Solve the trigonometric problem: If sin A = 4/5 and cos B = -12/13, with A in First Quadrant and B in Third Quadrant, find cos(A+B). Detailed solution

MHT-CET Trigonometric Problem: Find \( \cos(A + B) \)

Question: If \( \sin A = \frac{4}{5} \) and \( \cos B = -\frac{12}{13} \), where \( A \) lies in the first quadrant and \( B \) lies in the third quadrant, then \( \cos(A + B) \) is equal to:

Options:

a) \( \frac{56}{65} \)

b) \( -\frac{56}{65} \)

c) \( \frac{16}{65} \)

d) \( -\frac{16}{65} \)

Detailed Solution

Hello, student! Let’s solve this trigonometric problem together, step by step, so you can understand every part of the process. We’re given \( \sin A = \frac{4}{5} \) with \( A \) in the first quadrant, and \( \cos B = -\frac{12}{13} \) with \( B \) in the third quadrant. Our goal is to find \( \cos(A + B) \). Let’s get started!

Step 1: Understand the Formula

To find \( \cos(A + B) \), we use the **cosine addition formula**:

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

This means we need the values of \( \cos A \) and \( \sin B \), along with the given \( \sin A = \frac{4}{5} \) and \( \cos B = -\frac{12}{13} \). Let’s find the missing pieces.

Step 2: Calculate \( \cos A \)

Since \( A \) is in the **first quadrant**, both sine and cosine are positive. We can use the **Pythagorean identity** to find \( \cos A \):

\[ \sin^2 A + \cos^2 A = 1 \]

Substitute \( \sin A = \frac{4}{5} \):

\[ \left( \frac{4}{5} \right)^2 + \cos^2 A = 1 \] \[ \frac{16}{25} + \cos^2 A = 1 \] \[ \cos^2 A = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25} \] \[ \cos A = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

We take the positive root because \( A \) is in the first quadrant, where cosine is positive. So, \( \cos A = \frac{3}{5} \).

Step 3: Calculate \( \sin B \)

Since \( B \) is in the **third quadrant**, both sine and cosine are negative. We use the Pythagorean identity again to find \( \sin B \):

\[ \sin^2 B + \cos^2 B = 1 \]

Substitute \( \cos B = -\frac{12}{13} \):

\[ \sin^2 B + \left( -\frac{12}{13} \right)^2 = 1 \] \[ \sin^2 B + \frac{144}{169} = 1 \] \[ \sin^2 B = 1 - \frac{144}{169} = \frac{169}{169} - \frac{144}{169} = \frac{25}{169} \] \[ \sin B = \pm \sqrt{\frac{25}{169}} = \pm \frac{5}{13} \]

Since \( B \) is in the third quadrant, where sine is negative, we choose the negative value: \( \sin B = -\frac{5}{13} \).

Step 4: Compute \( \cos(A + B) \)

Now we have all the values we need:

- \( \sin A = \frac{4}{5} \) - \( \cos A = \frac{3}{5} \) - \( \cos B = -\frac{12}{13} \) - \( \sin B = -\frac{5}{13} \)

Plug these into the cosine addition formula:

\[ \cos(A + B) = \cos A \cos B - \sin A \sin B \] \[ = \left( \frac{3}{5} \right) \left( -\frac{12}{13} \right) - \left( \frac{4}{5} \right) \left( -\frac{5}{13} \right) \]

Calculate each term:

- First term: \( \frac{3}{5} \times -\frac{12}{13} = -\frac{36}{65} \) - Second term: \( \frac{4}{5} \times -\frac{5}{13} = -\frac{20}{65} \)

Now combine them:

\[ \cos(A + B) = -\frac{36}{65} - \left( -\frac{20}{65} \right) \] \[ = -\frac{36}{65} + \frac{20}{65} \] \[ = \frac{-36 + 20}{65} = \frac{-16}{65} \]

So, \( \cos(A + B) = -\frac{16}{65} \).

Step 5: Double-Check

Let’s make sure this makes sense. Since \( A \) is in Q1 (0° to 90°) and \( B \) is in Q3 (180° to 270°), \( A + B \) could be between 180° and 360°. A negative cosine suggests \( A + B \) is in the second or third quadrant, which is consistent with the possible angle range. The math checks out!

Final Answer

Comparing with the options:

- a) \( \frac{56}{65} \) - b) \( -\frac{56}{65} \) - c) \( \frac{16}{65} \) - d) \( -\frac{16}{65} \)

Our result, \( -\frac{16}{65} \), matches option d). So, the answer is:

Final Answer: \( \cos(A + B) = -\frac{16}{65} \) (option d)