Sum and Difference Formulas
Calculate trig values for sums and differences of angles
You might think that if you know how to find $\sin 45°$ and $\sin 30°$, then finding $\sin 75°$ should be easy. After all, $75° = 45° + 30°$, so maybe $\sin 75° = \sin 45° + \sin 30°$? It seems reasonable. Addition is addition, right?
Try it. You know that $\sin 45° = \frac{\sqrt{2}}{2} \approx 0.707$ and $\sin 30° = \frac{1}{2} = 0.5$. Adding these gives about $1.207$. But wait. The sine of any angle must be between $-1$ and $1$. So something has gone terribly wrong.
Here is the uncomfortable truth: $\sin(A + B) \neq \sin A + \sin B$. Trigonometric functions do not distribute over addition the way you might hope. If this seems unfair, know that you are not alone. Every student who has ever taken trigonometry has felt the same frustration. But here is the good news: there are formulas that tell you exactly how to handle sums and differences of angles. They are not obvious, but once you learn them, you gain the power to find exact values for angles like $15°$ and $75°$ that do not appear on the unit circle.
And here is the thing: if you already know the basic trig values for $30°$, $45°$, and $60°$, you are most of the way there. The sum and difference formulas just show you how to combine what you already know.
Core Concepts
Why the Simple Approach Fails
Before diving into the formulas, let us understand why $\sin(A + B) \neq \sin A + \sin B$.
Think about what sine actually measures: the $y$-coordinate of a point on the unit circle after rotating by a certain angle. When you rotate by angle $A$ and then by angle $B$, you are not simply adding $y$-coordinates. You are performing a geometric transformation where the second rotation builds on the first in a more complex way.
Consider a concrete example. We know:
- $\sin 30° = 0.5$
- $\sin 60° = \frac{\sqrt{3}}{2} \approx 0.866$
If sine distributed over addition, then $\sin 90°$ would equal $\sin 30° + \sin 60° \approx 0.5 + 0.866 = 1.366$. But $\sin 90° = 1$, not $1.366$. The simple approach simply does not match reality.
This is why we need the sum and difference formulas: they capture the actual geometry of how angles combine.
The Sum Formulas
Here are the formulas for the sum of two angles:
$$\sin(A + B) = \sin A \cos B + \cos A \sin B$$
$$\cos(A + B) = \cos A \cos B - \sin A \sin B$$
$$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$
Take a moment to look at these. Notice that each formula involves both angles $A$ and $B$, and it mixes sine and cosine together. This mixing is the key: when you combine rotations, what matters is not just the $y$-coordinates (sines) or just the $x$-coordinates (cosines), but how they interact.
A helpful way to remember the sine formula: it has a plus sign in the middle, matching the plus sign in $\sin(A + B)$. The formula has a symmetric structure: sine-cosine plus cosine-sine.
For cosine, notice the minus sign in the middle. This might seem counterintuitive for a “sum” formula, but it arises from the geometry of rotation.
The tangent formula can actually be derived from the sine and cosine formulas (since $\tan = \frac{\sin}{\cos}$), but it is useful enough to memorize on its own.
The Difference Formulas
The difference formulas are almost identical to the sum formulas, but with the signs flipped in the middle:
$$\sin(A - B) = \sin A \cos B - \cos A \sin B$$
$$\cos(A - B) = \cos A \cos B + \sin A \sin B$$
$$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$
Notice the pattern:
- For sine, the sign in the formula matches the operation: plus for sum, minus for difference
- For cosine, the sign in the formula is opposite: minus for sum, plus for difference
- For tangent, the numerator sign matches (plus or minus), while the denominator sign is opposite
If you can remember the sum formulas, you can quickly derive the difference formulas by just flipping the appropriate signs.
Finding Exact Values for Non-Standard Angles
One of the most powerful applications of these formulas is finding exact values for angles that are not on the standard unit circle. Angles like $15°$, $75°$, $105°$, and so on can be expressed as sums or differences of angles you already know.
