Geometry Quiz: Proving Angle Addition Subtraction Formulas

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Proving Angle Addition Subtraction FormulasQuestion 1 of 16

Given that $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$ , which of the following correctly represents $\cos(60° - 45°)$ using this formula?

\cos(60°)\cos(45°) - \sin(60°)\sin(45°)

\cos(60°)\cos(45°) + \sin(60°)\sin(45°)

\sin(60°)\cos(45°) - \cos(60°)\sin(45°)

\sin(60°)\cos(45°) + \cos(60°)\sin(45°)

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Geometry Quiz

Geometry Quiz: Proving Angle Addition Subtraction Formulas

Practice Proving Angle Addition Subtraction Formulas in Geometry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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 Geometry.

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.

All questions

Question 1

Given that $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$ , which of the following correctly represents $\cos(60° - 45°)$ using this formula?

$\cos(60°)\cos(45°) - \sin(60°)\sin(45°)$
$\cos(60°)\cos(45°) + \sin(60°)\sin(45°)$ (correct answer)
$\sin(60°)\cos(45°) - \cos(60°)\sin(45°)$
$\sin(60°)\cos(45°) + \cos(60°)\sin(45°)$

Explanation: The correct answer is B. Using the given cosine subtraction formula with

\alpha = 60°

and

\beta = 45°

, we get

\cos(60° - 45°) = \cos(60°)\cos(45°) + \sin(60°)\sin(45°)

. Choice A uses the cosine addition formula incorrectly. Choices C and D incorrectly use sine terms in the first position, which would be sine formulas.

Question 2

A student attempts to find $\tan(15°)$ by using $\tan(45° - 30°)$ and the formula $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$ . Given that $\tan(45°) = 1$ and $\tan(30°) = \frac{\sqrt{3}}{3}$ , which calculation should the student perform?

$\frac{1 - \frac{\sqrt{3}}{3}}{1 - 1 \cdot \frac{\sqrt{3}}{3}}$
$\frac{1 + \frac{\sqrt{3}}{3}}{1 - 1 \cdot \frac{\sqrt{3}}{3}}$
$\frac{1 \cdot \frac{\sqrt{3}}{3} - 1}{1 + 1 + \frac{\sqrt{3}}{3}}$
$\frac{1 - \frac{\sqrt{3}}{3}}{1 + 1 \cdot \frac{\sqrt{3}}{3}}$ (correct answer)

Explanation: When you encounter trigonometric expressions involving angle differences, the tangent difference formula is your key tool:

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

. To find

\tan(15°)

, you need to apply this formula to

\tan(45° - 30°)

. Here,

A = 45°

and

B = 30°

, so you substitute directly:

\tan(45° - 30°) = \frac{\tan(45°) - \tan(30°)}{1 + \tan(45°) \cdot \tan(30°)}

. Plugging in the given values

\tan(45°) = 1

and

\tan(30°) = \frac{\sqrt{3}}{3}

, you get:

\frac{1 - \frac{\sqrt{3}}{3}}{1 + 1 \cdot \frac{\sqrt{3}}{3}}

, which matches answer choice D. Let's examine the mistakes in other options: Choice A uses

1 - 1 \cdot \frac{\sqrt{3}}{3}

in the denominator, incorrectly applying subtraction instead of addition in the formula's denominator. Choice B has the wrong sign in the numerator, using

1 + \frac{\sqrt{3}}{3}

instead of

1 - \frac{\sqrt{3}}{3}

—this would be the tangent addition formula, not subtraction. Choice C completely scrambles the formula structure, placing the product term first in the numerator and adding an extra 1 in the denominator. Study tip: Always write out the difference formula completely before substituting values. The pattern is: numerator uses subtraction (

\tan A - \tan B

), denominator uses addition (

1 + \tan A \tan B

). Double-check that your signs match the formula exactly.

Question 3

A student claims that $\sin(x + y) + \sin(x - y) = 2\sin x \cos y$ . To verify this identity, which approach would be most direct?

