fails shown as a ladder of expressions, one row per algebraic move
\[ -\tfrac{1}{2} \neq 1 \]
Verify: check that the formula does give the right answer
Why: The formula gives cos 60 cos 60 minus sin 60 sin 60, which is one quarter minus three quarters, or negative one half — matching cos 120 degrees exactly. The failure of the shortcut and the success of the formula are two sides of the same test.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 949-949
Trap
\[ \sin(a+b) \]
Write it as the sum of the sines
Why: The function is treated like a multiplier.
\[ = \sin a+\sin b \quad \text{(wrong)} \]
At 30 and 45 degrees the left side is about 0.966 and the right about 1.207. They are not equal, and for many angles the right side even exceeds 1.
\[ \sin(a+b) = \sin a\cos b+\cos a\sin b \]
Use the formula, which mixes both functions
Why: Each term pairs one angle's sine with the other angle's cosine.
\[ \sin 75\degree = \tfrac{\sqrt6+\sqrt2}{4} \approx 0.966 \quad \checkmark \]
Sine is bounded by 1, so any expression claiming a sine of 1.207 is wrong before any formula is consulted.
Fill the middle
The key concept.
Fill in the blanks
\cos(a+b) = \cos a\cos b - \sin a\sin b
Why: The cosine formula reverses the sign, so a sum of angles gives a minus in the expansion. Testing at 60 and 60 degrees confirms it.
Matching
Watch the sign.
Match the pairs
Why: The two sine expansions share the same two products and differ only in sign, and so do the two cosine ones. What distinguishes sine from cosine is which functions get paired: sine mixes them and cosine keeps like with like.
Prediction
Commit before reasoning.
Predict first
Test cos(a + b) at a = b = 60 degrees with a plus sign instead of a minus. What goes wrong?
Correct: It gives 1, but cos 120 degrees is negative one half.
\[ \tfrac{1}{4}-\tfrac{3}{4} = -\tfrac{1}{2} = \cos 120\degree \quad \checkmark \]
Why: With a plus the expression is one quarter plus three quarters, which is 1 — the cosine of zero, not of 120 degrees. The minus sign is what lets the cosine become negative for obtuse angles, which it must. Testing a formula at a pair of angles you know is the fastest way to check a remembered sign, and it takes about ten seconds.
Section
Section 2
Concept
An angle that is not one of the special ones can often be written as a sum or difference of two that are. Applying the matching formula then gives an exact value rather than a decimal.
\[ 15\degree = 60\degree-45\degree \]
Every multiple of 15 degrees, and every twelfth of pi, can be built from 30, 45, 60 and 90 — which is exactly the range of angles the formulas make exact.
Figure (svg): Two awkward angles rewritten as sums or differences of familiar ones
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 949-949 — Evaluate a trigonometric expression
Picture it
Example 1: 15 degrees and 7 pi over 12.
Figure (svg): Two awkward angles rewritten as sums or differences of familiar ones
Fifteen degrees is 60 minus 45, and seven twelfths of pi is a third plus a quarter. Both land on angles whose values are known exactly.
Worked example
Example 1a.
\[ \text{Find the exact value of } \sin 15\degree. \]
Split the angle
Why: Sixty minus 45 gives 15, and both are special.
\[ \sin(60 - 45) \]
Apply the difference formula
Why: Sine matches the sign, so a minus.
\[ \sin 60 \cos 45 - \cos 60 \sin 45 \]
Substitute the known values
Why: Root 3 over 2, root 2 over 2, one half, root 2 over 2.
Simplify
Why: Root 6 over 4 minus root 2 over 4.
\[ \frac{\sqrt{6} - \sqrt{2}}{4} \]
Figure (svg): Two awkward angles rewritten as sums or differences of familiar ones
\[ \sin 15\degree = \frac{\sqrt6-\sqrt2}{4} \]
Verify: check against a decimal
Why: Root 6 is about 2.449 and root 2 about 1.414, so the answer is about 1.035 over 4, or 0.259 — and a calculator gives sin 15 degrees as 0.2588. The exact form is what the formula buys; the decimal only confirms it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 949-949
Sorting
Both pieces must be special angles.
