Problem 50 - 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{15}{17}, \sin B=\frac{4}{5}, \quad \frac{\pi}{2}

Step-by-Step Solution

Answer

(a) \(\sin(A+B) = -\frac{36}{85}\), (b) \(\sin(A-B) = \frac{84}{85}\), (c) \(\tan(A+B) = \frac{36}{77}\), (d) \(\tan(A-B) = -\frac{84}{13}\), (e) third/fourth quadrant, (f) first/second quadrant.

Step 1: Determine \\sin A

Since \( \cos A = -\frac{15}{17} \) and \( \frac{\pi}{2} < A < \pi \), \( A \) is in the second quadrant. In the second quadrant, \( \sin A > 0 \). To find \( \sin A \), use the identity \( \sin^2 A + \cos^2 A = 1 \):\[ \sin^2 A = 1 - \left(-\frac{15}{17}\right)^2 \] \[ \sin^2 A = 1 - \frac{225}{289} = \frac{64}{289} \] \[ \sin A = \sqrt{\frac{64}{289}} = \frac{8}{17} \]

Step 2: Determine \\cos B

Since \( \sin B = \frac{4}{5} \) and \( 0 < B < \frac{\pi}{2} \), \( B \) is in the first quadrant. In the first quadrant, \( \cos B > 0 \). To find \( \cos B \), use the identity \( \sin^2 B + \cos^2 B = 1 \):\[ \cos^2 B = 1 - \left(\frac{4}{5}\right)^2 \] \[ \cos^2 B = 1 - \frac{16}{25} = \frac{9}{25} \] \[ \cos B = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

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

To find \( \sin(A+B) \), use the identity \( \sin(A+B) = \sin A \cos B + \cos A \sin B \):\[ \sin(A+B) = \left(\frac{8}{17}\right) \left(\frac{3}{5}\right) + \left(-\frac{15}{17}\right)\left(\frac{4}{5}\right) \] \[ \sin(A+B) = \frac{24}{85} - \frac{60}{85} = -\frac{36}{85} \]

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

To find \( \sin(A-B) \), use the identity \( \sin(A-B) = \sin A \cos B - \cos A \sin B \):\[ \sin(A-B) = \left(\frac{8}{17}\right) \left(\frac{3}{5}\right) - \left(-\frac{15}{17}\right)\left(\frac{4}{5}\right) \] \[ \sin(A-B) = \frac{24}{85} + \frac{60}{85} = \frac{84}{85} \]

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

To find \( \tan(A+B) \), use the identity \( \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \). First, calculate \( \tan A \) and \( \tan B \):\[ \tan A = \frac{\sin A}{\cos A} = \frac{\frac{8}{17}}{-\frac{15}{17}} = -\frac{8}{15} \] \[ \tan B = \frac{\sin B}{\cos B} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3} \] \[ \tan(A+B) = \frac{-\frac{8}{15} + \frac{4}{3}}{1 - \left(-\frac{8}{15}\right)\left(\frac{4}{3}\right)} \] \[ \tan(A+B) = \frac{-\frac{8}{15} + \frac{20}{15}}{1 + \frac{32}{45}} = \frac{\frac{12}{15}}{\frac{77}{45}} = \frac{36}{77} \]

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

To find \( \tan(A-B) \), use the identity \( \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \):\[ \tan(A-B) = \frac{-\frac{8}{15} - \frac{4}{3}}{1 + \left(-\frac{8}{15}\right)\left(\frac{4}{3}\right)} \] \[ \tan(A-B) = \frac{-\frac{8}{15} - \frac{20}{15}}{1 - \frac{32}{45}} = \frac{-\frac{28}{15}}{\frac{13}{45}} = -\frac{84}{13} \]

Step 7: Determine Quadrants of A+B and A-B

Given \( \sin(A+B) = -\frac{36}{85} \), \( A+B \) must be in the third or fourth quadrant because the sine is negative. For \( \sin(A-B) = \frac{84}{85} \), \( A-B \) must be in the first or second quadrant since the sine is positive.

Key Concepts

Sine Addition Formula

The Sine Addition Formula helps us calculate the sine of the sum of two angles. The formula is given by: \[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]This identity simplifies the process of finding the sine of angle sums in trigonometry. It takes the individual sines and cosines of two angles, combines them, and provides the sine of their sum. For example, in the problem, to calculate \( \sin(A+B) \), we used the values of \( \sin A = \frac{8}{17} \), \( \cos A = -\frac{15}{17} \), \( \sin B = \frac{4}{5} \), and \( \cos B = \frac{3}{5} \) to substitute into the formula. This simple substitution helps us understand how trigonometric identities can transform complicated expressions into straightforward calculations.

Cosine Identity

Cosine identities are essential in solving trigonometric problems. For our exercise, we specifically used the Pythagorean Identity. It states:\[ \sin^2 A + \cos^2 A = 1 \]Using this identity, we determined the missing trigonometric function. When \( \cos A = -\frac{15}{17} \), and knowing the angle \( A \) is in the second quadrant, we found \( \sin A \) to be positive. Similarly, since \( \sin B = \frac{4}{5} \) in the first quadrant where both sine and cosine are positive, we deduced \( \cos B = \frac{3}{5} \). This use of the Pythagorean Identity is central when dealing with angles positioned in different quadrants.

Quadrant Analysis

Understanding the position of angles in their respective quadrants is crucial for interpreting trigonometric values. There are four quadrants:

• First Quadrant (\( 0 < \theta < \frac{\pi}{2} \)): both sine and cosine are positive.
• Second Quadrant (\( \frac{\pi}{2} < \theta < \pi \)): sine is positive, cosine is negative.
• Third Quadrant (\( \pi < \theta < \frac{3\pi}{2} \)): sine and cosine are negative.
• Fourth Quadrant (\( \frac{3\pi}{2} < \theta < 2\pi \)): sine is negative, cosine is positive.

In our exercise, due to the values of \( \sin(A+B) \) and \( \sin(A-B) \), we determined their resulting quadrants. Negative sine for \( A+B \) placed it in the third or fourth quadrant, while a positive sine for \( A-B \) meant it was in the first or second quadrant. Quadrant analysis is imperative to correctly interpret the sign of trigonometric functions and their resulting values.

Tangent Addition Formula

To find the tangent of the sum or difference of two angles, we use the Tangent Addition Formula. This is represented as:\[ \tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \]And for difference:\[ \tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]First, we calculate the tangent of each angle. For angle \( A \), \( \tan A = \frac{\sin A}{\cos A} \) and for angle \( B \), \( \tan B = \frac{\sin B}{\cos B} \). Using these values, we substitute into the formulas to get \( \tan(A+B) \) and \( \tan(A-B) \). In our exercise, by calculating these, we noticed how the formula simplifies complex calculations and avoids directly measuring angles, which might not be practical without calculators.