Use the product-to-sum formulas to simplify the right side first
Convert the right side to exponential form and compare with the left side
Expand both $\sin(x + y)$ and $\sin(x - y)$ using their respective formulas, then add the results (correct answer)
Apply the Pythagorean identity to both sides before expanding the angle formulas

Explanation: When you encounter trigonometric identities involving angle addition and subtraction, the most straightforward verification approach is to expand the complex expressions using fundamental formulas and see if they simplify to match. To verify this identity, you should expand the left side using the angle addition and subtraction formulas:

\sin(x + y) = \sin x \cos y + \cos x \sin y

and

\sin(x - y) = \sin x \cos y - \cos x \sin y

. Adding these expressions gives you:

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

The

\cos x \sin y

terms cancel out, leaving

2\sin x \cos y

, which matches the right side exactly. Answer C is correct because it takes the most direct path to verification. Answer A is backwards—product-to-sum formulas would complicate rather than simplify

2\sin x \cos y

. Answer B introduces unnecessary complexity; while exponential forms work, they're far more elaborate than needed for this straightforward identity. Answer D makes no sense because the Pythagorean identity (

\sin^2 θ + \cos^2 θ = 1

) doesn't apply to sums of different angles. Remember: when verifying trigonometric identities, always choose the path that directly applies the most relevant formulas. For expressions involving

\sin(x ± y)

, immediately think of the angle addition and subtraction formulas—they're usually your fastest route to a solution.

Question 4

A student claims that $\sin(75°)$ can be found using the angle addition formula by writing $75° = 45° + 30°$ . If the student correctly applies $\sin(A + B) = \sin A \cos B + \cos A \sin B$ , which expression represents the correct calculation?

$\sin(45°)\cos(30°) + \cos(45°)\sin(30°)$ (correct answer)
$\sin(45°)\sin(30°) + \cos(45°)\cos(30°)$
$\sin(45°)\cos(30°) - \cos(45°)\sin(30°)$
$\cos(45°)\cos(30°) - \sin(45°)\sin(30°)$

Explanation: The correct answer is A. Using the sine addition formula

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

with

A = 45°

and

B = 30°

, we get

\sin(75°) = \sin(45°)\cos(30°) + \cos(45°)\sin(30°)

. Choice B uses the cosine addition formula incorrectly. Choice C uses the sine subtraction formula. Choice D uses the cosine addition formula.

Question 5

Which reasoning supports the angle addition formula?

In the diagram, triangle $ABC$ has point $D$ on segment $AC$ . At vertex $A$ , ray $AD$ splits $\angle BAC$ into two angles labeled $\theta$ (between $AB$ and $AD$ ) and $\varphi$ (between $AD$ and $AC$ ), so $\angle BAC=\theta+\varphi$ . Segment $BD$ is drawn. A right-angle marker indicates $BD\perp AC$ at $D$ .

Which statement proves an identity for $\sin(\theta+\varphi)$ using this construction?

Because $\angle BAC=\theta+\varphi$ , we can write $\sin(\theta+\varphi)=\sin\theta+\sin\varphi$ directly from the diagram.
Decompose lengths using right triangles $\triangle ABD$ and $\triangle CBD$ to express the altitude $BD$ two ways, leading to $\sin(\theta+\varphi)=\sin\theta\cos\varphi+\cos\theta\sin\varphi$ . (correct answer)
Since $BD\perp AC$ , angles at $D$ are complementary, so $\sin(\theta+\varphi)=\sin\theta\cos\varphi-\cos\theta\sin\varphi$ .
Because $D$ lies on $AC$ , triangles $ABD$ and $ABC$ are similar, so $\sin(\theta+\varphi)=\sin\theta\cos\varphi$ .

