Problem 47 - A Graphical Approach to Precalculus with Limits

Question

Suppose that A and B are angles in standand position. Use the given information to find (a) \(\sin (A+B)\), (b) \(\sin (A-B)\), (c) \(\tan (A+B)\), (d) \(\tan (A-B)\), (e) the quadrant of \(A+B\), and ( \(f\) ) the quadrant of \(A-B\). Do not use a calculator. $$\cos A=\frac{3}{5}, \sin B=\frac{5}{13}, \quad 0

Step-by-Step Solution

Answer

(a) \( \frac{63}{65} \) (b) \( \frac{33}{65} \); (c) \( \frac{21}{16} \); (d) \( \frac{33}{56} \); (e) Quadrant 1; (f) Quadrant 1.

Step 1: Find \( \sin A \)

To find \( \sin A \), use the Pythagorean Identity: \( \sin^2 A + \cos^2 A = 1 \). We know \( \cos A = \frac{3}{5} \). Thus, \( \sin^2 A = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \). So, \( \sin A = \frac{4}{5} \), since \( A \) is in the first quadrant and sine is positive.

Step 2: Find \( \cos B \)

Using the Pythagorean Identity again: \( \sin^2 B + \cos^2 B = 1 \). We have \( \sin B = \frac{5}{13} \). Thus, \( \cos^2 B = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169} \). Therefore, \( \cos B = \frac{12}{13} \), as \( B \) is also in the first quadrant and cosine is positive.

Step 3: Calculate \( \sin(A+B) \)

Use the formula: \( \sin(A + B) = \sin A \cos B + \cos A \sin B \). Substitute the values: \( \sin(A + B) = \left( \frac{4}{5} \right) \left( \frac{12}{13} \right) + \left( \frac{3}{5} \right) \left( \frac{5}{13} \right) \).Calculate: \( \frac{48}{65} + \frac{15}{65} = \frac{63}{65} \).

Step 4: Calculate \( \sin(A-B) \)

Use the formula: \( \sin(A - B) = \sin A \cos B - \cos A \sin B \). Substitute the values: \( \sin(A - B) = \left( \frac{4}{5} \right) \left( \frac{12}{13} \right) - \left( \frac{3}{5} \right) \left( \frac{5}{13} \right) \).Calculate: \( \frac{48}{65} - \frac{15}{65} = \frac{33}{65} \).

Step 5: Calculate \( \tan A \) and \( \tan B \)

Use the identity \( \tan\theta = \frac{\sin\theta}{\cos\theta} \). We know \( \sin A = \frac{4}{5} \) and \( \cos A = \frac{3}{5} \), so \( \tan A = \frac{4}{3} \). Similarly, \( \sin B = \frac{5}{13} \) and \( \cos B = \frac{12}{13} \), so \( \tan B = \frac{5}{12} \).

Step 6: Calculate \( \tan(A+B) \)

Use the formula: \( \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \).Substitute the values: \( \tan(A + B) = \frac{\frac{4}{3} + \frac{5}{12}}{1 - \frac{4}{3} \cdot \frac{5}{12}} \).Calculate: The numerator becomes \( \frac{16}{12} + \frac{5}{12} = \frac{21}{12} \).The denominator becomes \( 1 - \frac{20}{36} = \frac{16}{36} = \frac{4}{9} \).Thus, \( \tan(A + B) = \frac{21}{12} \times \frac{9}{4} = \frac{63}{48} = \frac{21}{16} \).

Step 7: Calculate \( \tan(A-B) \)

Use the formula: \( \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \).Substitute the values: \( \tan(A - B) = \frac{\frac{4}{3} - \frac{5}{12}}{1 + \frac{4}{3} \cdot \frac{5}{12}} \).Calculate: The numerator becomes \( \frac{16}{12} - \frac{5}{12} = \frac{11}{12} \).The denominator becomes \( 1 + \frac{20}{36} = \frac{56}{36} = \frac{14}{9} \).Thus, \( \tan(A - B) = \frac{11}{12} \times \frac{9}{14} = \frac{99}{168} = \frac{33}{56} \).