For example:
- $75° = 45° + 30°$
- $15° = 45° - 30°$ (or $60° - 45°$)
- $105° = 60° + 45°$
This means that if you know the exact trig values for $30°$, $45°$, and $60°$, you can find exact values for many more angles.
Let us recall those values:
| Angle | $\sin$ | $\cos$ | $\tan$ |
|---|---|---|---|
| $30°$ | $\frac{1}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{3}}{3}$ |
| $45°$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | $1$ |
| $60°$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\sqrt{3}$ |
With these values and the sum and difference formulas, a whole new set of angles becomes accessible.
Working with Radians
Everything we have said applies equally to radians. Common combinations include:
- $\frac{\pi}{12} = \frac{\pi}{4} - \frac{\pi}{6}$ (equivalent to $15°$)
- $\frac{5\pi}{12} = \frac{\pi}{4} + \frac{\pi}{6}$ (equivalent to $75°$)
- $\frac{7\pi}{12} = \frac{\pi}{3} + \frac{\pi}{4}$ (equivalent to $105°$)
The formulas work the same way regardless of whether you measure angles in degrees or radians.
Simplifying Expressions
The sum and difference formulas can also simplify expressions involving trig functions. For instance, if you encounter $\sin(x + \frac{\pi}{2})$, you can expand it:
$$\sin\left(x + \frac{\pi}{2}\right) = \sin x \cos\frac{\pi}{2} + \cos x \sin\frac{\pi}{2}$$
Since $\cos\frac{\pi}{2} = 0$ and $\sin\frac{\pi}{2} = 1$:
$$= \sin x \cdot 0 + \cos x \cdot 1 = \cos x$$
This shows that $\sin(x + \frac{\pi}{2}) = \cos x$, a cofunction identity that you might already know. The sum formulas let you derive such identities rather than memorizing them separately.
Verifying Identities
These formulas are also essential for verifying trigonometric identities. When you need to prove that two expressions are equal, expanding sums or differences using these formulas often reveals the path to the solution.
Notation and Terminology
| Term | Meaning | Example |
|---|---|---|
| Sum formula | Formula for $f(A + B)$ | $\sin(A + B) = \sin A \cos B + \cos A \sin B$ |
| Difference formula | Formula for $f(A - B)$ | $\cos(A - B) = \cos A \cos B + \sin A \sin B$ |
| Cofunction | Trig functions related by complementary angles | $\sin \theta = \cos(90° - \theta)$ |
| Exact value | A value expressed with radicals, not decimals | $\sin 75° = \frac{\sqrt{6} + \sqrt{2}}{4}$ |
Examples
Find the exact value of $\sin 75°$ using the sum formula.
Solution:
Since $75° = 45° + 30°$, we can use the sum formula for sine:
$$\sin(A + B) = \sin A \cos B + \cos A \sin B$$
Let $A = 45°$ and $B = 30°$.
Step 1: Recall the needed values:
- $\sin 45° = \frac{\sqrt{2}}{2}$
- $\cos 45° = \frac{\sqrt{2}}{2}$
- $\sin 30° = \frac{1}{2}$
- $\cos 30° = \frac{\sqrt{3}}{2}$
Step 2: Substitute into the formula: $$\sin 75° = \sin 45° \cos 30° + \cos 45° \sin 30°$$ $$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$$
Step 3: Multiply: $$= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}$$
Step 4: Combine the fractions: $$= \frac{\sqrt{6} + \sqrt{2}}{4}$$
Answer: $\sin 75° = \frac{\sqrt{6} + \sqrt{2}}{4}$
Check: Using a calculator, $\frac{\sqrt{6} + \sqrt{2}}{4} \approx 0.966$, and $\sin 75° \approx 0.966$. It matches.
Find the exact value of $\cos 75°$.
Solution:
Since $75° = 45° + 30°$, we use the sum formula for cosine:
$$\cos(A + B) = \cos A \cos B - \sin A \sin B$$
Let $A = 45°$ and $B = 30°$.