Explanation: The skill focuses on proving the sine addition formula using triangle decompositions. The geometric setup features triangle ABC with point D on AC, ray AD splitting angle BAC into θ and φ, and BD perpendicular to AC. Angle BAC decomposes into θ between AB and AD plus φ between AD and AC, summing to θ + φ. Side relationships are tracked through right triangles ABD and CBD, expressing altitude BD in terms of opposite sides and angles. This justifies sin(θ + φ) = sinθ cosφ + cosθ sinφ by equating expressions for the height relative to the hypotenuse. A distractor misconception involves subtracting terms due to complementary angles, leading to an incorrect sign. Transfer strategy: think geometry before algebra by decomposing angles in triangles to build trig identities from basic definitions.

Question 6

A student is asked to prove that $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ . The student begins by writing $\tan(A + B) = \frac{\sin(A + B)}{\cos(A + B)}$ . What should be the student's next step?

Factor out common terms from the numerator and denominator before applying formulas
Convert the tangent function directly to the quotient form without using sine and cosine
Apply the Pythagorean identity to simplify the expression before substitution
Substitute the angle addition formulas for both sine and cosine in the numerator and denominator (correct answer)

Explanation: When proving trigonometric identities involving tangent addition formulas, you're working with the fundamental relationship that tangent equals sine divided by cosine. The student correctly started with

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

, which sets up the perfect foundation for using angle addition formulas. The next logical step is to substitute the angle addition formulas for both sine and cosine. You need to replace

\sin(A + B)

with

\sin A \cos B + \cos A \sin B

and

\cos(A + B)

with

\cos A \cos B - \sin A \sin B

. This gives you

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

. From here, you can divide both numerator and denominator by

\cos A \cos B

to eventually reach the desired form with individual tangent terms. Option A is incorrect because there are no common factors to extract before applying the formulas—you need the expanded forms first. Option B misses the point entirely since the student already chose the sine/cosine approach, which is perfectly valid and systematic. Option C is wrong because the Pythagorean identity

\sin^2 θ + \cos^2 θ = 1

doesn't directly help with angle addition—you need the specific addition formulas. Remember: when proving tangent addition formulas, always apply the sine and cosine addition formulas first, then manipulate algebraically to isolate the tangent terms. This systematic approach works reliably for all angle addition proofs.

Question 7

Which explanation correctly uses the geometry?

In triangle $ABC$ , $\angle A$ is labeled $\theta$ and $\angle B$ is labeled $\varphi$ . A segment from $C$ to $AB$ meets $AB$ at $D$ and is marked perpendicular to $AB$ (right-angle box at $D$ ). Thus $CD$ is an altitude.

Which statement gives a valid geometric justification path toward an identity involving $\sin(\theta+\varphi)$ (using that $\angle C = 180^\circ-(\theta+\varphi)$ ), without plugging in numbers or assuming the identity?

Use right triangles $\triangle ACD$ and $\triangle BCD$ to express $CD$ two ways with sines/cosines, then relate those expressions to an angle-sum through $\angle C$ . (correct answer)
Because $CD$ is an altitude, $\sin(\theta+\varphi)=\sin\theta+\sin\varphi$ .
Because $\angle C=180^\circ-(\theta+\varphi)$ , conclude $\sin(\theta+\varphi)=\sin\theta\sin\varphi$ .
Because the triangle is not to scale, the altitude cannot be used to form right triangles.

Explanation: The skill is proving the sine angle addition formula using an altitude in a triangle. The geometric setup involves triangle ABC with angles theta at A and phi at B, and altitude CD perpendicular to AB at D. This decomposes the supplementary angle at C as 180 degrees minus (theta + phi). Side relationships are tracked via CD expressed in right triangles ACD and BCD using sines and cosines of theta and phi. Relating these expressions through sin(180 - (theta + phi)) = sin(theta + phi) justifies the addition formula. A distractor misconception is adding sines directly due to the altitude, as in choice B, without considering the angle relations. To transfer this strategy, focus on geometric altitudes and right triangles before algebraic substitutions.

Question 8

Which statement proves the formula?

