Precalculus Quiz: Proving Angle Addition Subtraction Formulas

What this quiz covers

This quiz focuses on Proving Angle Addition Subtraction Formulas, giving you a quick way to practice the rules, question types, and explanations that matter most for Precalculus.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\cos(75°) = \cos(90° - 15°) = \cos(90°)\cos(15°) + \sin(90°)\sin(15°)$
2. $\cos(75°) = \cos(120° - 45°) = \cos(120°)\cos(45°) + \sin(120°)\sin(45°)$
3. $\cos(75°) = \cos(45° + 30°) = \cos(45°)\cos(30°) - \sin(45°)\sin(30°)$ (correct answer)
4. $\cos(75°) = \cos(135° - 60°) = \cos(135°)\cos(60°) + \sin(135°)\sin(60°)$

Explanation: When you encounter problems asking you to verify trigonometric values using angle formulas, look for angle combinations that break down into familiar reference angles (30°, 45°, 60°) whose exact trigonometric values you know by heart. Let's verify the given result by testing option C: $\cos(75°) = \cos(45° + 30°)$. Using the cosine addition formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$: $\cos(75°) = \cos(45°)\cos(30°) - \sin(45°)\sin(30°)$ $= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$ $= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$ This matches the given result perfectly. Option A uses $\cos(90° - 15°)$, but this becomes $\sin(15°)$ by the cofunction identity, not the cosine addition formula shown. Option B uses $\cos(120° - 45°)$ with angles whose values we know, but $\cos(120°) = -\frac{1}{2}$ and $\sin(120°) = \frac{\sqrt{3}}{2}$, which won't yield the correct result when calculated. Option D uses $\cos(135° - 60°)$, but $\cos(135°) = -\frac{\sqrt{2}}{2}$, and the negative value will prevent getting the positive result we need. Study tip: When verifying trigonometric identities, always choose angle combinations that use 30°, 45°, and 60° since you can calculate their exact values without a calculator. Avoid angles that introduce negative values unless the final result should be negative.

Question 2

1. $x = 90°$, because $\sin 90° = 1$ and $\cos 90° = 0$, eliminating terms and simplifying the expression
2. $x = 75°$, because this creates $\sin(75° + 60°) = \sin(135°)$, which has a known exact value
3. $x = 30°$, because $30°$ and $60°$ are complementary angles, which simplifies the trigonometric calculations
4. $x = 45°$, because both $\sin 45°$ and $\cos 45°$ have exact radical expressions that simplify calculations (correct answer)

Explanation: This question tests your understanding of angle addition formulas and strategic thinking in trigonometry. The given equation is actually the sine addition formula: $\sin(x + 60°) = \sin x \cos 60° + \cos x \sin 60°$. To find $\sin(105°)$, you need to choose an $x$ value that makes $x + 60° = 105°$, which means $x = 45°$. Choice D is correct because when $x = 45°$, the equation becomes $\sin(105°) = \sin 45° \cos 60° + \cos 45° \sin 60°$. Since $\sin 45° = \cos 45° = \frac{\sqrt{2}}{2}$, $\cos 60° = \frac{1}{2}$, and $\sin 60° = \frac{\sqrt{3}}{2}$, you can substitute these exact values to get $\sin(105°) = \frac{\sqrt{2}}{2} \cdot \frac{1}{2} + \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{2} + \sqrt{6}}{4}$. Choice A is incorrect because $x = 90°$ gives you $\sin(150°)$, not $\sin(105°)$. Choice B makes the same error—$x = 75°$ produces $\sin(135°)$, which isn't your target angle. Choice C is wrong because while $30°$ and $60°$ are complementary, using $x = 30°$ gives you $\sin(90°) = 1$, not $\sin(105°)$. Remember: when using addition formulas to find specific trigonometric values, always work backwards from your target angle to determine what substitution you need. Don't get distracted by angles that seem "nice" but don't lead to your desired result.

