7.2 Sum and Difference Identities

Sum and Difference Identities

Trigonometric sum and difference identities let you find exact values of trig functions at non-standard angles by rewriting them as combinations of angles you already know (like 30°, 45°, 60°). They're also essential for simplifying complex trig expressions and proving other identities.

These identities are the foundation for double-angle and half-angle formulas you'll see next, so getting comfortable with them now pays off quickly.

more resources to help you study

practice questions

Sum and Difference Identities

Cosine sum and difference applications

The cosine identities are:

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

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

Notice the pattern: the cosine formulas use cosine-cosine and sine-sine products. The sum formula has a minus sign between them, and the difference formula has a plus sign. This is the opposite of what you might expect, so watch out.

Example: Find the exact value of $\cos(75°)$

You can write 75° as 45° + 30°, then apply the sum formula:

1. Set $A = 45°$ and $B = 30°$

2. Substitute: $\cos(75°) = \cos 45° \cos 30° - \sin 45° \sin 30°$

3. Plug in known values: $= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$

4. Simplify: $= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$

These formulas work the same way with radian measure. For instance, $\cos\!\left(\frac{5\pi}{12}\right)$ can be split into $\frac{\pi}{4} + \frac{\pi}{6}$.

Sine sum and difference usage

The sine identities are:

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

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

Here the pattern is: sine-cosine and cosine-sine products, and the sign between them matches the operation (sum gives plus, difference gives minus). This is the more intuitive one.

Example: Find the exact value of $\sin(15°)$

Write 15° as 45° − 30°:

1. Set $A = 45°$ and $B = 30°$

2. Substitute: $\sin(15°) = \sin 45° \cos 30° - \cos 45° \sin 30°$

3. Plug in known values: $= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$

4. Simplify: $= \frac{\sqrt{6} - \sqrt{2}}{4}$

The key skill is recognizing how to decompose an angle into sums or differences of special angles (multiples of 30° and 45°, or in radians, multiples of $\frac{\pi}{6}$ and $\frac{\pi}{4}$).

Tangent sum and difference simplification

The tangent identities are:

$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$

$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$

These are fractions, which makes them a bit messier to work with. The sign in the numerator matches the operation, while the sign in the denominator is flipped.

Example: Find the exact value of $\tan(75°)$

Write 75° as 45° + 30°:

1. $\tan(75°) = \frac{\tan 45° + \tan 30°}{1 - \tan 45° \tan 30°}$

2. Plug in: $= \frac{1 + \frac{\sqrt{3}}{3}}{1 - 1 \cdot \frac{\sqrt{3}}{3}} = \frac{\frac{3 + \sqrt{3}}{3}}{\frac{3 - \sqrt{3}}{3}} = \frac{3 + \sqrt{3}}{3 - \sqrt{3}}$

3. Rationalize if needed by multiplying by $\frac{3 + \sqrt{3}}{3 + \sqrt{3}}$

One thing to watch: the tangent formula is undefined when $\tan A \tan B = 1$ (because the denominator becomes zero). This happens when $A + B = 90°$, which makes sense since $\tan 90°$ is undefined.

Cofunction Identities and Proof Techniques

Cofunction identities with sum-difference formulas

Cofunction identities state that a trig function of an angle equals the co-function of its complement. Two angles are complementary when they add to 90° (or $\frac{\pi}{2}$).

• $\sin(90° - \theta) = \cos \theta$
• $\cos(90° - \theta) = \sin \theta$
• $\tan(90° - \theta) = \cot \theta$
• $\cot(90° - \theta) = \tan \theta$
• $\sec(90° - \theta) = \csc \theta$
• $\csc(90° - \theta) = \sec \theta$

You can actually prove these using the sum and difference formulas. For example, to show $\sin(90° - \theta) = \cos \theta$:

1. Apply the sine difference formula: $\sin(90° - \theta) = \sin 90° \cos \theta - \cos 90° \sin \theta$

2. Substitute known values: $= (1)\cos \theta - (0)\sin \theta$

3. Simplify: $= \cos \theta$

This connection between cofunction identities and sum/difference formulas shows up frequently in identity proofs, so it's worth understanding rather than just memorizing.

Sum-difference formulas in identity proofs

When you need to verify a trig identity, the general approach is to pick the more complex side and transform it into the simpler side.

Steps for proving identities using sum-difference formulas:

1. Identify which side looks more complicated and start there

2. Look for expressions like $\sin(A + B)$ or $\cos(A - B)$ that you can expand

3. Apply the appropriate sum or difference formula

4. Simplify using known values, Pythagorean identities, or algebraic techniques (factoring, combining fractions)

5. Continue simplifying until you reach the other side of the equation

Common proof strategies:

• Direct manipulation: Work one side until it matches the other. This is the most common approach.
• Work both sides independently: Simplify each side separately until they reach the same expression. (Some instructors require you to work only one side, so check your class expectations.)

Example: Prove that $\sin(A + B)\sin(A - B) = \sin^2 A - \sin^2 B$

1. Expand the left side using both sine formulas: $(\sin A \cos B + \cos A \sin B)(\sin A \cos B - \cos A \sin B)$

2. This is a difference of squares pattern: $= \sin^2 A \cos^2 B - \cos^2 A \sin^2 B$

3. Replace $\cos^2 B$ with $1 - \sin^2 B$ and $\cos^2 A$ with $1 - \sin^2 A$: $= \sin^2 A(1 - \sin^2 B) - (1 - \sin^2 A)\sin^2 B$

4. Distribute: $= \sin^2 A - \sin^2 A \sin^2 B - \sin^2 B + \sin^2 A \sin^2 B$

5. The middle terms cancel: $= \sin^2 A - \sin^2 B$

Connection to Double-Angle and Half-Angle Formulas

The sum and difference identities lead directly to other important formulas:

• Double-angle formulas come from setting $B = A$ in the sum formulas. For instance, $\sin(A + A) = \sin A \cos A + \cos A \sin A = 2\sin A \cos A$, giving you $\sin(2A) = 2\sin A \cos A$.
• Half-angle formulas are then derived from the double-angle formulas by solving for $\sin\!\left(\frac{\theta}{2}\right)$ or $\cos\!\left(\frac{\theta}{2}\right)$.

You'll work with these in upcoming sections, but understanding that they're built from sum and difference identities helps you reconstruct them if you forget.