A right triangle $\triangle OAB$ has right angle at $A$ , with $OA$ along the positive $x$ -axis and $AB$ vertical (right-angle box at $A$ ). The angle at $O$ between $OA$ and $OB$ is labeled $\theta+\varphi$ . A ray from $O$ inside the triangle meets $AB$ at point $C$ and splits the angle at $O$ into $\angle AOC=\theta$ and $\angle COB=\varphi$ .

Which reasoning correctly leads to an identity for $\tan(\theta+\varphi)$ using this decomposition?

Express slopes: $\tan\theta=\frac{AC}{OA}$ and $\tan\varphi=\frac{CB}{OC}$ , then add to get $\tan(\theta+\varphi)=\tan\theta+\tan\varphi$ .
Relate similar right triangles formed by dropping a perpendicular from $C$ to $OA$ , then write the overall slope $\frac{AB}{OA}$ in terms of the two smaller slopes to obtain $\tan(\theta+\varphi)=\frac{\tan\theta+\tan\varphi}{1-\tan\theta\tan\varphi}$ . (correct answer)
Because the angle is split, $\tan(\theta+\varphi)=\tan\theta\tan\varphi$ by multiplying the two tangent ratios.
Since $\triangle OAB$ is right, conclude $\tan(\theta+\varphi)=\frac{AB}{OB}$ and replace $OB$ by $OA+AB$ .

Explanation: The skill is proving the tangent angle addition formula using a right triangle decomposition. The geometric setup features right triangle OAB with right angle at A, angle at O as theta + phi, and a ray OC splitting it into theta and phi, intersecting AB at C. This splits the angle at O into adjacent parts theta and phi. Side relationships are tracked via slopes, with tan theta as AC/OA and tan phi as CB/OC, and the overall slope AB/OA. By relating similar right triangles from a perpendicular at C, we justify tan(theta + phi) = (tan theta + tan phi)/(1 - tan theta tan phi). A distractor misconception is simply adding the tangents, as in choice A, disregarding the denominator adjustment. To transfer this strategy, focus on geometric slope decompositions in triangles before algebraic computations.

Question 9

A unit circle is centered at $O$ . Point $P$ corresponds to angle $\theta$ from the positive $x$ -axis, and point $Q$ corresponds to angle $\varphi$ from the positive $x$ -axis. The chord $PQ$ is drawn. Which reasoning supports the angle addition formula by relating the dot product $\overrightarrow{OP}\cdot\overrightarrow{OQ}$ to a cosine of a difference?

Use that the dot product of unit vectors equals the cosine of the angle between them.

Since $|OP|=|OQ|=1$ , $\overrightarrow{OP}\cdot\overrightarrow{OQ}=\cos(\theta-\varphi)$ , and expanding the dot product using coordinates gives $\cos\theta\cos\varphi+\sin\theta\sin\varphi$ . (correct answer)
Since $|OP|=|OQ|=1$ , $\overrightarrow{OP}\cdot\overrightarrow{OQ}=\cos(\theta+\varphi)$ , and expanding gives $\cos\theta\cos\varphi-\sin\theta\sin\varphi$ .
Since chord $PQ$ is drawn, $\overrightarrow{OP}\cdot\overrightarrow{OQ}=|PQ|$ , so $\cos(\theta-\varphi)=\cos\theta-\cos\varphi$ .
Since the diagram is a circle, $\overrightarrow{OP}\cdot\overrightarrow{OQ}=\sin(\theta-\varphi)$ , so $\sin(\theta-\varphi)=\cos\theta\cos\varphi+\sin\theta\sin\varphi$ .

Explanation: This question uses the dot product of unit vectors to derive angle subtraction formulas, specifically for cosine. Points P and Q on the unit circle correspond to angles θ and φ respectively, making vectors OP and OQ unit vectors. The dot product of two unit vectors equals the cosine of the angle between them, which is |θ-φ|. Expanding the dot product using coordinates gives OP·OQ = cosθ·cosφ + sinθ·sinφ = cos(θ-φ). Choice B incorrectly relates the dot product to cos(θ+φ) instead of cos(θ-φ), while choice C wrongly equates the dot product to the chord length |PQ|. The geometric insight is that the dot product naturally encodes the cosine of the angle between vectors, and coordinate expansion reveals the addition formula. This approach elegantly connects vector algebra to trigonometric identities through geometric interpretation.

