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 $A = B = 30^\circ$, so $A + B = 60^\circ$.

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 $15^\circ$, $75^\circ$, or $105^\circ$? These aren't on the unit circle — yet.

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

• $15^\circ = 45^\circ - 30^\circ$
• $75^\circ = 45^\circ + 30^\circ$
• $105^\circ = 60^\circ + 45^\circ$

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 ($30^\circ, 45^\circ, 60^\circ$).

What the Identities Unlock

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

1. Find exact values like $\cos 15^\circ$ and $\tan 75^\circ$ (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 six core formulas — two each for cosine, sine, and tangent. We'll build them one family at a time, starting with cosine in Part 2.

🔑 Roadmap: cosine first (Part 2), because the sine and tangent formulas are derived from it.

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 the Law of Cosines (angle between them is $A - B$): $PQ^2 = 1^2 + 1^2 - 2(1)(1)\cos(A - B) = 2 - 2\cos(A - B)$

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 $\cos 75^\circ = \cos(45^\circ + 30^\circ) = \cos 45^\circ\cos 30^\circ - \sin 45^\circ\sin 30^\circ$. Enter each special value as a decimal rounded to 3 places.

1) $\cos 30^\circ = \,?$ 2) $\sin 30^\circ = \,?$ 3) $\cos 75^\circ$ overall $= \,?$ (3 decimals)

First Exact Value: $\cos 15^\circ$

Write $15^\circ = 45^\circ - 30^\circ$ and use the difference formula:

$\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ$

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 $\cos 15^\circ \approx 0.966$. ✓

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: $\sin 75^\circ$

Write $75^\circ = 45^\circ + 30^\circ$ and use the sum formula:

$\sin 75^\circ = \sin 45^\circ\cos 30^\circ + \cos 45^\circ\sin 30^\circ$

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

💡 Cofunction sanity check: $\sin 75^\circ = \cos 15^\circ$ because $75^\circ$ and $15^\circ$ 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: $\sin 15^\circ$

Write $15^\circ = 45^\circ - 30^\circ$ and use the difference formula:

$\sin 15^\circ = \sin 45^\circ\cos 30^\circ - \cos 45^\circ\sin 30^\circ$

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

✅ Check: $\dfrac{\sqrt6 - \sqrt2}{4} \approx \dfrac{2.449 - 1.414}{4} \approx 0.259$, and $\sin 15^\circ \approx 0.259$. ✓

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) $\sin 75^\circ = \dfrac{\sqrt6 \;?\; \sqrt2}{4}$ 2) $\sin 15^\circ = \dfrac{\sqrt6 \;?\; \sqrt2}{4}$ 3) $\cos 15^\circ = \dfrac{\sqrt6 \;?\; \sqrt2}{4}$

🔺 Sum and Difference Identities

Part 4 of 7 — The Tangent Formulas

🔑 The two tangent identities: $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}$ $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A\tan B}$

Where It Comes From

Since $\tan = \dfrac{\sin}{\cos}$, divide the sine sum formula by the cosine sum formula:

$\tan(A+B) = \frac{\sin A\cos B + \cos A\sin B}{\cos A\cos B - \sin A\sin B}$

Now divide every term (top and bottom) by $\cos A\cos B$:

$= \frac{\dfrac{\sin A}{\cos A} + \dfrac{\sin B}{\cos B}}{1 - \dfrac{\sin A}{\cos A}\cdot\dfrac{\sin B}{\cos B}} = \frac{\tan A + \tan B}{1 - \tan A\tan B}$

⚠️ Sign pattern (note the swap): the numerator sign matches the inside, but the denominator sign is the opposite. So $\tan(A+B)$ has $+$ on top and $-$ on the bottom; $\tan(A-B)$ has $-$ on top and $+$ on the bottom.

Concept Check 🎯

Worked Example: $\tan 15^\circ$

Write $15^\circ = 45^\circ - 30^\circ$. Recall $\tan 45^\circ = 1$ and $\tan 30^\circ = \dfrac{1}{\sqrt3} = \dfrac{\sqrt3}{3}$.

