Sum and Difference Identities

🔺 Sum and Difference Identities

Part 1 of 7 — Why We Need Them

Topics in This Part

🔑 Key Concept: Trig functions do not distribute over addition. $\sin(A + B) \ne \sin A + \sin B$. The sum and difference identities are the correct rules for splitting an angle into pieces you already know.

The "Distributive" Trap

A tempting but wrong move is to treat $\sin$ like a multiplier:

$\sin(A + B) \;\overset{?}{=}\; \sin A + \sin B \quad \color{red}{\textbf{WRONG}}$

A Quick Counterexample

Let $A = B = \frac{\pi}{6}$, so $A + B = \frac{\pi}{3}$.

The two columns disagree ($0.866 \ne 1$), so the "distribute" rule is false.

⚠️ Never write $\sin(A+B) = \sin A + \sin B$ or $\cos(A+B) = \cos A + \cos B$. That is the single most common error on this whole topic.

Concept Check 🎯

Special vs. Non-Special Angles

From the unit circle you already know the special angles exactly:

But what about $\frac{\pi}{12}$, $\frac{5\pi}{12}$, or $\frac{7\pi}{12}$? These aren't on the unit circle — yet.

💡 The trick: any of these can be built by adding or subtracting special angles:

• $\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6}$
• $\frac{5\pi}{12} = \frac{\pi}{4} + \frac{\pi}{6}$
• $\frac{7\pi}{12} = \frac{\pi}{3} + \frac{\pi}{4}$

The sum and difference identities turn that decomposition into an exact answer.

Decompose the Angle 🔽

Express each angle as a sum or difference of two special angles ($\frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}$).

What the Identities Unlock

By the end of this lesson you'll be able to:

1. Find exact values like $\cos \frac{\pi}{12}$ and $\sin \frac{5\pi}{12}$ (no calculator).
2. Simplify messy expressions such as $\sin x \cos 2x + \cos x \sin 2x$ into a single term.
3. Prove new identities and evaluate composite expressions like $\sin(\arcsin a + \arccos b)$.

There are four core formulas — two each for cosine and sine. We'll build them one family at a time, starting with cosine in Part 2.

🔑 Roadmap: cosine first (Part 2), because the sine formulas are derived from it. Part 4 then uses the sine sum formula to rewrite $a\sin x + b\cos x$ as a single sinusoid.

Why It Matters 🎯

🔺 Sum and Difference Identities

Part 2 of 7 — The Cosine Formulas

🔑 The two cosine identities: $\cos(A - B) = \cos A \cos B + \sin A \sin B$ $\cos(A + B) = \cos A \cos B - \sin A \sin B$

Reading the Pattern

Both cosine formulas have the same pieces — only the middle sign and the structure differ:

$\cos(A \pm B) = \cos A \cos B \;\mp\; \sin A \sin B$

⚠️ The sign FLIPS. A plus inside ($A+B$) gives a minus in the formula; a minus inside ($A-B$) gives a plus. This "opposite sign" rule is unique to cosine.

How to remember the terms

• Cosine formulas are cosine–cosine, sine–sine (the functions match within each product).
• "Cosine keeps company": $\cos\cos$ comes first, then $\sin\sin$.

Concept Check 🎯

Where It Comes From (the difference formula)

You don't need to memorize the proof, but seeing it builds trust. Place points $P = (\cos A, \sin A)$ and $Q = (\cos B, \sin B)$ on the unit circle. The distance $PQ$ can be computed two ways:

By the distance formula: $PQ^2 = (\cos A - \cos B)^2 + (\sin A - \sin B)^2 = 2 - 2(\cos A\cos B + \sin A\sin B)$

By rotating the picture: rotate both points clockwise by angle $B$. Rotation does not change distances, $Q$ lands on $(1, 0)$, and $P$ lands on $(\cos(A - B), \sin(A - B))$. So $PQ^2 = (\cos(A - B) - 1)^2 + \sin^2(A - B) = 2 - 2\cos(A - B)$

Setting the two equal and cancelling:

$\cos(A - B) = \cos A\cos B + \sin A\sin B$

💡 Replace $B$ with $-B$ (and use $\cos(-B)=\cos B$, $\sin(-B)=-\sin B$) to get the $\cos(A+B)$ version.