Question 10

A unit circle is centered at $O$ . From point $A=(1,0)$ , rotate by $\theta$ to reach point $B$ on the circle. Then rotate by $\varphi$ to reach point $C$ on the circle. A perpendicular from $C$ to the $x$ -axis meets it at $H$ . Which argument justifies $\sin(\theta+\varphi)$ by expressing the height $CH$ using components from the intermediate point $B$ ?

Use the idea that the second rotation mixes the $x$ - and $y$ -components of $B$ .

Since $CH=\sin(\theta+\varphi)$ , rotating $(\cos\theta,\sin\theta)$ by $\varphi$ gives $CH=\sin\theta\cos\varphi+\cos\theta\sin\varphi$ . (correct answer)
Since $CH=\sin(\theta+\varphi)$ , the height equals $\sin\theta+\sin\varphi$ because rotations add vertical changes.
Since $CH=\sin(\theta+\varphi)$ , the height equals $\sin\theta\cos\varphi-\cos\theta\sin\varphi$ because both angles are counterclockwise.
Since $CH=\sin(\theta+\varphi)$ , the height equals $\cos\theta\cos\varphi-\sin\theta\sin\varphi$ because $CH$ is adjacent to $\theta+\varphi$ .

Explanation: This question derives the sine addition formula by tracking the height of a point after two successive rotations on a unit circle. Starting at A=(1,0), we rotate by θ to reach B=(cosθ, sinθ), then rotate by φ to reach C. The height CH represents sin(θ+φ), and we can express it by considering how the second rotation transforms B's coordinates. The y-coordinate of C equals the y-component of B times cosφ plus the x-component of B times sinφ, giving sin(θ+φ) = sinθcosφ + cosθsinφ. Choice B incorrectly assumes rotations add heights linearly, while choice C has the wrong sign, suggesting subtraction. The geometric principle is that each rotation mixes the existing x and y components according to the rotation angle, with both components contributing to the final height. This mixing effect is fundamental to understanding why trigonometric addition formulas involve products of different functions.

Question 11

Which conclusion follows from the diagram?

Triangle $OXY$ is a right triangle with right angle at $X$ (right-angle box at $X$ ). Ray $OZ$ lies inside $\angle XOY$ and splits it into two angles: $\angle XOZ=\theta$ and $\angle ZOY=\varphi$ . From point $Z$ on ray $OZ$ , perpendiculars are dropped to $OX$ and to $XY$ meeting them at $M$ and $N$ (right-angle boxes at $M$ and $N$ ). No other information is marked.

Which statement correctly describes a geometric strategy that can lead to a sum identity (rather than assuming it)?

Use right triangles $\triangle OZM$ and $\triangle ZXN$ to relate projections along $OX$ and $XY$ , then combine to express a projection corresponding to $\cos(\theta+\varphi)$ . (correct answer)
Assume $\cos(\theta+\varphi)$ equals $\cos\theta\cos\varphi-\sin\theta\sin\varphi$ and label the segments to match.
Because $\theta$ and $\varphi$ share vertex $O$ , conclude $\cos(\theta+\varphi)=\cos\theta+\cos\varphi$ .
Because the diagram is not to scale, no trigonometric identity can be derived from it.

Explanation: The skill focuses on proving cosine sum identities through projections in a right triangle. The geometric setup is right triangle OXY with right angle at X, ray OZ splitting angle at O into theta and phi, and perpendiculars from Z to OX and XY at M and N. This decomposes the angle at O as theta + phi. Side relationships are tracked via projections in right triangles OZM and ZXN along OX and XY. Combining these projections justifies cos(theta + phi) = cos theta cos phi - sin theta sin phi. A distractor misconception is adding cosines directly due to shared vertex, as in choice C, without projection adjustments. To transfer this strategy, visualize geometric perpendiculars and projections in triangles before algebraic work.