Question 3

1. $\sin(A+B)=\sin(A)+\sin(B)$
2. $\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)$ (correct answer)
3. $\sin(A+B)=\sin(A)\sin(B)+\cos(A)\cos(B)$
4. $\sin(A+B)=\sin(A)\cos(B)-\cos(A)\sin(B)$

Explanation: This question tests understanding of the angle addition and subtraction formulas for sine. The sine addition formula is sin(A + B) = sin(A)cos(B) + cos(A)sin(B), which shows that sine of a sum is NOT simply sin(A) + sin(B) but requires cross terms involving both sine and cosine of each angle. The formula sin(A + B) = sin(A)cos(B) + cos(A)sin(B) directly matches the required addition formula. Choice B is correct because it correctly states the formula with proper signs. Choice A incorrectly claims sin(A + B) = sin(A) + sin(B), missing the essential cross terms that make the actual formula sin(A + B) = sin(A)cos(B) + cos(A)sin(B). Key to angle addition formulas: memorize that both sine and cosine formulas involve four terms (two cross products), with sine having + for addition and - for subtraction, while cosine has - for addition and + for subtraction (opposite signs). Common error: students try sin(A + B) = sin(A) + sin(B), but you can quickly verify this is wrong by trying A = B = 45°: sin(90°) = 1 but sin(45°) + sin(45°) = √2/2 + √2/2 = √2 ≠ 1.

Question 4

1. $\sin(2\theta)=\sin^2(\theta)+\cos^2(\theta)$
2. $\sin(2\theta)=\sin(\theta)\cos(\theta)$
3. $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ (correct answer)
4. $\sin(2\theta)=2\sin(\theta)+2\cos(\theta)$

Explanation: This question tests understanding of the angle addition formulas for sine. The sine addition formula is sin(A + B) = sin(A)cos(B) + cos(A)sin(B), which shows that sine of a sum is NOT simply sin(A) + sin(B) but requires cross terms involving both sine and cosine of each angle. Applying the sine addition formula with A = B = θ gives sin(2θ) = sin(θ + θ) = sin(θ)cos(θ) + cos(θ)sin(θ) = 2 sin(θ) cos(θ). Choice C is correct because it correctly states the formula with proper signs. Choice B makes an arithmetic error in the evaluation, omitting the factor of 2 when combining the identical terms. These formulas are fundamental: they cannot be derived from simpler trig properties but must be proven using geometry, the unit circle, or other methods, and they serve as the foundation for proving many other trig identities including double angle and half angle formulas. Key to angle addition formulas: memorize that both sine and cosine formulas involve four terms (two cross products), with sine having + for addition and - for subtraction, while cosine has - for addition and + for subtraction (opposite signs).

Question 5

1. $\dfrac{\sqrt{6}+\sqrt{2}}{4}$ (correct answer)
2. $\dfrac{\sqrt{6}-\sqrt{2}}{4}$
3. $\dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}$
4. $\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}$

Explanation: This question tests understanding of the angle subtraction formula for cosine. The cosine addition formula is cos(A + B) = cos(A)cos(B) - sin(A)sin(B), and the subtraction formula is cos(A - B) = cos(A)cos(B) + sin(A)sin(B), with the key difference being the sign between the two product terms (minus for addition, plus for subtraction). Using cos(A - B) = cos(A)cos(B) + sin(A)sin(B) with A = 45°, B = 30°, we calculate cos(45°)cos(30°) + sin(45°)sin(30°) = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6 + √2)/4. Choice A is correct because it correctly applies the cosine subtraction formula with the proper positive sign between terms. Choice B incorrectly has (√6 - √2)/4, which would result from using the cosine addition formula instead of subtraction, mixing up the sign convention. Key to angle addition formulas: memorize that both sine and cosine formulas involve four terms (two cross products), with sine having + for addition and - for subtraction, while cosine has - for addition and + for subtraction (opposite signs).

Question 6

1. $2\sin(x)\cos\left(\dfrac{\pi}{4}\right)$ (correct answer)
2. $2\cos(x)\sin\left(\dfrac{\pi}{4}\right)$
3. $\sin(x)+\sin\left(\dfrac{\pi}{4}\right)$
4. $2\sin(x)\sin\left(\dfrac{\pi}{4}\right)$

