Let's solve this trigonometry problem step-by-step to find the value of \(A + B\) given the values of \( \sin A \) and \( \cos B \). We are provided with:

• \( \sin A = \frac{1}{2} \)
• \( \cos B = \frac{1}{2} \)

Finding the Value of Angle A

We are given \( \sin A = \frac{1}{2} \). To find the value of angle \(A\), we need to determine which angle has a sine value of \( \frac{1}{2} \). We recall the standard trigonometric values for common angles:

• \( \sin 0^\circ = 0 \)
• \( \sin 30^\circ = \frac{1}{2} \)
• \( \sin 45^\circ = \frac{1}{\sqrt{2}} \)
• \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)
• \( \sin 90^\circ = 1 \)

Comparing the given value \( \sin A = \frac{1}{2} \) with the standard values, we see that \( \sin 30^\circ = \frac{1}{2} \). Therefore, the angle \(A\) is \(30^\circ\).

So, \( A = 30^\circ \).

Finding the Value of Angle B

Next, we are given \( \cos B = \frac{1}{2} \). To find the value of angle \(B\), we need to determine which angle has a cosine value of \( \frac{1}{2} \). We recall the standard trigonometric values for common angles:

• \( \cos 0^\circ = 1 \)
• \( \cos 30^\circ = \frac{\sqrt{3}}{2} \)
• \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)
• \( \cos 60^\circ = \frac{1}{2} \)
• \( \cos 90^\circ = 0 \)

Comparing the given value \( \cos B = \frac{1}{2} \) with the standard values, we see that \( \cos 60^\circ = \frac{1}{2} \). Therefore, the angle \(B\) is \(60^\circ\).

So, \( B = 60^\circ \).

Calculating A + B

Now that we have found the values of \(A\) and \(B\), we can find their sum \(A + B\). We have \(A = 30^\circ\) and \(B = 60^\circ\).

\(A + B = 30^\circ + 60^\circ\)

\(A + B = 90^\circ\)

Thus, the value of \(A + B\) is \(90^\circ\).

Summary of Findings

Given \( \sin A = \frac{1}{2} \) and \( \cos B = \frac{1}{2} \), we determined:

• \( A = 30^\circ \)
• \( B = 60^\circ \)

Therefore, \( A + B = 30^\circ + 60^\circ = 90^\circ \).

Revision Table: Standard Trigonometric Values

Additional Information: Inverse Trigonometric Functions

The process of finding an angle when the value of a trigonometric ratio is known involves inverse trigonometric functions. For example:

• If \( \sin A = x \), then \( A = \sin^{-1}(x) \) (read as "A is the inverse sine of x" or "A is the angle whose sine is x").
• If \( \cos B = y \), then \( B = \cos^{-1}(y) \) (read as "B is the inverse cosine of y" or "B is the angle whose cosine is y").

In this problem:

• \( A = \sin^{-1}\left(\frac{1}{2}\right) = 30^\circ \)
• \( B = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \)

Inverse trigonometric functions are essential for solving trigonometric equations and finding unknown angles in various problems.