Question 12

Which conclusion follows from the diagram?

In triangle $ABC$ , point $D$ lies on segment $AC$ . Segment $BD$ is drawn. The angle at $B$ is split by ray $BD$ into two adjacent angles: $\angle ABD=\theta$ and $\angle DBC=\varphi$ . A perpendicular from $D$ to line $AB$ meets $AB$ at $E$ , and a perpendicular from $D$ to line $BC$ meets $BC$ at $F$ (right-angle boxes at $E$ and $F$ ). No other special relationships are marked.

Using only this construction, which statement provides a valid geometric route to an angle-addition identity?

Write $\sin(\theta+\varphi)=\sin\theta+\sin\varphi$ because the angles are adjacent at $B$ .
Use right triangles $\triangle BDE$ and $\triangle BDF$ to express the same altitude $BD\sin\theta$ and $BD\sin\varphi$ , then combine projections to relate $\sin(\theta+\varphi)$ to $\sin\theta$ and $\cos\varphi$ . (correct answer)
Since $BD$ bisects $\angle ABC$ into $\theta$ and $\varphi$ , conclude $\theta=\varphi$ and simplify $\sin(\theta+\varphi)$ .
Assume $\sin(\theta+\varphi)=\sin\theta\cos\varphi+\cos\theta\sin\varphi$ and note the diagram is consistent with it.

Explanation: The skill is proving the sine angle addition formula through geometric decomposition in a triangle. The geometric setup includes triangle ABC with point D on AC, and BD splitting the angle at B into adjacent angles theta and phi, with perpendiculars from D to AB and BC at E and F. This construction decomposes the total angle at B as theta plus phi, using BD as the splitting ray. Side relationships are tracked via the altitudes in right triangles BDE and BDF, where BD relates to projections involving sin theta and sin phi along the sides. By combining these projections and equating expressions for the same lengths, we justify the formula sin(theta + phi) = sin theta cos phi + cos theta sin phi. A distractor misconception is simply adding the sines of adjacent angles, as in choice A, which overlooks the necessary cosine adjustments for proper projection. To transfer this strategy, prioritize visualizing the geometric splitting and perpendiculars in triangles before resorting to algebraic manipulations.

Question 13

Which statement proves the formula?

A unit circle centered at $O$ is shown with point $A$ on the positive $x$ -axis. Point $P$ is on the circle with $\angle AOP=\theta$ . From ray $OP$ , rotate clockwise by angle $\varphi$ to ray $OQ$ (the clockwise arc is labeled $\varphi$ ), so $\angle AOQ=\theta-\varphi$ . A perpendicular from $Q$ meets the $x$ -axis at $N$ .

Which claim correctly justifies a formula for $\cos(\theta-\varphi)$ from this rotation idea?

Clockwise rotation changes only the sign of sine, so $\cos(\theta-\varphi)=\cos\theta+\cos\varphi$ .
Using rotation of the point $(\cos\theta,\sin\theta)$ by $-\varphi$ , the new $x$ -coordinate is $\cos\theta\cos\varphi+\sin\theta\sin\varphi$ . (correct answer)
Because $\angle AOQ=\theta-\varphi$ , we can read off $\cos(\theta-\varphi)=ON=\cos\theta-\cos\varphi$ .
Since $Q$ lies on the unit circle, $\cos(\theta-\varphi)=\cos\theta\cos\varphi$ .

Explanation: The skill focuses on deriving the cosine subtraction formula using clockwise rotations on a unit circle. The geometric setup features a unit circle centered at O, point P at angle θ from OA on the x-axis, and Q obtained by rotating clockwise by φ from OP, resulting in angle AOQ as θ - φ. The angle θ - φ decomposes through the clockwise rotation from θ. Side relationships are tracked by x-coordinates, with the rotation by -φ affecting projections. This justifies cos(θ - φ) = cosθ cosφ + sinθ sinφ via the rotation matrix applied to point P. A distractor misconception assumes subtraction like cos(θ - φ) = cosθ - cosφ, ignoring the additive sine terms. Transfer strategy: think geometry before algebra by incorporating directionality in rotations to derive difference formulas.