Sort into buckets
Sort each proposed split of 75 degrees.
Several splits usually work, and they all give the same exact value — 45 plus 30 is simply the tidiest of the three here.
Worked example
Example 1b and Guided Practice 3.
\[ \text{Find the exact values of } \tan\frac{7\pi}{12} \text{ and } \tan\frac{5\pi}{12}. \]
Split the first angle
Why: A third plus a quarter is seven twelfths.
\[ \tan(\frac{\pi}{3} + \frac{\pi}{4}) \]
Apply the sum formula
Why: Root 3 plus 1, over 1 minus root 3.
Rationalise
Why: Multiply above and below by 1 plus root 3.
\[ -2 - \sqrt{3} \]
Repeat for the second
Why: A sixth plus a quarter is five twelfths.
\[ 2 + \sqrt{3} \]
Figure (svg): Two awkward angles rewritten as sums or differences of familiar ones
\[ -2-\sqrt3; \qquad 2+\sqrt3 \]
Verify: check the signs against the quadrants
Why: Seven twelfths of pi is just past a right angle, in the second quadrant, where tangent is negative — and negative 2 minus root 3 is about negative 3.73. Five twelfths of pi is just under a right angle, in the first quadrant, where tangent is positive and large: 2 plus root 3 is about 3.73. Both signs match their quadrants.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 949-950
Error analysis
A student tries to find the exact value of the sine of 15 degrees.
Annotate
On: \( \sin 15\degree = \sin(10\degree+5\degree) = \sin 10\degree\cos 5\degree+\cos 10\degree\sin 5\degree \)
The formula only helps when both pieces are special angles. Choosing the split is the real decision, and there are usually only one or two that work.
Fill the middle
Example 1a.
Fill in the blanks
15\degree = 60\degree-45\degree
Why: Sixty minus 45 is 15, and both are angles whose exact values are known. Thirty plus negative 15 would not help, since 15 is what is being found.
Matching
Split, expand, simplify.
Match the pairs
Why: Two pairs share a value: sin 15 equals cos 75 and sin 105 equals cos 15, since sine and cosine of complementary angles agree. That cofunction relationship is itself a consequence of the difference formula, as the fourth idea shows.
Prediction
Commit before reasoning.
Predict first
Which of 15, 20, 75 and 105 degrees can these formulas evaluate exactly?
Correct: 15, 75 and 105, since each is a sum or difference of special angles; 20 is not.
\[ 15\degree \text{ steps: } 15, 30, 45, 60, 75, 90, \dots \]
Why: Fifteen is 45 minus 30, 75 is 45 plus 30, and 105 is 60 plus 45. But 20 cannot be built from 30, 45, 60 and 90 by adding or subtracting, so no split lands on known values. The formulas extend the exact-value list from multiples of 30 and 45 to all multiples of 15 — and no further, at least not until the half-angle formulas of the next lesson.
Section
Section 3
Concept
When one function value and an interval are given for each angle, the Pythagorean identity supplies the missing value's size and the quadrant supplies its sign. Then the formula can be applied.
\[ \sin^2\theta+\cos^2\theta = 1 \]
Skipping the quadrant step is the usual error, because the identity is happy to return a positive value that the interval forbids.
Figure (svg): Two angles located by quadrant so that the missing function values can be signed
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 950-950 — Use a difference formula
Picture it
Example 2: one angle in the third quadrant and one in the first.
Figure (svg): Two angles located by quadrant so that the missing function values can be signed
The third-quadrant angle has both functions negative, so its sine is negative three fifths rather than positive. The first-quadrant angle keeps both positive.