$\tan 15^\circ = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ\tan 30^\circ} = \frac{1 - \frac{1}{\sqrt3}}{1 + 1\cdot\frac{1}{\sqrt3}} = \frac{\;\frac{\sqrt3 - 1}{\sqrt3}\;}{\;\frac{\sqrt3 + 1}{\sqrt3}\;} = \frac{\sqrt3 - 1}{\sqrt3 + 1}$

Rationalize by multiplying top and bottom by $(\sqrt3 - 1)$:

$= \frac{(\sqrt3 - 1)^2}{(\sqrt3 + 1)(\sqrt3 - 1)} = \frac{3 - 2\sqrt3 + 1}{3 - 1} = \frac{4 - 2\sqrt3}{2} = 2 - \sqrt3$

✅ Check: $2 - \sqrt3 \approx 2 - 1.732 = 0.268$, and $\tan 15^\circ \approx 0.268$. ✓

Build the Tangent Formula 🔽

You are expanding $\tan(60^\circ + 45^\circ)$ to find $\tan 105^\circ$. Choose each piece.

Two Tangents Worth Memorizing

The two most common tangent results from this method are clean:

$\tan 15^\circ = 2 - \sqrt3 \qquad \tan 75^\circ = 2 + \sqrt3$

They are reciprocals of each other, since $15^\circ$ and $75^\circ$ are complementary and $\tan(90^\circ - \theta) = \cot\theta = \dfrac{1}{\tan\theta}$.

💡 Indeed $(2-\sqrt3)(2+\sqrt3) = 4 - 3 = 1$, confirming they multiply to $1$.

Decimal Check 🧮

Use $\tan 45^\circ = 1$, $\tan 30^\circ = \frac{1}{\sqrt3} \approx 0.577$, and $\tan 60^\circ = \sqrt3 \approx 1.732$.

1) $\tan 15^\circ = 2 - \sqrt3 = \,?$ (3 decimals) 2) $\tan 75^\circ = 2 + \sqrt3 = \,?$ (3 decimals)

🔺 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/cos/tan) and write the formula.
3. Substitute unit-circle values.
4. Simplify (and rationalize if it's a tangent).

Worked Example: $\cos 105^\circ$

Step 1 — Decompose: $105^\circ = 60^\circ + 45^\circ$.

Step 2 — Formula: cosine of a sum flips the sign: $\cos 105^\circ = \cos 60^\circ\cos 45^\circ - \sin 60^\circ\sin 45^\circ$

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: $\cos 105^\circ = \frac{\sqrt2 - \sqrt6}{4}$

✅ Check: $\dfrac{\sqrt2 - \sqrt6}{4} \approx \dfrac{1.414 - 2.449}{4} \approx -0.259$. Since $105^\circ$ is in Quadrant II, cosine is negative — consistent. ✓

Apply the Method 🎯

Worked Example: $\sin 105^\circ$

Decompose: $105^\circ = 60^\circ + 45^\circ$. Use the sine sum formula (sine keeps its sign):

$\sin 105^\circ = \sin 60^\circ\cos 45^\circ + \cos 60^\circ\sin 45^\circ$

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

💡 Notice: $\sin 105^\circ = \frac{\sqrt6+\sqrt2}{4} = \sin 75^\circ$. That's because $\sin(180^\circ - \theta) = \sin\theta$, and $105^\circ = 180^\circ - 75^\circ$. 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) $\cos 105^\circ = \dfrac{\sqrt2 - \sqrt6}{4} = \,?$ 2) $\sin 15^\circ = \dfrac{\sqrt6 - \sqrt2}{4} = \,?$ 3) $\tan 15^\circ = 2 - \sqrt3 = \,?$

Don't Forget the Quadrant

Angles like $105^\circ$, $165^\circ$, and $195^\circ$ 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 $\sin 165^\circ$. 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

$\sin 50^\circ\cos 20^\circ - \cos 50^\circ\sin 20^\circ = \sin(50^\circ - 20^\circ) = \sin 30^\circ = \frac{1}{2}$

💡 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) $\sin 70^\circ\cos 25^\circ - \cos 70^\circ\sin 25^\circ = \sin(\,?^\circ)$ — enter the angle in degrees. 2) That value $\sin 45^\circ = \,?$ (3 decimals) 3) $\cos 100^\circ\cos 40^\circ + \sin 100^\circ\sin 40^\circ = \cos(\,?^\circ)$ — enter the angle in degrees.

🔺 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. Tangent's denominator sign is opposite the inside sign.

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.