Step 8: Determine the Quadrant of \( A+B \) and \( A-B \)

Since both \( A \) and \( B \) are in the first quadrant \( (0 < A, B < \frac{\pi}{2}) \), their sum \( A+B \) will be wholly in the first quadrant \((0 < A+B < \pi)\). For \( A-B \), as long as \( B \) is less than \( A \), \( A-B \) will remain within the first quadrant as well.

Key Concepts

Angle Addition

In trigonometry, the angle addition formulas are particularly useful for finding the sine, cosine, and tangent of the sum of two angles. These formulas allow us to express trigonometric functions of sum of angles in terms of functions of individual angles. For sine, the formula is:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]
This formula indicates how the sine function of the sum of two angles A and B can be represented using their individual sine and cosine values.

• When using this formula, it's important to correctly substitute the values of \(\sin A\), \(\cos B\), \(\cos A\), and \(\sin B\).
• Add the two resulting products together to find \(\sin(A + B)\).

Applying this approach, with \(\sin A = \frac{4}{5}\), \(\cos B = \frac{12}{13}\), \(\cos A = \frac{3}{5}\), and \(\sin B = \frac{5}{13}\), we expressed \(\sin(A + B)\) as \(\frac{63}{65}\). This illustrates how angle addition helps in computing trigonometric values without relying on calculators.

Angle Subtraction

Angle subtraction formulas work similarly to addition formulas but are used when you need to find the sine, cosine, or tangent of the difference between two angles. The subtraction formula for sine is given by:
\[ \sin(A - B) = \sin A \cos B - \cos A \sin B \]
This method helps in determining the sine of the difference of angles by subtracting the product of their cosine and sine values.

• Using this formula requires the same preparatory steps as with angle addition.
• You substitute the calculated sine and cosine values and then subtract the products accordingly.

With known values such as \(\sin A = \frac{4}{5}\), \(\cos B = \frac{12}{13}\), \(\cos A = \frac{3}{5}\), and \(\sin B = \frac{5}{13}\), you find \(\sin(A - B)\) equals \(\frac{33}{65}\). This demonstrates how the angle subtraction formula eases computation of complex trigonometric expressions.

Pythagorean Identity

The Pythagorean Identity is foundational in trigonometry. It relates the square of sine and cosine functions:
\[ \sin^2\theta + \cos^2\theta = 1 \]
This identity shows that for any angle \(\theta\), the sum of the squares of sine and cosine will always equal one. It's particularly useful when one trigonometric function value is known.

• This relationship was crucial in this exercise to find alternate trigonometric values not provided directly, like \(\sin A\) or \(\cos B\).
• To find \(\sin A\) from \(\cos A\), substitute in the identity: \(\sin^2 A = 1 - \cos^2 A\).

Here, using \(\cos A = \frac{3}{5}\), allowed calculation of \(\sin A\) as \(\frac{4}{5}\), and \(\sin B = \frac{5}{13}\) enabled finding \(\cos B = \frac{12}{13}\). This method highlights how the Pythagorean Identity facilitates finding unknown trigonometric values simply.

Sine and Cosine Functions

Understanding sine and cosine functions is essential for mastering trigonometry. The sine function, represented as \(\sin\), describes the ratio of the opposite side to the hypotenuse in a right triangle, while the cosine function \(\cos\) represents the ratio of the adjacent side to the hypotenuse.
In the context of this problem:

• The exercise utilized the sine values \(\sin A\) and \(\sin B\) calculated using fundamental trigonometric identities and existing cosine values.
• These calculations were essential for applying angle addition and subtraction formulas to find the final results.

By understanding these functions and how they work together to represent angles and sides of triangles, you can apply them to solve complex trigonometric identities, such as finding \(\sin(A+B)\) or \(\sin(A-B)\), accurately and efficiently. This forms the basis for solving most trigonometric problems.