Worked example
Example 2.
\[ \text{Find } \cos(a-b) \text{ given } \cos a = -\tfrac{4}{5} \text{ with } \pi < a < \tfrac{3\pi}{2}, \text{ and } \sin b = \tfrac{5}{13} \text{ with } 0 < b < \tfrac{\pi}{2}. \]
Find the missing value for a
Why: The identity gives three fifths; the third quadrant makes it negative.
\[ \sin a = -\frac{3}{5} \]
Find the missing value for b
Why: The identity gives twelve thirteenths; the first quadrant keeps it positive.
\[ \cos b = \frac{12}{13} \]
Apply the difference formula
Why: Cosine reverses the sign, so a plus.
\[ \cos a \cos b + \sin a \sin b \]
Substitute and simplify
Why: Negative 48 over 65 plus negative 15 over 65.
\[ -\frac{63}{65} \]
Figure (svg): Two angles located by quadrant so that the missing function values can be signed
\[ \cos(a-b) = -\frac{63}{65} \]
Verify: check the answer is a valid cosine
Why: Negative 63 over 65 is about negative 0.969, comfortably within negative 1 and 1 — as any cosine must be. A result outside that range would signal an arithmetic slip, and it is the fastest check available for these problems.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 950-950
Sorting
Sine is positive above the axis; cosine to the right of it.
Sort into buckets
Sort each combination.
Sine tracks the vertical coordinate and cosine the horizontal one, so the signs follow directly from which quadrant the angle lands in.
Worked example
Guided Practice 5.
\[ \text{Find } \sin(a-b) \text{ given } \sin a = \tfrac{8}{17} \text{ with } 0 < a < \tfrac{\pi}{2}, \text{ and } \cos b = -\tfrac{24}{25} \text{ with } \pi < b < \tfrac{3\pi}{2}. \]
Find cos a
Why: The identity gives fifteen seventeenths; first quadrant, so positive.
\[ \cos a = \frac{15}{17} \]
Find sin b
Why: The identity gives seven twenty-fifths; third quadrant, so negative.
\[ \sin b = -\frac{7}{25} \]
Apply the formula
Why: Sine matches the sign, so a minus.
\[ \sin a \cos b - \cos a \sin b \]
Substitute and simplify
Why: Negative 192 over 425 plus 105 over 425.
\[ -\frac{87}{425} \]
Figure (svg): The solution to Worked example a sine of a difference shown as a ladder of expressions, one row per algebraic move
\[ \sin(a-b) = -\frac{87}{425} \]
Verify: check the second term's sign carefully
Why: The second term is minus cos a sin b, and sin b is itself negative — so subtracting a negative adds, giving plus 105 over 425. Two negatives in one term is where these problems most often go wrong, and writing the substitution with brackets before simplifying prevents it.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 950-950
Trap
\[ \cos a = -\tfrac{4}{5}, \; \pi < a < \tfrac{3\pi}{2} \]
Take the positive root from the identity
Why: The Pythagorean identity is applied and the positive value kept.
\[ \sin a = \tfrac{3}{5} \quad \text{(wrong)} \]
The interval places a in the third quadrant, where the sine is negative. The identity gives the size, never the sign.
\[ \sin a = -\tfrac{3}{5} \]
Use the interval to choose the sign
Why: Each quadrant fixes the sign of every function.
\[ \text{QIII: both sine and cosine negative} \]
The stated interval is not decoration. Sketching the angle, or recalling which functions are positive in each quadrant, settles it in seconds.
Fill the middle
Example 2.
Fill in the blanks
\cos a = -\tfrac-3/5___, \; \text___ \;\Longrightarrow\; \sin a = ___
Why: The identity gives a size of three fifths and the third quadrant makes it negative. Taking the positive root would place the angle in the second quadrant instead.
Ranking
Evaluating a sum or difference from given values.
Put in order
Why: Steps two and three could be swapped in thought but not in effect: the identity never produces a sign, so the quadrant has to supply it before the substitution. Step five costs nothing and catches an arithmetic slip immediately.
Prediction
Commit before reasoning.
Predict first
Given that cos a is negative four fifths, how many values could sin a take?
