2cos (A+B) = 1 = 2sin (AB) and if A, B are acute angles, show that A = 45°,

B = 15"

We are given two trigonometric equations:

\[
2 \cos(A + B) = 1
\]
\[
2 \sin(AB) = 1
\]

We need to show that for acute angles \(A\) and \(B\), the solution is \(A = 45^\circ\) and \(B = 15^\circ\).

### Step 1: Solve \(2 \cos(A + B) = 1\)

\[
2 \cos(A + B) = 1
\]
Dividing both sides by 2:
\[
\cos(A + B) = \frac{1}{2}
\]
Now, \(\cos(A + B) = \frac{1}{2}\) implies that \(A + B = 60^\circ\), since \(\cos 60^\circ = \frac{1}{2}\).

### Step 2: Solve \(2 \sin(AB) = 1\)

\[
2 \sin(AB) = 1
\]
Dividing both sides by 2:
\[
\sin(AB) = \frac{1}{2}
\]
The sine of an angle is \(\frac{1}{2}\) when that angle is \(30^\circ\). So, we have:
\[
AB = 30^\circ
\]

### Step 3: Solve the system of equations

We now have the following two equations:

1. \(A + B = 60^\circ\)
2. \(AB = 30^\circ\)

Let’s assume that \(A = 45^\circ\) and \(B = 15^\circ\). We can check if these values satisfy both equations.

1. \(A + B = 45^\circ + 15^\circ = 60^\circ\) (True)
2. \(AB = 45^\circ \times 15^\circ = 30^\circ\) (True)

Thus, the values \(A = 45^\circ\) and \(B = 15^\circ\) satisfy both equations, confirming the solution.

### Conclusion

For the given conditions, \(A = 45^\circ\) and \(B = 15^\circ\) satisfy both equations.

Do you want further details or have any questions?

Here are some related questions for practice:

1. Solve \(2 \sin(2A) = 1\) for \(A\).
2. Show that if \(\cos(2A) = 0\), \(A = 45^\circ\).
3. What is the value of \(A\) if \(2 \cos(3A) = 1\)?
4. Prove that \(\sin(90^\circ - A) = \cos(A)\) for any angle \(A\).
5. Find \(A\) if \(\sin(2A) = \cos(A)\).

**Tip**: When solving trigonometric equations, always check for all possible angle values within the given range.

2cos(A + B) = 1 = 2sin(AB) and if A, B are acute angles, show that A = 45°, B = 15°

This solution explains how to solve the trigonometric equations 2cos(A + B) = 1 and 2sin(AB) = 1 for acute angles A and B. Using key trigonometric identities, we show that A = 45° and B = 15°. Suitable for students in Grades 10-12.

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