Explanation: This question tests understanding of the angle addition and subtraction formulas for sine. The sine addition formula is sin(A + B) = sin(A)cos(B) + cos(A)sin(B), which shows that sine of a sum is NOT simply sin(A) + sin(B) but requires cross terms involving both sine and cosine of each angle. Expanding sin(x + π/4) + sin(x - π/4) gives [sin(x)cos(π/4) + cos(x)sin(π/4)] + [sin(x)cos(π/4) - cos(x)sin(π/4)] = 2 sin(x) cos(π/4), since the cos(x) terms cancel. Choice A is correct because it properly substitutes the angle values and simplifies accurately. Choice C incorrectly claims sin(A + B) = sin(A) + sin(B), missing the essential cross terms that make the actual formula sin(A + B) = sin(A)cos(B) + cos(A)sin(B). Key to angle addition formulas: memorize that both sine and cosine formulas involve four terms (two cross products), with sine having + for addition and - for subtraction, while cosine has - for addition and + for subtraction (opposite signs). To use these formulas for exact values, break unfamiliar angles into sums or differences of standard angles (like 75° = 45° + 30° or 15° = 45° - 30°), then apply the formula with the known exact trig values.

Question 7

1. $\dfrac{\sqrt{6}-\sqrt{2}}{4}$ (correct answer)
2. $\dfrac{\sqrt{6}+\sqrt{2}}{4}$
3. $\dfrac{\sqrt{2}}{4}$
4. $\dfrac{1}{2}$

Explanation: This question tests understanding of the angle addition and subtraction formulas for sine. The sine addition formula is sin(A + B) = sin(A)cos(B) + cos(A)sin(B), which shows that sine of a sum is NOT simply sin(A) + sin(B) but requires cross terms involving both sine and cosine of each angle. To find sin(15°) where 15° = 45° - 30°, we substitute into sin(45° - 30°) = sin(45°)cos(30°) - cos(45°)sin(30°), giving (√2/2)(√3/2) - (√2/2)(1/2) = √6/4 - √2/4 = (√6 - √2)/4. Choice A is correct because it properly substitutes the angle values and simplifies accurately. Choice B has the wrong sign between the terms, using plus when the formula for sine subtraction requires minus. To use these formulas for exact values, break unfamiliar angles into sums or differences of standard angles (like 75° = 45° + 30° or 15° = 45° - 30°), then apply the formula with the known exact trig values. Key to angle addition formulas: memorize that both sine and cosine formulas involve four terms (two cross products), with sine having + for addition and - for subtraction, while cosine has - for addition and + for subtraction (opposite signs).

Question 8

1. $\tan(A+B)=\dfrac{\tan(A)+\tan(B)}{1+\tan(A)\tan(B)}$
2. $\tan(A+B)=\dfrac{\tan(A)-\tan(B)}{1-\tan(A)\tan(B)}$
3. $\tan(A+B)=\dfrac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$ (correct answer)
4. $\tan(A+B)=\tan(A)+\tan(B)$

Explanation: This question tests understanding of the angle addition and subtraction formulas for tangent. The tangent addition formula is tan(A + B) = (tan(A) + tan(B))/(1 - tan(A)tan(B)), which shows that tangent of a sum involves both a numerator (sum of tangents) and a denominator (1 minus their product). Applying tan(A + B) = (tan(A) + tan(B))/(1 - tan(A)tan(B)), we see that the denominator uses a minus sign for the addition formula, distinguishing it from the subtraction version. Choice C is correct because it correctly states the formula with proper signs. Choice D incorrectly claims tan(A + B) = tan(A) + tan(B), missing the essential denominator that makes the actual formula tan(A + B) = (tan(A) + tan(B))/(1 - tan(A)tan(B)). For tangent, remember the formulas have fractions: tan(A + B) = (tan(A) + tan(B))/(1 - tan(A)tan(B)), with the sign in the denominator opposite to the numerator (minus for addition, plus for subtraction). Common error: students try tan(A + B) = tan(A) + tan(B), but you can quickly verify this is wrong by trying A = B = 45°: tan(90°) is undefined but tan(45°) + tan(45°) = 1 + 1 = 2, which is finite.