Substitute the Values 🧮

You're expanding $\cos \frac{5\pi}{12} = \cos(\frac{\pi}{4} + \frac{\pi}{6}) = \cos \frac{\pi}{4}\cos \frac{\pi}{6} - \sin \frac{\pi}{4}\sin \frac{\pi}{6}$. Enter each special value as a decimal rounded to 3 places.

1) $\cos \frac{\pi}{6} = \,?$ 2) $\sin \frac{\pi}{6} = \,?$ 3) $\cos \frac{5\pi}{12}$ overall $= \,?$ (3 decimals)

First Exact Value: $\cos \frac{\pi}{12}$

Write $\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6}$ and use the difference formula:

$\cos \frac{\pi}{12} = \cos(\frac{\pi}{4} - \frac{\pi}{6}) = \cos \frac{\pi}{4}\cos \frac{\pi}{6} + \sin \frac{\pi}{4}\sin \frac{\pi}{6}$

Substitute special values:

$= \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} + \frac{\sqrt2}{2}\cdot\frac{1}{2} = \frac{\sqrt6}{4} + \frac{\sqrt2}{4} = \frac{\sqrt6 + \sqrt2}{4}$

✅ Check: $\dfrac{\sqrt6 + \sqrt2}{4} \approx \dfrac{2.449 + 1.414}{4} \approx 0.966$, and a calculator gives $\cos \frac{\pi}{12} \approx 0.966$. ✓

Expand Each Cosine 🔽

Match each expression to its correct expansion.

🔺 Sum and Difference Identities

Part 3 of 7 — The Sine Formulas

🔑 The two sine identities: $\sin(A + B) = \sin A \cos B + \cos A \sin B$ $\sin(A - B) = \sin A \cos B - \cos A \sin B$

Reading the Pattern

The sine formulas use mixed products ($\sin\cos$ and $\cos\sin$), and the sign matches the inside:

$\sin(A \pm B) = \sin A\cos B \;\pm\; \cos A\sin B$

⚠️ Sine keeps its sign — a plus inside gives a plus, a minus inside gives a minus. This is the opposite behavior from cosine (which flips). Don't mix them up!

Memory hook

Sine is the "mixed, same-sign" formula:

• Mixed functions in each product: $\sin\cos + \cos\sin$.
• Same sign as inside the parentheses.

Concept Check 🎯

Worked Example: $\sin \frac{5\pi}{12}$

Write $\frac{5\pi}{12} = \frac{\pi}{4} + \frac{\pi}{6}$ and use the sum formula:

$\sin \frac{5\pi}{12} = \sin \frac{\pi}{4}\cos \frac{\pi}{6} + \cos \frac{\pi}{4}\sin \frac{\pi}{6}$

$= \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} + \frac{\sqrt2}{2}\cdot\frac{1}{2} = \frac{\sqrt6}{4} + \frac{\sqrt2}{4} = \frac{\sqrt6 + \sqrt2}{4}$

💡 Cofunction sanity check: $\sin \frac{5\pi}{12} = \cos \frac{\pi}{12}$ because $\frac{5\pi}{12}$ and $\frac{\pi}{12}$ are complementary. We got the same value $\frac{\sqrt6+\sqrt2}{4}$ in Part 2 — consistent! ✓

Expand Each Sine 🔽

Match each sine expression to its expansion.

Worked Example: $\sin \frac{\pi}{12}$

Write $\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6}$ and use the difference formula:

$\sin \frac{\pi}{12} = \sin \frac{\pi}{4}\cos \frac{\pi}{6} - \cos \frac{\pi}{4}\sin \frac{\pi}{6}$

$= \frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} - \frac{\sqrt2}{2}\cdot\frac{1}{2} = \frac{\sqrt6}{4} - \frac{\sqrt2}{4} = \frac{\sqrt6 - \sqrt2}{4}$

✅ Check: $\dfrac{\sqrt6 - \sqrt2}{4} \approx \dfrac{2.449 - 1.414}{4} \approx 0.259$, and $\sin \frac{\pi}{12} \approx 0.259$. ✓

Fill the Numerator 🧮

Each exact value has the form $\dfrac{\sqrt6 \;\square\; \sqrt2}{4}$. Enter + or - for the box (type a plus or minus sign).