Correct: Two — positive and negative three fifths — until the quadrant decides.
\[ \sin^2 a = \tfrac{9}{25} \;\Longrightarrow\; \sin a = \pm\tfrac{3}{5} \]
Why: The identity gives sin squared a equals nine twenty-fifths, and a square root has two signs. Both are genuinely possible: the second quadrant would give positive three fifths and the third negative three fifths, and both have the same cosine. The interval is what picks between them, which is why a problem stating one function value must also state a quadrant.
Section
Section 4
Concept
When one of the two angles is a known one such as pi or pi over 2, expanding with a formula leaves terms whose known factors evaluate to 0, 1 or negative 1, collapsing the expression.
\[ \cos(x+\pi) = -\cos x \]
Every reduction identity — the cofunction relations and the shifts by pi and by a right angle — comes out of these six formulas the same way.
Figure (svg): Two expressions simplified by expanding with a formula and evaluating the known angle
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 950-950 — Simplify an expression
Picture it
Example 3 and the cofunction identity of lesson exercise 11.
Figure (svg): Two expressions simplified by expanding with a formula and evaluating the known angle
In both, one factor became 1 and another became 0, so a two-term expansion reduced to a single function.
Worked example
Example 3.
\[ \text{Simplify } \cos(x+\pi). \]
Apply the sum formula for cosine
Why: Cosine reverses the sign.
\[ \cos x \cos \pi - \sin x \sin \pi \]
Evaluate the known angle
Why: Cosine of pi is negative 1; sine of pi is 0.
\[ (\cos x) (-1) - (\sin x) (0) \]
Simplify
Why: The second term vanishes.
\[ -\cos x \]
Interpret
Why: Shifting by pi reflects the cosine curve.
Figure (svg): Two expressions simplified by expanding with a formula and evaluating the known angle
\[ \cos(x+\pi) = -\cos x \]
Verify: check at a value
Why: At x equal to zero the left side is the cosine of pi, which is negative 1, and the right side is negative cos 0, also negative 1. At x equal to pi over 2 both sides give zero. The identity holds for every x, and checking two values is enough to catch a sign slip.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 950-950
Fill the middle
Example 3.
Fill in the blanks
\cos x\cos\pi-\sin x\sin\pi = (\cos x)(-1)-(\sin x)(0)
Why: The cosine of pi is negative 1, so the first term becomes negative cos x. The sine of pi is zero, which removes the second term entirely.
Worked example
Lesson exercise 11.
\[ \text{Derive } \sin\left(\tfrac{\pi}{2}-\theta\right) = \cos\theta \text{ using the difference formula.} \]
Apply the difference formula for sine
Why: Sine matches the sign, so a minus.
\[ \sin(\frac{\pi}{2}) \cos t - \cos(\frac{\pi}{2}) \sin t \]
Evaluate the known angle
Why: Sine of a right angle is 1; cosine of it is 0.
\[ (1) (\cos t) - (0) (\sin t) \]
Simplify
Why: The second term vanishes again.
State the identity
Why: The sine of the complement is the cosine.
Figure (svg): Two expressions simplified by expanding with a formula and evaluating the known angle
\[ \sin\left(\tfrac{\pi}{2}-\theta\right) = \cos\theta \]
Verify: check with complementary angles
Why: Sine of 30 degrees is one half and cosine of 60 degrees is also one half; sine of 15 degrees and cosine of 75 degrees both equal root 6 minus root 2, over 4. The identity explains why the exact-value table has each entry appearing twice, which is worth knowing rather than treating as a coincidence.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 952-952
Error analysis
A student simplifies the cosine of x plus pi.
Annotate
On: \( \cos(x+\pi) = \cos x\cos\pi-\sin x\sin\pi = (\cos x)(1)-(\sin x)(0) = \cos x \)
Checking at a single value catches it: at x equal to zero the expression is cos pi, which is negative 1, and the claimed answer gives positive 1.
Matching
Expand, then evaluate the known angle.