Question 9

1. The distributive property, since we can factor out the common factor of $2$ from both terms
2. The commutative property, since $\sin\theta\cos\theta = \cos\theta\sin\theta$, allowing us to combine like terms (correct answer)
3. The associative property, since we can group the terms $\sin\theta\cos\theta$ and $\cos\theta\sin\theta$ together
4. The identity property, since adding $\sin\theta\cos\theta$ to itself preserves the original expression

Explanation: When working with trigonometric identities and algebraic manipulations, you need to identify which fundamental properties of real numbers justify each step in your reasoning. Let's examine what happens in the final step: $\sin\theta\cos\theta + \cos\theta\sin\theta = 2\sin\theta\cos\theta$. The key insight is recognizing that $\sin\theta\cos\theta$ and $\cos\theta\sin\theta$ are actually the same expression. Since multiplication of real numbers is commutative, $\sin\theta\cos\theta = \cos\theta\sin\theta$. This means you're adding two identical terms: $\sin\theta\cos\theta + \sin\theta\cos\theta$, which equals $2\sin\theta\cos\theta$. The commutative property is what allows you to see these as like terms that can be combined. Choice A incorrectly describes the distributive property, which would involve factoring out a common factor from different terms—but we're not factoring here, we're combining like terms. Choice C mentions the associative property, which deals with how we group terms in addition or multiplication, not with recognizing that two products are equal. Choice D refers to the identity property, but adding a term to itself doesn't preserve the original expression—it doubles it. The correct answer is B because the commutative property of multiplication is what allows us to recognize that $\cos\theta\sin\theta = \sin\theta\cos\theta$, making them like terms. Study tip: When simplifying algebraic expressions involving products, always check if the commutative property reveals hidden like terms that can be combined.

Question 10

1. $\dfrac{36}{65}$
2. $\dfrac{56}{65}$ (correct answer)
3. $\dfrac{16}{65}$
4. $\dfrac{9}{25}+\dfrac{25}{169}$

Explanation: This question tests understanding of the angle addition and subtraction formulas for sine. The sine addition formula is sin(A + B) = sin(A)cos(B) + cos(A)sin(B), which shows that sine of a sum is NOT simply sin(A) + sin(B) but requires cross terms involving both sine and cosine of each angle. To find sin(A + B) where sin(A) = 3/5, cos(A) = 4/5, sin(B) = 5/13, cos(B) = 12/13, we substitute into sin(A + B) = sin(A)cos(B) + cos(A)sin(B), giving (3/5)(12/13) + (4/5)(5/13) = 36/65 + 20/65 = 56/65. Choice B is correct because it properly substitutes the angle values and simplifies accurately. Choice A makes an arithmetic error in the evaluation, calculating only the first product 36/65 instead of adding both products for 56/65. Key to angle addition formulas: memorize that both sine and cosine formulas involve four terms (two cross products), with sine having + for addition and - for subtraction, while cosine has - for addition and + for subtraction (opposite signs). Common error: students try sin(A + B) = sin(A) + sin(B), but you can quickly verify this is wrong by trying A = B = 45°: sin(90°) = 1 but sin(45°) + sin(45°) = √2/2 + √2/2 = √2 ≠ 1.

Question 11

1. $\cos(A-B)=\cos(A)\cos(B)-\sin(A)\sin(B)$
2. $\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)$ (correct answer)
3. $\cos(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)$
4. $\cos(A-B)=\cos(A)-\cos(B)$

Explanation: This question tests understanding of the angle subtraction formula for cosine. The cosine addition formula is cos(A + B) = cos(A)cos(B) - sin(A)sin(B), and the subtraction formula is cos(A - B) = cos(A)cos(B) + sin(A)sin(B), with the key difference being the sign between the two product terms (minus for addition, plus for subtraction). For cosine subtraction specifically, we have cos(A - B) = cos(A)cos(B) + sin(A)sin(B), where the plus sign between the terms is crucial. Choice B is correct because it correctly states the formula with proper signs: cos(A - B) = cos(A)cos(B) + sin(A)sin(B). Choice A incorrectly has the wrong sign between the terms, using cos(A)cos(B) - sin(A)sin(B) which is actually the cosine addition formula for cos(A + B), not the subtraction formula. Remember the pattern: sin formulas have sin·cos + cos·sin (same functions in each term), while cos formulas have cos·cos ∓ sin·sin (same functions in each term but different from numerator function), and the signs are opposite between sin and cos formulas.