1) $\sin \frac{5\pi}{12} = \dfrac{\sqrt6 \;?\; \sqrt2}{4}$ 2) $\sin \frac{\pi}{12} = \dfrac{\sqrt6 \;?\; \sqrt2}{4}$ 3) $\cos \frac{\pi}{12} = \dfrac{\sqrt6 \;?\; \sqrt2}{4}$

🔺 Sum and Difference Identities

Part 4 of 7 — Combining Sine and Cosine Into One Sinusoid

🔑 Key Concept: Read the sine sum identity backwards and an expression like $a\sin x + b\cos x$ becomes a single sinusoid $R\sin(x + C)$. That equivalent form shows the amplitude and phase shift at a glance and turns hard equations into easy ones.

Tangent sum and difference identities are not assessed on the AP Precalculus exam, so this lesson uses only the sine and cosine identities.

Forward First: Expand a Shifted Sine

Expand $2\sin\!\left(x + \dfrac{\pi}{6}\right)$ with the sine sum identity:

$2\sin\!\left(x + \frac{\pi}{6}\right) = 2\left(\sin x\cos\frac{\pi}{6} + \cos x\sin\frac{\pi}{6}\right) = 2\left(\frac{\sqrt3}{2}\sin x + \frac{1}{2}\cos x\right) = \sqrt3\sin x + \cos x$

So $\sqrt3\sin x + \cos x$ is really a sine wave with amplitude $2$, shifted left $\frac{\pi}{6}$.

Backward: Match the Coefficients

In general,

$R\sin(x + C) = (R\cos C)\sin x + (R\sin C)\cos x$

To write $a\sin x + b\cos x$ in this form, match $a = R\cos C$ and $b = R\sin C$:

Concept Check 🎯

Worked Example: Rewrite, Then Solve

Solve $\sin x + \sqrt3\cos x = 1$ on $[0, 2\pi)$.

Step 1 — Rewrite. $a = 1$, $b = \sqrt3$, so $R = \sqrt{1 + 3} = 2$. Then $\cos C = \frac12$ and $\sin C = \frac{\sqrt3}{2}$, so $C = \frac{\pi}{3}$:

$\sin x + \sqrt3\cos x = 2\sin\!\left(x + \frac{\pi}{3}\right)$

Step 2 — Solve the simpler equation. $2\sin\!\left(x + \frac{\pi}{3}\right) = 1$ means $\sin u = \frac12$ with $u = x + \frac{\pi}{3}$. Since $0 \le x < 2\pi$, the new angle runs over $\frac{\pi}{3} \le u < \frac{7\pi}{3}$. In that window, $\sin u = \frac12$ at $u = \frac{5\pi}{6}$ and $u = \frac{13\pi}{6}$.

Step 3 — Undo the shift. $x = u - \frac{\pi}{3}$:

$x = \frac{5\pi}{6} - \frac{\pi}{3} = \frac{\pi}{2} \qquad x = \frac{13\pi}{6} - \frac{\pi}{3} = \frac{11\pi}{6}$

✅ Check: at $x = \frac{\pi}{2}$, $1 + \sqrt3\cdot 0 = 1$ ✓. At $x = \frac{11\pi}{6}$, $-\frac12 + \sqrt3\cdot\frac{\sqrt3}{2} = -\frac12 + \frac32 = 1$ ✓.

⚠️ Shift the interval along with the angle. Solving $\sin u = \frac12$ on $[0, 2\pi)$ instead would give $u = \frac{\pi}{6}$, which leads to $x = -\frac{\pi}{6}$, outside the interval.

Read the Features 🔽

Use $y = \sin x + \sqrt3\cos x = 2\sin\!\left(x + \dfrac{\pi}{3}\right)$.

When a Coefficient Is Negative

Rewrite $\sin x - \cos x$. Here $a = 1$, $b = -1$, so $R = \sqrt2$, $\cos C = \frac{1}{\sqrt2}$, and $\sin C = -\frac{1}{\sqrt2}$. Cosine positive and sine negative puts $C$ in Quadrant IV: $C = -\frac{\pi}{4}$.

$\sin x - \cos x = \sqrt2\sin\!\left(x - \frac{\pi}{4}\right)$

Verify by expanding: $\sqrt2\left(\sin x\cos\frac{\pi}{4} - \cos x\sin\frac{\pi}{4}\right) = \sqrt2\left(\frac{\sqrt2}{2}\sin x - \frac{\sqrt2}{2}\cos x\right) = \sin x - \cos x$ ✓

💡 Always confirm a rewrite by expanding it with the sum or difference identity. It takes one line and catches sign errors in $C$.

