Proving Angle Addition/Subtraction Formulas

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Pre-Calculus › Proving Angle Addition/Subtraction Formulas

Questions 1 - 10
1

Using the cosine addition formula, what is the exact value of $\cos(105^\circ)$ if you rewrite it as $\cos(60^\circ+45^\circ)$?

$\dfrac{\sqrt{6}-\sqrt{2}}{4}$

$\dfrac{\sqrt{6}+\sqrt{2}}{4}$

$-\dfrac{\sqrt{6}-\sqrt{2}}{4}$

$\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}$

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).

2

Which formula correctly gives the cosine subtraction identity $\cos(A-B)$ in terms of $\sin$ and $\cos$?

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

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

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

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

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.

3

Using the sine angle addition formula, if $\sin(A)=\dfrac{3}{5}$ and $\cos(A)=\dfrac{4}{5}$, and $\sin(B)=\dfrac{5}{13}$ and $\cos(B)=\dfrac{12}{13}$ (with $A$ and $B$ in Quadrant I), what is $\sin(A+B)$?

$\dfrac{16}{65}$

$\dfrac{36}{65}$

$\dfrac{9}{25}+\dfrac{25}{169}$

$\dfrac{56}{65}$

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.

4

Which formula correctly gives $\cos(A-B)$? (Be careful: the sign pattern for cosine subtraction is the opposite of what many students expect.)

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

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

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

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

Explanation

This question tests understanding of the angle addition and subtraction 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). Using cos(A - B) = cos(A)cos(B) + sin(A)sin(B), we see that the subtraction formula has a positive sign for the sin(A)sin(B) term, distinguishing it from the addition formula. Choice B is correct because it correctly states the formula with proper signs. Choice A has the wrong sign between the terms, using minus when the formula for cosine subtraction requires plus. 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). 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.

5

Use the cosine angle addition formula to find the exact value of $\cos(105^\circ)$ by writing it as $\cos(60^\circ+45^\circ)$.

$\dfrac{\sqrt{6}-\sqrt{2}}{4}$

$\dfrac{\sqrt{6}+\sqrt{2}}{4}$

$-\dfrac{\sqrt{6}-\sqrt{2}}{4}$

$-\dfrac{\sqrt{6}+\sqrt{2}}{4}$

Explanation

This question tests understanding of the angle addition and subtraction 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). Using cos(A + B) = cos(A)cos(B) - sin(A)sin(B) with A = 60°, B = 45°, we calculate cos(60°)cos(45°) - sin(60°)sin(45°) = (1/2)(√2/2) - (√3/2)(√2/2) = √2/4 - √6/4 = (√2 - √6)/4 = - (√6 - √2)/4. Choice D is correct because it properly substitutes the angle values and simplifies accurately, accounting for the negative value in the second quadrant. Choice C has the wrong sign between the terms, using the larger positive value when the formula for cosine addition requires subtraction leading to a negative result. 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).

6

Use the sine angle subtraction formula to find the exact value of $\sin(15^\circ)$ by writing it as $\sin(45^\circ-30^\circ)$.

$\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}$

$\dfrac{\sqrt{6}-\sqrt{2}}{4}$

$\dfrac{\sqrt{6}+\sqrt{2}}{4}$

$\dfrac{\sqrt{3}}{2}$

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 $\sin(15^\circ)$, we recognize $15^\circ = 45^\circ - 30^\circ$, so $\sin(45^\circ - 30^\circ) = \sin(45^\circ)\cos(30^\circ) - \cos(45^\circ)\sin(30^\circ) = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4}$. Choice B is correct because it properly substitutes the angle values and simplifies accurately. Choice A has the wrong sign between the terms, using addition when the formula for sine subtraction requires subtraction. To use these formulas for exact values, break unfamiliar angles into sums or differences of standard angles (like $75^\circ = 45^\circ + 30^\circ$ or $15^\circ = 45^\circ - 30^\circ$), 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).

7

Use angle addition formulas to simplify the expression $\sin\left(x+\dfrac{\pi}{4}\right)+\sin\left(x-\dfrac{\pi}{4}\right)$.

$\sqrt{2}\cos(x)$

$2\cos(x)$

$2\sin(x)$

$\sqrt{2}\sin(x)$

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) = 2 sin(x) (√2/2) = √2 sin(x). Choice A is correct because it properly substitutes the angle values and simplifies accurately. Choice B confuses the sine formula with the cosine formula, using a cosine result when sine requires sin(A)cos(B) + cos(A)sin(B) for addition and subtraction accordingly. 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, then apply the formula with the known exact trig values.

8

Using the angle addition formula, which statement correctly explains why $\sin(A+B)\ne \sin(A)+\sin(B)$ in general?

Because $\sin(A+B)$ always equals $\sin(A)\sin(B)$ for any angles $A$ and $B$.

Because $\sin(A)+\sin(B)$ is always between $-1$ and $1$, but $\sin(A+B)$ is not.

Because $\sin(A+B)=\sin(A)\cos(B)-\cos(A)\sin(B)$, so the middle sign is always negative.

Because $\sin(A+B)$ must include cross terms $\sin(A)\cos(B)$ and $\cos(A)\sin(B)$, not just a sum of sines.

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 explains why cross terms are necessary. Choice D has the wrong sign between the terms, using minus when the formula for sine addition requires plus. 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).

9

Using the tangent angle addition formula, what is $\tan(A+B)$ in terms of $\tan(A)$ and $\tan(B)$?

$\tan(A+B)=\tan(A)+\tan(B)$

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

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

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

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.

10

If $\sin(A)=\dfrac{3}{5}$ and $\cos(A)=\dfrac{4}{5}$, and $\sin(B)=\dfrac{5}{13}$ and $\cos(B)=\dfrac{12}{13}$ (with $A$ and $B$ in Quadrant I), what is $\sin(A+B)$ using the sine addition formula?

$\dfrac{63}{65}$

$\dfrac{16}{65}$

$\dfrac{56}{65}$

$\dfrac{9}{65}$

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 $\sin(A) = \dfrac{3}{5}$, $\cos(A) = \dfrac{4}{5}$, $\sin(B) = \dfrac{5}{13}$, and $\cos(B) = \dfrac{12}{13}$, we substitute into $\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$, giving $\dfrac{3}{5} \times \dfrac{12}{13} + \dfrac{4}{5} \times \dfrac{5}{13} = \dfrac{36}{65} + \dfrac{20}{65} = \dfrac{56}{65}$. Choice A is correct because it properly substitutes the angle values and simplifies accurately. Choice B makes an arithmetic error in the evaluation, calculating something like $\dfrac{3}{5} \times \dfrac{12}{13} + \dfrac{4}{5} \times \dfrac{12}{13}$ instead of using $\sin(B)$ in the second term. 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). Remember the pattern: sin formulas have $\sin \cdot \cos + \cos \cdot \sin$ (same functions in each term), while cos formulas have $\cos \cdot \cos \mp \sin \cdot \sin$ (same functions in each term but different from numerator function), and the signs are opposite between sin and cos formulas.

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