Question 12

1. $\frac{33}{65}$
2. $\frac{56}{65}$
3. $\frac{63}{65}$ (correct answer)
4. $\frac{39}{65}$

Explanation: Using the sine addition formula: $\sin(x + y) = \sin x \cos y + \cos x \sin y$. First, find the missing trigonometric values. Since $\sin x = \frac{3}{5}$ and $x$ is in quadrant I, $\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$. Since $\cos y = \frac{5}{13}$ and $y$ is in quadrant I, $\sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - \frac{25}{169}} = \frac{12}{13}$. Therefore: $\sin(x + y) = \frac{3}{5} \cdot \frac{5}{13} + \frac{4}{5} \cdot \frac{12}{13} = \frac{15}{65} + \frac{48}{65} = \frac{63}{65}$. Option A results from calculation errors. Option B comes from incorrectly using cosine addition formula. Option D results from sign errors in the Pythagorean theorem applications.

Question 13

1. $\sin(x + h) = \sin h \cos x + \cos h \sin x$, which rearranges terms to facilitate the differentiation process
2. $\sin(x + h) = \sin x \sin h + \cos x \cos h$, which enables the use of product rule on each term separately
3. $\sin(x + h) = \cos x \cos h - \sin x \sin h$, which directly provides the derivative through term identification
4. $\sin(x + h) = \sin x \cos h + \cos x \sin h$, which allows separation of terms involving $x$ and $h$ for differentiation (correct answer)

Explanation: When working with trigonometric identities and derivatives, the sine addition formula is a fundamental tool that breaks down complex expressions into manageable parts. The formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$ allows you to expand any sine of a sum into terms that can be differentiated separately. For $\sin(x + h)$, you need to identify $A = x$ and $B = h$, then apply the formula directly. This gives you $\sin(x + h) = \sin x \cos h + \cos x \sin h$. This expansion is crucial for differentiation because it separates the variable $x$ (which you're differentiating with respect to) from the parameter $h$, making it possible to apply derivative rules to each term individually. Choice A reverses the order of terms in the addition formula, writing $\sin h \cos x + \cos h \sin x$ instead of the correct $\sin x \cos h + \cos x \sin h$. While mathematically equivalent due to commutativity, this doesn't match the standard application of the sine addition formula. Choice B incorrectly uses $\sin x \sin h + \cos x \cos h$, which is actually the cosine addition formula $\cos(x - h)$, not the sine addition formula. Choice C gives $\cos x \cos h - \sin x \sin h$, which is the cosine addition formula for $\cos(x + h)$, completely wrong for expanding $\sin(x + h)$. Remember: always match the trigonometric function in your expansion to the function you're working with. Sine addition formulas expand sines, cosine addition formulas expand cosines.

Question 14

1. $\dfrac{\sqrt{6}+\sqrt{2}}{4}$
2. $\dfrac{\sqrt{6}-\sqrt{2}}{4}$
3. $-\dfrac{\sqrt{6}-\sqrt{2}}{4}$ (correct answer)
4. $\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}$

Explanation: This question tests understanding of the angle addition formulas for cosine. The cosine addition formula is cos(A + B) = cos(A)cos(B) - sin(A)sin(B), and the subtraction formula is cos(A - B) = cos(A)cos(B) + sin(A)sin(B), with the key difference being the sign between the two product terms (minus for addition, plus for subtraction). To find cos(105°) where 105° = 60° + 45°, we substitute into cos(60° + 45°) = cos(60°)cos(45°) - sin(60°)sin(45°), giving (1/2)(√2/2) - (√3/2)(√2/2) = √2/4 - √6/4 = (√2 - √6)/4 = - (√6 - √2)/4. Choice C is correct because it properly substitutes the angle values and simplifies accurately, including the negative sign due to the quadrant. Choice A has the wrong sign between the terms, using plus when the formula for cosine addition requires minus. To use these formulas for exact values, break unfamiliar angles into sums or differences of standard angles (like 75° = 45° + 30° or 15° = 45° - 30°), then apply the formula with the known exact trig values. Key to angle addition formulas: memorize that both sine and cosine formulas involve four terms (two cross products), with sine having + for addition and - for subtraction, while cosine has - for addition and + for subtraction (opposite signs).

Question 15

1. Because $\sin(A+B)$ must include the cross terms $\sin(A)\cos(B)$ and $\cos(A)\sin(B)$, not just a sum of sines. (correct answer)
2. Because $\sin(A+B)=\sin(A)\sin(B)+\cos(A)\cos(B)$ for all angles.
3. Because $\sin(A+B)=\sin(A)-\sin(B)$ for all angles.
4. Because $\sin(A)+\sin(B)$ is always between $-1$ and $1$, but $\sin(A+B)$ is not.

