9.2 Sum and Difference Identities

Sum and Difference Identities

Sum and difference identities let you find exact values of trig functions at angles that aren't on the standard unit circle (like $75°$ or $\frac{7\pi}{12}$) by breaking them into angles you already know. They also serve as the foundation for double-angle formulas, half-angle formulas, and solving more complex trig equations.

Sum and Difference Formulas

The six core formulas

Notice the sign patterns here. For sine, the formula keeps the same sign as the operation (sum stays "+", difference stays "−"). For cosine, the sign flips (sum gets "−", difference gets "+"). For tangent, the numerator matches the operation's sign while the denominator flips.

Sine:

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

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

Cosine:

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

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

Tangent:

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

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

A quick way to remember: Sine formulas use a mix of Sine and Cosine ($\sin \cdot \cos$), while Cosine formulas pair like with like ($\cos \cdot \cos$ and $\sin \cdot \sin$).

Finding exact values with these formulas

The whole point is to decompose an unfamiliar angle into two familiar ones. For example, to find $\cos(75°)$:

1. Rewrite $75°$ as $45° + 30°$.

2. Apply the cosine sum formula: $\cos(45° + 30°) = \cos 45° \cos 30° - \sin 45° \sin 30°$

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

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

Common decompositions to keep in your back pocket:

• $75° = 45° + 30°$ and $15° = 45° - 30°$
• $\frac{7\pi}{12} = \frac{\pi}{3} + \frac{\pi}{4}$ and $\frac{\pi}{12} = \frac{\pi}{3} - \frac{\pi}{4}$
• $105° = 60° + 45°$ and $\frac{5\pi}{12} = \frac{\pi}{4} + \frac{\pi}{6}$

Cofunction identities from the difference formulas

Cofunction identities are actually a special case of the difference formulas. When you set $A = \frac{\pi}{2}$ and apply the difference formula, you get:

• $\sin\left(\frac{\pi}{2} - \theta\right) = \cos \theta$ and $\cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta$
• $\tan\left(\frac{\pi}{2} - \theta\right) = \cot \theta$ and $\cot\left(\frac{\pi}{2} - \theta\right) = \tan \theta$
• $\sec\left(\frac{\pi}{2} - \theta\right) = \csc \theta$ and $\csc\left(\frac{\pi}{2} - \theta\right) = \sec \theta$

Each pair involves complementary angles (angles that add to $\frac{\pi}{2}$, or $90°$). When you see an expression like $\sin\left(\frac{\pi}{2} - x\right)$ inside a larger problem, swap it for $\cos x$ right away to simplify your work.

Verification of trigonometric identities

To verify an identity means to show that both sides of an equation are equivalent for all values where they're defined. The strategy:

1. Pick the more complicated side (usually the LHS).
2. Expand any sum/difference expressions using the formulas above.
3. Apply cofunction identities or Pythagorean identities as needed.
4. Simplify using algebra (factor, combine fractions, cancel) until it matches the other side.

Example: Verify that $\sin(x + y) + \sin(x - y) = 2\sin x \cos y$.

1. Expand the LHS using sum and difference formulas:

   • $\sin(x+y) = \sin x \cos y + \cos x \sin y$
   • $\sin(x-y) = \sin x \cos y - \cos x \sin y$

2. Add them: $(\sin x \cos y + \cos x \sin y) + (\sin x \cos y - \cos x \sin y)$

3. The $\cos x \sin y$ terms cancel, leaving $2\sin x \cos y$. This matches the RHS. ✓

Solving equations with identities

When a trig equation contains compound angles, use sum/difference formulas to rewrite everything in terms of a single angle.

1. Apply the appropriate sum or difference formula to expand compound-angle terms.

2. Substitute cofunction identities if any $\frac{\pi}{2} - \theta$ expressions appear.

3. Collect terms and solve using algebra (factoring, quadratic formula, etc.).

4. Find all solutions in the given domain. Remember: sine and cosine repeat every $2\pi$, tangent and cotangent repeat every $\pi$.

5. Check your answers by plugging them back in. Factoring steps can sometimes introduce false solutions.

Simplification of complex expressions

When simplifying, look for these clues that a sum/difference formula applies:

• You see $\sin A \cos B \pm \cos A \sin B$ (that's a sine formula in disguise).
• You see $\cos A \cos B \mp \sin A \sin B$ (that's a cosine formula).
• You see a fraction with $\frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$ (tangent formula).

Working backward from the expanded form back to the compact form is just as important as expanding. Many simplification problems ask you to recognize the pattern and condense.

Relationship to the unit circle

These identities come from the geometry of the unit circle. If you place two angles $A$ and $B$ on the unit circle, the cosine difference formula can be derived by computing the distance between the two corresponding points in two different ways. This geometric origin is why the formulas work for any angle, not just acute ones.

Applying Sum and Difference Identities

Solving equations with identities

Example: Solve $\sin(x + \frac{\pi}{6}) = \frac{1}{2}$ for $0 \leq x < 2\pi$.

1. You know $\sin(\theta) = \frac{1}{2}$ when $\theta = \frac{\pi}{6}$ or $\theta = \frac{5\pi}{6}$ (plus multiples of $2\pi$).

2. Set $x + \frac{\pi}{6} = \frac{\pi}{6}$, giving $x = 0$.

3. Set $x + \frac{\pi}{6} = \frac{5\pi}{6}$, giving $x = \frac{5\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{3}$.

4. Solutions: $x = 0$ and $x = \frac{2\pi}{3}$.

Simplification of complex expressions

Example: Simplify $\tan\!\left(x + \frac{\pi}{4}\right) - \tan\!\left(x - \frac{\pi}{4}\right)$.

1. Apply the tangent sum and difference formulas:

   • $\tan\!\left(x + \frac{\pi}{4}\right) = \frac{\tan x + 1}{1 - \tan x}$
   • $\tan\!\left(x - \frac{\pi}{4}\right) = \frac{\tan x - 1}{1 + \tan x}$

2. Subtract and find a common denominator $(1 - \tan x)(1 + \tan x) = 1 - \tan^2 x$: $\frac{(\tan x + 1)(1 + \tan x) - (\tan x - 1)(1 - \tan x)}{1 - \tan^2 x}$

3. Expand the numerator:

   • $(\tan x + 1)^2 = \tan^2 x + 2\tan x + 1$
   • $(\tan x - 1)(1 - \tan x) = -(tan x - 1)^2 = -(\tan^2 x - 2\tan x + 1)$
   • Numerator: $\tan^2 x + 2\tan x + 1 + \tan^2 x - 2\tan x + 1 = 2\tan^2 x + 2 = 2(\tan^2 x + 1) = 2\sec^2 x$

4. Full expression: $\frac{2\sec^2 x}{1 - \tan^2 x}$