Step 1: Recall the needed values:
- $\cos 45° = \frac{\sqrt{2}}{2}$
- $\cos 30° = \frac{\sqrt{3}}{2}$
- $\sin 45° = \frac{\sqrt{2}}{2}$
- $\sin 30° = \frac{1}{2}$
Step 2: Substitute into the formula: $$\cos 75° = \cos 45° \cos 30° - \sin 45° \sin 30°$$ $$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$$
Step 3: Multiply: $$= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$$
Step 4: Combine the fractions: $$= \frac{\sqrt{6} - \sqrt{2}}{4}$$
Answer: $\cos 75° = \frac{\sqrt{6} - \sqrt{2}}{4}$
Notice: Compare this with Example 1. The answers for $\sin 75°$ and $\cos 75°$ look similar, but one has a plus and one has a minus. This makes sense because $75°$ is in the first quadrant where sine is larger than cosine (since $75° > 45°$), and indeed $\frac{\sqrt{6} + \sqrt{2}}{4} > \frac{\sqrt{6} - \sqrt{2}}{4}$.
Find the exact value of $\sin 15°$.
Solution:
We can write $15°$ as a difference: $15° = 45° - 30°$ (or $15° = 60° - 45°$; either works).
Using $15° = 45° - 30°$ and the difference formula for sine:
$$\sin(A - B) = \sin A \cos B - \cos A \sin B$$
Let $A = 45°$ and $B = 30°$.
Step 1: Recall the needed values:
- $\sin 45° = \frac{\sqrt{2}}{2}$
- $\cos 45° = \frac{\sqrt{2}}{2}$
- $\sin 30° = \frac{1}{2}$
- $\cos 30° = \frac{\sqrt{3}}{2}$
Step 2: Substitute into the formula: $$\sin 15° = \sin 45° \cos 30° - \cos 45° \sin 30°$$ $$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$$
Step 3: Multiply: $$= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$$
Step 4: Combine: $$= \frac{\sqrt{6} - \sqrt{2}}{4}$$
Answer: $\sin 15° = \frac{\sqrt{6} - \sqrt{2}}{4}$
Observation: Notice that $\sin 15° = \cos 75°$. This is not a coincidence! Since $15°$ and $75°$ are complementary angles (they add to $90°$), their sine and cosine values are swapped. The sum and difference formulas naturally produce this relationship.
Simplify the expression $\sin\left(x + \frac{\pi}{2}\right)$.
Solution:
We apply the sum formula for sine:
$$\sin(A + B) = \sin A \cos B + \cos A \sin B$$
Let $A = x$ and $B = \frac{\pi}{2}$.
Step 1: Recall the values for $\frac{\pi}{2}$ (which is $90°$):
- $\sin\frac{\pi}{2} = 1$
- $\cos\frac{\pi}{2} = 0$
Step 2: Substitute into the formula: $$\sin\left(x + \frac{\pi}{2}\right) = \sin x \cos\frac{\pi}{2} + \cos x \sin\frac{\pi}{2}$$ $$= \sin x \cdot 0 + \cos x \cdot 1$$
Step 3: Simplify: $$= 0 + \cos x = \cos x$$
Answer: $\sin\left(x + \frac{\pi}{2}\right) = \cos x$
What this means: Shifting a sine function to the left by $\frac{\pi}{2}$ (or $90°$) gives you a cosine function. This is the cofunction relationship in action. Graphically, the sine curve shifted left by $90°$ perfectly overlaps the cosine curve.
Verify that $\sin(x + y) + \sin(x - y) = 2\sin x \cos y$.
Solution:
To verify an identity, we start with one side and show it equals the other. Let us expand the left side using the sum and difference formulas.
Step 1: Expand $\sin(x + y)$ using the sum formula: $$\sin(x + y) = \sin x \cos y + \cos x \sin y$$
Step 2: Expand $\sin(x - y)$ using the difference formula: $$\sin(x - y) = \sin x \cos y - \cos x \sin y$$
Step 3: Add these two expressions: $$\sin(x + y) + \sin(x - y)$$ $$= (\sin x \cos y + \cos x \sin y) + (\sin x \cos y - \cos x \sin y)$$
Step 4: Combine like terms: $$= \sin x \cos y + \cos x \sin y + \sin x \cos y - \cos x \sin y$$
Notice that $\cos x \sin y$ and $-\cos x \sin y$ cancel: $$= \sin x \cos y + \sin x \cos y$$ $$= 2\sin x \cos y$$
Conclusion: We have shown that $\sin(x + y) + \sin(x - y) = 2\sin x \cos y$.