Match the pairs
Why: Shifting by pi negates both sine and cosine; reflecting in a right angle swaps them. Both families of reduction identity fall out of the six formulas by the same two-line argument, which is why they never need separate memorising.
Sorting
Evaluate at the special angle.
Sort into buckets
Sort each factor by its value.
Every simplification in this idea works because one factor becomes zero and kills a term while the other becomes 1 or negative 1 and leaves a single function behind.
Prediction
Commit before reasoning.
Predict first
Why is there no need to memorise the identity for the sine of x plus pi separately?
Correct: Because expanding it with the sum formula and evaluating at pi derives it in two lines.
\[ \sin(x+\pi) = \sin x(-1)+\cos x(0) = -\sin x \]
Why: Expanding gives sin x cos pi plus cos x sin pi, which is negative sin x plus zero, or negative sin x. Every shift by pi, by a right angle, or by a full turn yields to the same two-step treatment. Six formulas therefore replace a long table of reduction identities, which is a much better thing to carry in memory.
Section
Section 5
Concept
Expanding a sum or difference often makes terms cancel, leaving an equation in a single trigonometric function that can be solved by the methods of Lesson 14.4.
\[ \sin\left(x+\tfrac{\pi}{3}\right)+\sin\left(x-\tfrac{\pi}{3}\right) = \sin x \]
In a model the same expansion turns two waves of different form into one tangent equation, which then gives every solution at once.
Figure (svg): Daylight hours for two cities plotted across a year, crossing twice
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 950-951 — Solve a trigonometric equation
Picture it
Example 5: daylight hours in Dallas and Anchorage.
Figure (svg): Daylight hours for two cities plotted across a year, crossing twice
The two curves cross near days 97 and 279 — April 8 and October 7 — which the expansion finds exactly rather than by reading the graph.
Worked example
Example 4.
\[ \text{Solve } \sin\left(x+\tfrac{\pi}{3}\right)+\sin\left(x-\tfrac{\pi}{3}\right) = 1 \text{ for } 0 \leq x < 2\pi. \]
Expand both terms
Why: One sum formula and one difference formula.
Collect
Why: The cosine terms have opposite signs and cancel.
Simplify
Why: One half sin x plus one half sin x.
\[ \sin x = 1 \]
Solve on the interval
Why: The sine reaches 1 once in a full turn.
\[ x = \frac{\pi}{2} \]
Figure (svg): The solution to Worked example solve by expanding shown as a ladder of expressions, one row per algebraic move
\[ x = \frac{\pi}{2} \]
Verify: substitute back
Why: At x equal to pi over 2, the two terms are the sines of five sixths of pi and of a sixth of pi, both one half — and one half plus one half is 1. The cancellation of the cosine terms is what made a two-angle equation into a one-angle one, and it happened because the two shifts were equal and opposite.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 950-950
Fill the middle
Example 4.
Fill in the blanks
\tfracsin x___\sin x+\tfrac______\cos x+\tfrac______\sin x-\tfrac______\cos x = ___
Why: The two cosine terms are equal and opposite, so they cancel, and the two half-sines add to a full sin x. The whole two-angle equation collapses to one angle.
Worked example
Example 5.
\[ \text{Solve } 2\sin\left(\tfrac{\pi t}{182}-1.352\right)+12.1 = -6\cos\left(\tfrac{\pi t}{182}\right)+12.1. \]
Cancel the constants and halve
Why: Both sides carry 12.1, and the left has a factor of 2.
\[ \sin(u - 1.352) = -3 \cos u \]
Expand the difference
Why: Sine matches the sign, so a minus.
\[ \sin u(0.219) - \cos u(0.976) = -3 \cos u \]
Collect and divide by cosine
Why: Move the cosine terms together, then divide.
\[ \tan u = -9.242 \]
Solve and convert to days
Why: Take the inverse tangent and add multiples of pi.