Explanation: This question tests understanding of why the angle addition formula for sine requires cross terms. The sine addition formula is sin(A + B) = sin(A)cos(B) + cos(A)sin(B), which shows that sine of a sum is NOT simply sin(A) + sin(B) but requires cross terms involving both sine and cosine of each angle. If sin(A + B) simply equaled sin(A) + sin(B), then sin(30° + 60°) = sin(90°) = 1 would equal sin(30°) + sin(60°) = 1/2 + √3/2 ≈ 1.37, which is false, demonstrating that the formula requires the cross terms sin(A)cos(B) + cos(A)sin(B) instead. Choice A is correct because it explains why cross terms are necessary: the sine of a sum must include the cross terms sin(A)cos(B) and cos(A)sin(B), not just a sum of sines. Choice D incorrectly focuses on the range of values rather than the structural reason why the formula needs cross terms - the issue isn't about bounds but about the fundamental trigonometric relationship. Common error: students try sin(A + B) = sin(A) + sin(B), but you can quickly verify this is wrong by trying A = B = 45°: sin(90°) = 1 but sin(45°) + sin(45°) = √2/2 + √2/2 = √2 ≠ 1.

Question 16

1. Because $\sin(A+B)$ must include cross terms $\sin(A)\cos(B)$ and $\cos(A)\sin(B)$, not just $\sin(A)$ and $\sin(B)$. (correct answer)
2. Because $\sin(A+B)=\sin(A)\sin(B)+\cos(A)\cos(B)$ for all angles.
3. Because $\sin(A+B)$ always equals $1$ whenever $A$ and $B$ are acute.
4. Because $\sin(A)+\sin(B)$ is undefined whenever $A+B=90^\circ$.

Explanation: This question tests understanding of the angle addition formulas for sine. The sine addition formula is sin(A + B) = sin(A)cos(B) + cos(A)sin(B), which shows that sine of a sum is NOT simply sin(A) + sin(B) but requires cross terms involving both sine and cosine of each angle. If sin(A + B) simply equaled sin(A) + sin(B), then sin(30° + 60°) = sin(90°) = 1 would equal sin(30°) + sin(60°) = 1/2 + √3/2 ≈ 1.37, which is false, demonstrating that the formula requires the cross terms sin(A)cos(B) + cos(A)sin(B) instead. Choice A is correct because it explains why cross terms are necessary. Choice B confuses the sine formula with the cosine formula, using sin(A)sin(B) + cos(A)cos(B) when sine requires sin(A)cos(B) + cos(A)sin(B). Common error: students try sin(A + B) = sin(A) + sin(B), but you can quickly verify this is wrong by trying A = B = 45°: sin(90°) = 1 but sin(45°) + sin(45°) = √2/2 + √2/2 = √2 ≠ 1. Key to angle addition formulas: memorize that both sine and cosine formulas involve four terms (two cross products), with sine having + for addition and - for subtraction, while cosine has - for addition and + for subtraction (opposite signs).

Question 17

1. $\dfrac{\sqrt{6}-\sqrt{2}}{4}$
2. $\dfrac{\sqrt{6}+\sqrt{2}}{4}$ (correct answer)
3. $\dfrac{\sqrt{3}+1}{2}$
4. $\dfrac{\sqrt{2}}{2}+\dfrac{1}{2}$

Explanation: This question tests understanding of the angle addition formula for sine. The sine addition formula is sin(A + B) = sin(A)cos(B) + cos(A)sin(B), which shows that sine of a sum is NOT simply sin(A) + sin(B) but requires cross terms involving both sine and cosine of each angle. To find sin(75°), we recognize 75° = 45° + 30°, so sin(75°) = sin(45°)cos(30°) + cos(45°)sin(30°) = (√2/2)(√3/2) + (√2/2)(1/2) = √6/4 + √2/4 = (√6 + √2)/4. Choice B is correct because it properly substitutes the angle values and simplifies accurately to (√6 + √2)/4. Choice A incorrectly has (√6 - √2)/4, which would result from using the wrong sign between the terms or confusing this with a cosine formula. Remember the pattern: sin formulas have sin·cos + cos·sin (same functions in each term), while cos formulas have cos·cos ∓ sin·sin (same functions in each term but different from numerator function), and the signs are opposite between sin and cos formulas.