Rewrite and Solve 🧮

Write $\sqrt3\sin x - \cos x$ as $R\sin(x + C)$.

1) $R = \,?$ 2) The maximum value of $y = \sqrt3\sin x - \cos x$ is $\,?$ 3) Solve $\sqrt3\sin x - \cos x = 2$ on $[0, 2\pi)$. Enter $x$ as a multiple of pi, like 3pi/4.

🔺 Sum and Difference Identities

Part 5 of 7 — Exact Values: A Repeatable Method

🔑 The 4-step recipe for any "find the exact value" problem:

1. Decompose the angle into two special angles (sum or difference).
2. Pick the right family (sin or cos) and write the formula.
3. Substitute unit-circle values.
4. Simplify, then check the sign against the quadrant.

Worked Example: $\cos \frac{7\pi}{12}$

Step 1 — Decompose: $\frac{7\pi}{12} = \frac{\pi}{3} + \frac{\pi}{4}$.

Step 2 — Formula: cosine of a sum flips the sign: $\cos \frac{7\pi}{12} = \cos \frac{\pi}{3}\cos \frac{\pi}{4} - \sin \frac{\pi}{3}\sin \frac{\pi}{4}$

Step 3 — Substitute: $= \frac{1}{2}\cdot\frac{\sqrt2}{2} - \frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2} = \frac{\sqrt2}{4} - \frac{\sqrt6}{4}$

Step 4 — Simplify: $\cos \frac{7\pi}{12} = \frac{\sqrt2 - \sqrt6}{4}$

✅ Check: $\dfrac{\sqrt2 - \sqrt6}{4} \approx \dfrac{1.414 - 2.449}{4} \approx -0.259$. Since $\frac{7\pi}{12}$ is in Quadrant II, cosine is negative — consistent. ✓

Apply the Method 🎯

Worked Example: $\sin \frac{7\pi}{12}$

Decompose: $\frac{7\pi}{12} = \frac{\pi}{3} + \frac{\pi}{4}$. Use the sine sum formula (sine keeps its sign):

$\sin \frac{7\pi}{12} = \sin \frac{\pi}{3}\cos \frac{\pi}{4} + \cos \frac{\pi}{3}\sin \frac{\pi}{4}$

$= \frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2} + \frac{1}{2}\cdot\frac{\sqrt2}{2} = \frac{\sqrt6}{4} + \frac{\sqrt2}{4} = \frac{\sqrt6 + \sqrt2}{4}$

💡 Notice: $\sin \frac{7\pi}{12} = \frac{\sqrt6+\sqrt2}{4} = \sin \frac{5\pi}{12}$. That's because $\sin(\pi - \theta) = \sin\theta$, and $\frac{7\pi}{12} = \pi - \frac{5\pi}{12}$. A nice built-in check.

Decimal Check 🧮

Compute each exact value, then enter its decimal rounded to 3 decimal places. (Use a calculator to confirm after you set up the exact form.)

1) $\cos \frac{7\pi}{12} = \dfrac{\sqrt2 - \sqrt6}{4} = \,?$ 2) $\sin \frac{\pi}{12} = \dfrac{\sqrt6 - \sqrt2}{4} = \,?$ 3) $\cos \frac{\pi}{12} = \dfrac{\sqrt6 + \sqrt2}{4} = \,?$

Don't Forget the Quadrant

Angles like $\frac{7\pi}{12}$, $\frac{11\pi}{12}$, and $\frac{13\pi}{12}$ live outside Quadrant I, so the sign of your answer matters. After you decompose and expand, sanity-check the sign against the quadrant.

⚠️ A correct expansion with a sign that contradicts the quadrant means an arithmetic slip — go back and check.

Set Up the Right Expansion 🔽

You want the exact value of $\sin \frac{11\pi}{12}$. Choose the best setup.

🔺 Sum and Difference Identities

Part 6 of 7 — Simplifying & Proving Identities

The formulas run both directions. Reading them right-to-left lets you collapse a long expression into a single trig function.

🔑 Recognition is everything: if you spot the shape $\sin A\cos B + \cos A\sin B$, you can instantly rewrite it as $\sin(A+B)$.

Collapsing Expressions

Example

$\sin \frac{5\pi}{18}\cos \frac{\pi}{9} - \cos \frac{5\pi}{18}\sin \frac{\pi}{9} = \sin(\frac{5\pi}{18} - \frac{\pi}{9}) = \sin \frac{\pi}{6} = \frac{1}{2}$

💡 This is how a "scary" expression becomes a single known value. Look for the mixed-product, opposite-order pattern that signals sine, or the matched-product pattern that signals cosine.