\[ t\text{ about } 97\text{ and } 279 \]
Figure (svg): Daylight hours for two cities plotted across a year, crossing twice
\[ t \approx 97 \text{ and } 279 \]
Verify: check both against the graph
Why: At day 97 both models give about 12.7 hours and at day 279 both give about 11.5 — the two crossings visible on the graph. And the answers are about six months apart, near the equinoxes, which is exactly when two cities at different latitudes should agree on daylight.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 951-951
Trap
\[ 0.219\sin u = -2.024\cos u \]
Divide both sides by cosine and continue
Why: The division is performed without comment.
\[ \tan u = -9.242 \quad \text{(incomplete)} \]
If the cosine were zero the division would be invalid, and any solution with cosine zero would be lost.
\[ \text{check } \cos u = 0 \text{ separately} \]
Test whether cosine can vanish here
Why: Substituting cosine equal to zero forces sine to be zero too, which is impossible.
\[ \text{no solutions lost, so the division is safe} \]
Dividing by a variable expression always deserves this check, exactly as it did for rational equations in Lesson 8.6.
Ranking
Solving an equation with shifted angles.
Put in order
Why: Step three is where the payoff arrives: the whole point of expanding is that terms cancel or combine. If nothing cancels after expanding, the split was probably not the one the problem intended.
Comparison
Fill the blanks. Expand, then collect.
Comparison matrix
| Question | Example 4 | Example 5 |
|---|---|---|
| What is expanded | both shifted sines | the sine of a difference |
| What cancels or collects | the two cosine terms cancel | the cosine terms gather on one side |
| What is left | sin x = 1 | tan u = -9.242 |
| Solutions | x = pi/2 only | two days, about 97 and 279 |
In both cases the expansion converted an equation nobody could solve directly into one from Lesson 14.4, which is the whole reason for the formulas here.
Prediction
Commit before reasoning.
Predict first
The tangent equation gives solutions separated by 182 days. Why does that make sense?
Correct: Because the tangent has period pi, and pi in this model corresponds to half a year.
\[ \frac{\pi(t+182)}{182} = \frac{\pi t}{182}+\pi \]
Why: The argument of the tangent is pi t over 182, so t increasing by 182 increases the argument by pi — exactly one tangent period. Half of 365 is about 182, so consecutive solutions fall about six months apart. That places them near the two equinoxes, which is when cities at different latitudes really do share daylight, so the mathematics and the astronomy agree.
Comparison
Fill the blanks. Sine and tangent match; cosine reverses.
Comparison matrix
| Function | Sum | Difference |
|---|---|---|
| sine | sin a cos b + cos a sin b | sin a cos b - cos a sin b |
| cosine | cos a cos b - sin a sin b | cos a cos b + sin a sin b |
| tangent | (tan a + tan b)/(1 - tan a tan b) | (tan a - tan b)/(1 + tan a tan b) |
| The pattern | sine and tangent keep the sign | cosine reverses it |
The tangent formulas reverse the sign in the denominator rather than the numerator, which is the same reversal appearing in a different place.
Pattern
Identify what is given, then choose the formula.
Every answer that is a sine or cosine must lie between negative 1 and 1, which is the fastest check available.
OpenStax Algebra and Trigonometry 2e, §9.2 Sum and Difference Identities §9.2
Check
Split into special angles.
Check your understanding
What is the exact value of sin 15 degrees?
Answer: A
Why: Splitting as 60 minus 45 and using the difference formula for sine gives this.
Check
The quadrant fixes the sign.
Check your understanding
Given cos a = -4/5 with a in the third quadrant and sin b = 5/13 with b in the first, what is cos(a - b)?
Answer: A
Why: With sin a = -3/5 and cos b = 12/13, the formula gives -48/65 plus -15/65.
Check
Expand, then evaluate the known angle.
Check your understanding
Simplify cos(x + pi).
Answer: A
Why: The cosine of pi is -1 and the sine of pi is 0, leaving -cos x.