Question 14

Which of the following is equivalent to the expression $\cos(x)\cos(y) - \sin(x)\sin(y)$ ?

$\cos(x - y)$
$\cos(x + y)$ (correct answer)
$\sin(x + y)$
$\sin(x - y)$

Explanation: The correct answer is B. The expression

\cos(x)\cos(y) - \sin(x)\sin(y)

matches the cosine addition formula:

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

. Choice A would be the cosine subtraction formula with a plus sign instead of minus. Choices C and D are sine formulas, which have different structures involving sine and cosine terms in different positions.

Question 15

Given the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$ , which statement about this formula is correct?

The formula can be derived by substituting $-B$ for $B$ in the sine addition formula (correct answer)
The formula is only valid when both $A$ and $B$ are acute angles
The formula produces the same result as $\cos(A + B)$ when $A = 90°$
The formula requires that $A > B$ for the result to be positive

Explanation: The correct answer is A. The sine subtraction formula can be derived from the sine addition formula

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

by replacing

B

with

-B

and using the facts that

\sin(-B) = -\sin B

and

\cos(-B) = \cos B

. Choice B is incorrect because the formula works for all real angles. Choice C is incorrect because

\sin(90° - B) = \cos B

while

\cos(90° + B) = -\sin B

. Choice D is incorrect because the sign of the result depends on the quadrant of

A - B

, not just whether

A > B

Question 16

To prove that $\cos(2\theta) = \cos^2 \theta - \sin^2 \theta$ , a student decides to use the angle addition formula for cosine. Which substitution should the student make?

Let $A = \theta$ and $B = -\theta$ in $\cos(A - B) = \cos A \cos B + \sin A \sin B$
Let $A = 2\theta$ and $B = 0$ in $\cos(A + B) = \cos A \cos B - \sin A \sin B$
Let $A = \theta$ and $B = \theta$ in $\cos(A + B) = \cos A \cos B - \sin A \sin B$ (correct answer)
Let $A = \frac{\theta}{2}$ and $B = \frac{3\theta}{2}$ in $\cos(A + B) = \cos A \cos B - \sin A \sin B$

Explanation: When you encounter double angle formulas like

\cos(2\theta)

, the most direct approach is recognizing that

2\theta = \theta + \theta

. This allows you to apply the angle addition formula by setting both angles equal to

\theta

. Using the angle addition formula

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

with

A = \theta

and

B = \theta

, you get:

\cos(\theta + \theta) = \cos \theta \cos \theta - \sin \theta \sin \theta = \cos^2 \theta - \sin^2 \theta

This directly proves the desired formula, making choice C correct. Let's examine why the other options don't work: Choice A uses subtraction (

A - B

) instead of addition, and with

A = \theta

and

B = -\theta

, you'd get

\cos(\theta - (-\theta)) = \cos(2\theta)

, but the subtraction formula gives

\cos \theta \cos(-\theta) + \sin \theta \sin(-\theta)

, which simplifies to

\cos^2 \theta + \sin^2 \theta = 1

, not the target formula. Choice B substitutes

A = 2\theta

and

B = 0

, giving

\cos(2\theta + 0) = \cos(2\theta)

. While this equals

\cos(2\theta)

, it doesn't help prove the formula since you're starting with what you're trying to prove. Choice D creates

\cos(\frac{\theta}{2} + \frac{3\theta}{2}) = \cos(2\theta)

, but the resulting expression involves

\cos(\frac{\theta}{2})

and

\cos(\frac{3\theta}{2})

, which won't simplify to the desired form. Study tip: For double angle proofs, always think "double = single + single" and use the appropriate addition formula with both angles equal to the single angle.