Why this identity is useful: This is one of the “sum-to-product” identities that lets you convert a sum of sines into a product. Such identities are valuable in calculus and signal processing.
If $\sin A = \frac{3}{5}$ and $\cos B = \frac{12}{13}$, where both $A$ and $B$ are in Quadrant I, find $\sin(A + B)$.
Solution:
To use the sum formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$, we need all four values: $\sin A$, $\cos A$, $\sin B$, and $\cos B$. We are given $\sin A$ and $\cos B$, so we need to find $\cos A$ and $\sin B$.
Step 1: Find $\cos A$.
Since $A$ is in Quadrant I, $\cos A$ is positive. Using the Pythagorean identity: $$\sin^2 A + \cos^2 A = 1$$ $$\left(\frac{3}{5}\right)^2 + \cos^2 A = 1$$ $$\frac{9}{25} + \cos^2 A = 1$$ $$\cos^2 A = 1 - \frac{9}{25} = \frac{16}{25}$$ $$\cos A = \frac{4}{5}$$ (positive because Quadrant I)
Step 2: Find $\sin B$.
Since $B$ is in Quadrant I, $\sin B$ is positive. Using the Pythagorean identity: $$\sin^2 B + \cos^2 B = 1$$ $$\sin^2 B + \left(\frac{12}{13}\right)^2 = 1$$ $$\sin^2 B + \frac{144}{169} = 1$$ $$\sin^2 B = 1 - \frac{144}{169} = \frac{25}{169}$$ $$\sin B = \frac{5}{13}$$ (positive because Quadrant I)
Step 3: Apply the sum formula.
$$\sin(A + B) = \sin A \cos B + \cos A \sin B$$ $$= \frac{3}{5} \cdot \frac{12}{13} + \frac{4}{5} \cdot \frac{5}{13}$$ $$= \frac{36}{65} + \frac{20}{65}$$ $$= \frac{56}{65}$$
Answer: $\sin(A + B) = \frac{56}{65}$
Note: You might recognize $3$-$4$-$5$ and $5$-$12$-$13$ as Pythagorean triples. When you see these ratios in trigonometry problems, you can often use the third number from the triple instead of going through the full calculation.
Key Properties and Rules
The Six Formulas
Sum Formulas: $$\sin(A + B) = \sin A \cos B + \cos A \sin B$$ $$\cos(A + B) = \cos A \cos B - \sin A \sin B$$ $$\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$$
Difference Formulas: $$\sin(A - B) = \sin A \cos B - \cos A \sin B$$ $$\cos(A - B) = \cos A \cos B + \sin A \sin B$$ $$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$
Memory Aids
- Sine formulas have a mix of sine and cosine: “sine-cosine, cosine-sine”
- Cosine formulas keep functions together: “cosine-cosine, sine-sine”
- The sign in the middle of sine matches the operation (sum has plus, difference has minus)
- The sign in the middle of cosine is opposite the operation (sum has minus, difference has plus)
Common Mistakes to Avoid
- Forgetting to use the formula at all: $\sin(45° + 30°) \neq \sin 45° + \sin 30°$. Always use the sum formula.
- Mixing up the signs: Remember that cosine formulas have the opposite sign from what you might expect.
- Forgetting a term: The sine formula has two terms ($\sin A \cos B + \cos A \sin B$). Do not leave one out.
- Wrong angle in the denominator: For tangent, the denominator has the product $\tan A \tan B$, not a sum.