Question 18

1. $\sin(A+B)=\sin(A)\cos(B)+\cos(A)\sin(B)$ (correct answer)
2. $\sin(A+B)=\sin(A)\sin(B)+\cos(A)\cos(B)$
3. $\sin(A+B)=\sin(A)\cos(B)-\cos(A)\sin(B)$
4. $\sin(A+B)=\sin(A)+\sin(B)$

Explanation: This question tests understanding of the angle addition and subtraction formulas for sine. The sine addition formula is sin(A + B) = sin(A)cos(B) + cos(A)sin(B), which shows that sine of a sum is NOT simply sin(A) + sin(B) but requires cross terms involving both sine and cosine of each angle. If sin(A + B) simply equaled sin(A) + sin(B), then sin(30° + 60°) = sin(90°) = 1 would equal sin(30°) + sin(60°) = 1/2 + √3/2 ≈ 1.37, which is false, demonstrating that the formula requires the cross terms sin(A)cos(B) + cos(A)sin(B) instead. Choice A is correct because it correctly states the formula with proper signs. Choice D incorrectly claims sin(A + B) = sin(A) + sin(B), missing the essential cross terms that make the actual formula sin(A + B) = sin(A)cos(B) + cos(A)sin(B). Common error: students try sin(A + B) = sin(A) + sin(B), but you can quickly verify this is wrong by trying A = B = 45°: sin(90°) = 1 but sin(45°) + sin(45°) = √2/2 + √2/2 = √2 ≠ 1. Remember the pattern: sin formulas have sin·cos + cos·sin (same functions in each term), while cos formulas have cos·cos ∓ sin·sin (same functions in each term but different from numerator function), and the signs are opposite between sin and cos formulas.

Question 19

1. Student 1 is correct because substituting $-B$ for $B$ immediately gives the desired formula without additional justification
2. Student 2 is correct because the substitution requires using the even-odd properties: $\cos(-B) = \cos B$ and $\sin(-B) = -\sin B$ (correct answer)
3. Student 1 is correct because the cosine subtraction formula is just the negative of the cosine addition formula
4. Student 2 is correct because the derivation requires proving that $\cos(A + (-B)) = \cos(A - B)$ using angle measurement properties

Explanation: Student 2 is correct. While substituting $-B$ for $B$ in $\cos(A + B) = \cos A \cos B - \sin A \sin B$ gives $\cos(A + (-B)) = \cos A \cos(-B) - \sin A \sin(-B)$, this requires using the even-odd properties of trigonometric functions to simplify: $\cos(-B) = \cos B$ (cosine is even) and $\sin(-B) = -\sin B$ (sine is odd). This yields $\cos A \cos B - \sin A(-\sin B) = \cos A \cos B + \sin A \sin B$. Option A ignores these necessary steps. Option C is incorrect about the relationship. Option D overcomplicates the required justification.

Question 20

1. $\tan(A+B)=\dfrac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$ (correct answer)
2. $\tan(A+B)=\dfrac{\tan(A)+\tan(B)}{1+\tan(A)\tan(B)}$
3. $\tan(A+B)=\dfrac{\tan(A)-\tan(B)}{1-\tan(A)\tan(B)}$
4. $\tan(A+B)=\tan(A)+\tan(B)$

Explanation: This question tests understanding of the angle addition formula for tangent. The tangent addition formula is tan(A + B) = (tan(A) + tan(B))/(1 - tan(A)tan(B)), which shows that tangent of a sum involves both a numerator (sum of tangents) and a denominator (1 minus their product). The formula has a fraction structure where the numerator contains the sum tan(A) + tan(B), and the denominator contains 1 - tan(A)tan(B), with a minus sign being crucial for the addition formula. Choice A is correct because it correctly states the formula with proper signs: tan(A + B) = (tan(A) + tan(B))/(1 - tan(A)tan(B)). Choice B incorrectly has the wrong sign in the denominator, using 1 + tan(A)tan(B) when the formula for tan(A + B) requires the denominator to be 1 - tan(A)tan(B). For tangent, remember the formulas have fractions: tan(A + B) = (tan(A) + tan(B))/(1 - tan(A)tan(B)), with the sign in the denominator opposite to the numerator (minus for addition, plus for subtraction).