Real world
Two speakers play the same pure tone, but one signal arrives slightly later, so the combined sound at a listener is sin(2 pi f t) plus sin(2 pi f t + d), where d is the phase difference.
Discussion prompt
Expand the second term and combine, then say what happens when d is pi and when d is zero.
Hint: Expand the sine of a sum and collect the like terms.
Answer:
\[ \sin(u)+\sin(u+d) = \sin u+\sin u\cos d+\cos u\sin d \]
\[ = (1+\cos d)\sin u+(\sin d)\cos u \]
When d is zero the expression becomes 2 sin u — the two signals reinforce and the sound is twice as loud. When d is pi, cos d is negative 1 and sin d is zero, so the whole thing becomes zero: the signals cancel completely and the listener hears silence.
This is the principle behind noise-cancelling headphones, which measure incoming sound and emit the same wave shifted by pi. It also explains why moving a few centimetres in a room can make a bass note vanish: the path difference from two speakers changes d, and at half a wavelength the waves cancel. The whole analysis is one application of the sine sum formula, and the two extreme cases fall out by evaluating cos d and sin d at the two angles.
Commit first
Answer, then rate your confidence honestly.
Predict first
Is the cosine of a plus b equal to cos a cos b plus sin a sin b?
Correct: No — the cosine formula reverses the sign, so a sum needs a minus.
\[ \cos(a+b) = \cos a\cos b-\sin a\sin b \]
Why: Testing at a and b both 60 degrees settles it: the version with a plus gives one quarter plus three quarters, which is 1, while cos 120 degrees is negative one half. The version with a minus gives one quarter minus three quarters, which is negative one half — correct. Sine and tangent keep the sign that appears in the angle and cosine flips it, and that asymmetry is the single most common source of error in the lesson. Ten seconds spent testing at two known angles recovers the right sign whenever memory fails.
Explain it
They keep writing the sine of a sum as the sum of the sines.
Discussion prompt
In four sentences or fewer, explain why that is wrong and what to do instead.
Hint: Try it with two angles you both know.
Answer:
Try it with 30 and 45 degrees. The sines add to about 1.21, but a sine can never be more than 1, so the answer must be wrong.
Sine does not spread across a sum the way multiplication spreads across addition. Instead there is a formula: the sine of a sum is sin a cos b plus cos a sin b, mixing the two functions rather than just adding them.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the sign, test at two 60-degree angles and see which version gives negative one half. For splits, write out the special angles and look for a pair that hits your target. For quadrants, sketch the angle before taking any root. For equations, expand fully before collecting, and expect the cross terms to cancel.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a formulas page. Top: write all six formulas in a two-column table, circling the three signs that reverse, and beside them the test at 60 and 60 degrees that recovers the cosine sign. Middle left: find three exact values of your own by splitting angles, showing the split and the simplification for each. Middle right: work Example 2 in full, marking where the quadrant supplied a sign the identity could not. Bottom left: derive two reduction identities from the formulas, showing which factor became zero each time. Bottom right: solve Example 4, circling the two terms that cancelled and writing one sentence on why they did.
If any of your exact values exceeds 1 in size for a sine or cosine, recheck the arithmetic before anything else — that alone is impossible.
Recap
Five things, from six formulas.
| If you see | Then |
|---|---|
| A sine or tangent of a sum | Keep the sign the angle carries |
| A cosine of a sum or difference | Reverse the sign |
| An angle like 15 or 75 degrees | Split it into two special angles |
| A given value with an interval | Identity for the size, quadrant for the sign |
| A known angle inside the expression | Expand; its factors become 0, 1 or -1 |
| Shifted angles in an equation | Expand and expect terms to cancel |
Lesson 14.7 takes the special case where the two angles are equal, giving the double-angle formulas, and then runs them backwards for half angles.
McDougal Littell Algebra 2 (Texas Edition), Ch. 14 Trigonometric Graphs, Identities, and Equations — Lesson 14.6 Apply Sum and Difference Formulas §14.6, pp. 949-953 — everything on these slides traces back here
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