Derived Identities
From the sum and difference formulas, you can derive many other identities:
Cofunction identities: $$\sin\left(\frac{\pi}{2} - \theta\right) = \cos \theta$$ $$\cos\left(\frac{\pi}{2} - \theta\right) = \sin \theta$$
Sum-to-product formulas (like the one verified in Example 5): $$\sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}$$ $$\sin x - \sin y = 2\cos\frac{x+y}{2}\sin\frac{x-y}{2}$$
Real-World Applications
Phase Shifts in Waves
When physicists and engineers describe waves, they often write them as $A\sin(\omega t + \phi)$, where $\phi$ is a phase shift. Using the sum formula, this can be rewritten as: $$A\sin(\omega t + \phi) = A(\sin \omega t \cos \phi + \cos \omega t \sin \phi)$$ $$= (A\cos\phi)\sin\omega t + (A\sin\phi)\cos\omega t$$
This shows that any phase-shifted sine wave can be expressed as a combination of a sine and cosine wave, which is useful for analyzing circuits and signals.
Combining Sound Waves (Interference)
When two sound waves of slightly different frequencies combine, you get “beats.” The mathematical analysis uses sum-to-product formulas derived from the sum and difference identities. If you hear two notes close in pitch, the resulting volume fluctuation is described by: $$\sin(2\pi f_1 t) + \sin(2\pi f_2 t) = 2\sin\left(2\pi\frac{f_1+f_2}{2}t\right)\cos\left(2\pi\frac{f_1-f_2}{2}t\right)$$
The cosine factor creates the “beating” effect you hear when tuning a musical instrument.
Rotation Matrices in Graphics
In computer graphics and game development, rotating a point $(x, y)$ by an angle $\theta$ uses the formulas: $$x’ = x\cos\theta - y\sin\theta$$ $$y’ = x\sin\theta + y\cos\theta$$
These are exactly the sum formulas in disguise! The new coordinates come from treating the original coordinates as sine and cosine values and applying the rotation.
AC Circuit Analysis
In electrical engineering, alternating current (AC) circuits involve voltages and currents that vary sinusoidally. When analyzing circuits with components that shift the phase between voltage and current, the sum and difference formulas are essential for calculating power and understanding interference between signals.
Self-Test Problems
Problem 1: Find the exact value of $\sin 105°$.
Show Answer
Write $105° = 60° + 45°$ and apply the sum formula:
$$\sin 105° = \sin 60° \cos 45° + \cos 60° \sin 45°$$ $$= \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} + \frac{1}{2} \cdot \frac{\sqrt{2}}{2}$$ $$= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}$$ $$= \frac{\sqrt{6} + \sqrt{2}}{4}$$
Problem 2: Find the exact value of $\cos 15°$.
Show Answer
Write $15° = 45° - 30°$ and apply the difference formula:
$$\cos 15° = \cos 45° \cos 30° + \sin 45° \sin 30°$$ $$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$$ $$= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}$$ $$= \frac{\sqrt{6} + \sqrt{2}}{4}$$
Problem 3: Simplify $\cos(x - \frac{\pi}{2})$.
Show Answer
Using the difference formula for cosine:
$$\cos\left(x - \frac{\pi}{2}\right) = \cos x \cos\frac{\pi}{2} + \sin x \sin\frac{\pi}{2}$$
Since $\cos\frac{\pi}{2} = 0$ and $\sin\frac{\pi}{2} = 1$:
$$= \cos x \cdot 0 + \sin x \cdot 1 = \sin x$$
Problem 4: Find $\tan 75°$ using the sum formula.
Show Answer
Write $75° = 45° + 30°$ and use the sum formula for tangent:
$$\tan 75° = \frac{\tan 45° + \tan 30°}{1 - \tan 45° \tan 30°}$$
Since $\tan 45° = 1$ and $\tan 30° = \frac{\sqrt{3}}{3}$:
$$= \frac{1 + \frac{\sqrt{3}}{3}}{1 - 1 \cdot \frac{\sqrt{3}}{3}}$$ $$= \frac{\frac{3 + \sqrt{3}}{3}}{\frac{3 - \sqrt{3}}{3}}$$ $$= \frac{3 + \sqrt{3}}{3 - \sqrt{3}}$$
Rationalize by multiplying by $\frac{3 + \sqrt{3}}{3 + \sqrt{3}}$:
$$= \frac{(3 + \sqrt{3})^2}{(3 - \sqrt{3})(3 + \sqrt{3})} = \frac{9 + 6\sqrt{3} + 3}{9 - 3} = \frac{12 + 6\sqrt{3}}{6} = 2 + \sqrt{3}$$
Problem 5: Verify that $\cos(x + y) - \cos(x - y) = -2\sin x \sin y$.