Recognize the Pattern 🎯

Proving an Identity

Prove: $\sin\!\left(x + \dfrac{\pi}{2}\right) = \cos x$.

Expand with the sine sum formula:

$\sin\!\left(x + \frac{\pi}{2}\right) = \sin x\cos\frac{\pi}{2} + \cos x\sin\frac{\pi}{2}$

Substitute $\cos\dfrac{\pi}{2} = 0$ and $\sin\dfrac{\pi}{2} = 1$:

$= \sin x\cdot 0 + \cos x\cdot 1 = \cos x \qquad \blacksquare$

💡 Phase shifts as identities: this is exactly why shifting a sine graph left by $\frac{\pi}{2}$ produces the cosine graph. The algebra and the picture agree.

Complete the Proof 🔽

Prove that $\cos(\pi - x) = -\cos x$. Fill each step.

The Reverse Skill in One Sentence

When you see two products being added or subtracted, ask: are the functions matched or mixed?

• Matched ($\cos\cos \pm \sin\sin$) → it's a cosine of $(A \mp B)$.
• Mixed ($\sin\cos \pm \cos\sin$) → it's a sine of $(A \pm B)$.

Then read off the two inner angles and combine them. The whole expression collapses to one term you can often evaluate exactly.

Collapse, Then Evaluate 🧮

Each expression collapses to a single trig value. Enter the exact value (use a/b for fractions; a decimal is fine where noted).

1) $\sin \frac{5\pi}{12}\cos \frac{\pi}{6} - \cos \frac{5\pi}{12}\sin \frac{\pi}{6} = \sin(\,?\cdot\pi)$, where the angle is the difference of the two given angles (between $0$ and $\frac{\pi}{2}$) — enter the coefficient of $\pi$ (e.g. 1/6). 2) That value $= \,?$ (3 decimals) 3) $\cos \frac{7\pi}{12}\cos \frac{\pi}{4} + \sin \frac{7\pi}{12}\sin \frac{\pi}{4} = \cos(\,?\cdot\pi)$, again with the angle between $0$ and $\frac{\pi}{2}$ — enter the coefficient of $\pi$.

🔺 Sum and Difference Identities

Part 7 of 7 — Applications, Mixed Practice & Exit Quiz

The hardest AP-style problems give you $\sin$ and $\cos$ of two angles without the angles themselves, and ask for the sine or cosine of their sum.

Worked Example: Given Sines & Cosines

Suppose $\sin A = \dfrac{3}{5}$ with $A$ in Quadrant I, and $\cos B = \dfrac{5}{13}$ with $B$ in Quadrant I. Find $\sin(A + B)$.

Step 1 — Find the missing pieces using Pythagorean triples:

• $A$ in QI with $\sin A = \frac35 \Rightarrow \cos A = \frac45$ (3-4-5 triangle).
• $B$ in QI with $\cos B = \frac{5}{13} \Rightarrow \sin B = \frac{12}{13}$ (5-12-13 triangle).

Step 2 — Apply the sine sum formula: $\sin(A+B) = \sin A\cos B + \cos A\sin B = \frac{3}{5}\cdot\frac{5}{13} + \frac{4}{5}\cdot\frac{12}{13}$

$= \frac{15}{65} + \frac{48}{65} = \frac{63}{65}$

⚠️ Watch the quadrants. If an angle were in QII–QIV, one of $\sin$ or $\cos$ would be negative. Always assign the sign from the quadrant before multiplying.

Your Turn (Given Values) 🧮

Use $\sin A = \dfrac{3}{5}$ ($A$ in QI) and $\cos B = \dfrac{5}{13}$ ($B$ in QI), so $\cos A = \dfrac{4}{5}$ and $\sin B = \dfrac{12}{13}$.

1) $\cos(A + B) = \cos A\cos B - \sin A\sin B = \,?$ (enter as a fraction like a/b) 2) $\sin(A - B) = \sin A\cos B - \cos A\sin B = \,?$ (fraction)

Master Reference

🔑 Three rules to never forget:

1. Cosine flips the sign; sine keeps it.
2. Cosine uses matched products; sine uses mixed.
3. Assign each missing sine or cosine its sign from the quadrant before you multiply.

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.