Show Answer
Expand both terms:
$$\cos(x + y) = \cos x \cos y - \sin x \sin y$$ $$\cos(x - y) = \cos x \cos y + \sin x \sin y$$
Subtract: $$\cos(x + y) - \cos(x - y)$$ $$= (\cos x \cos y - \sin x \sin y) - (\cos x \cos y + \sin x \sin y)$$ $$= \cos x \cos y - \sin x \sin y - \cos x \cos y - \sin x \sin y$$ $$= -2\sin x \sin y$$
Problem 6: If $\cos A = \frac{5}{13}$ with $A$ in Quadrant I, and $\sin B = \frac{4}{5}$ with $B$ in Quadrant II, find $\cos(A + B)$.
Show Answer
First, find the missing values.
For angle $A$ (Quadrant I, so $\sin A > 0$): $$\sin^2 A = 1 - \cos^2 A = 1 - \frac{25}{169} = \frac{144}{169}$$ $$\sin A = \frac{12}{13}$$
For angle $B$ (Quadrant II, so $\cos B < 0$): $$\cos^2 B = 1 - \sin^2 B = 1 - \frac{16}{25} = \frac{9}{25}$$ $$\cos B = -\frac{3}{5}$$ (negative in Quadrant II)
Now apply the sum formula: $$\cos(A + B) = \cos A \cos B - \sin A \sin B$$ $$= \frac{5}{13} \cdot \left(-\frac{3}{5}\right) - \frac{12}{13} \cdot \frac{4}{5}$$ $$= -\frac{15}{65} - \frac{48}{65}$$ $$= -\frac{63}{65}$$
Problem 7: Find the exact value of $\sin\frac{7\pi}{12}$.
Show Answer
Note that $\frac{7\pi}{12} = \frac{3\pi}{12} + \frac{4\pi}{12} = \frac{\pi}{4} + \frac{\pi}{3}$.
Using the sum formula: $$\sin\frac{7\pi}{12} = \sin\frac{\pi}{4}\cos\frac{\pi}{3} + \cos\frac{\pi}{4}\sin\frac{\pi}{3}$$ $$= \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2}$$ $$= \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}$$ $$= \frac{\sqrt{2} + \sqrt{6}}{4}$$
Summary
-
The naive approach fails: $\sin(A + B) \neq \sin A + \sin B$. Trig functions do not distribute over addition.
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Sum formulas tell you how to find trig values for $(A + B)$:
- $\sin(A + B) = \sin A \cos B + \cos A \sin B$
- $\cos(A + B) = \cos A \cos B - \sin A \sin B$
- $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
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Difference formulas are the same but with flipped signs in the middle:
- $\sin(A - B) = \sin A \cos B - \cos A \sin B$
- $\cos(A - B) = \cos A \cos B + \sin A \sin B$
- $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$
-
Finding exact values: Express non-standard angles as sums or differences of $30°$, $45°$, or $60°$, then apply the formulas. For example, $15° = 45° - 30°$ and $75° = 45° + 30°$.
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Simplifying expressions: The formulas can reduce expressions like $\sin(x + \frac{\pi}{2})$ to simpler forms (in this case, $\cos x$).
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Verifying identities: Many trig identities can be proven by expanding sums and differences using these formulas.
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Real-world applications include analyzing phase shifts in waves, understanding sound interference, performing rotations in computer graphics, and calculating AC circuit behavior.
The sum and difference formulas are among the most frequently used tools in trigonometry. They open up a world of exact values beyond the standard unit circle angles and provide the foundation for many other identities you will encounter. Master these six formulas, and you will have a powerful toolkit for solving a wide variety of problems.