Final answer:
The problem involves solving for acute angles A and B given two equations involving trigonometric functions. While cos(A−B) equates to a possible angle due to a perfect square root value, sin(A+B) corresponds to an impossible sine value exceeding 1, indicating a possible error in the provided question.
Step-by-step explanation:
If √2 cos(A−B)=1 and 2sin(A+B)=3, and A and B are acute angles, we must solve for A and B.
From the first equation, √2 cos(A−B)=1, we can solve for cos(A−B) which gives us cos(A−B) = √2/2. This simplifies to cos(A−B) = 1/√2, which indicates that A−B must be 45 degrees, because cos(45°) = 1/√2.
From the second equation, 2sin(A+B)=3, we can solve for sin(A+B) which gives us sin(A+B) = 3/2. However, since the sine of an angle cannot be greater than 1, there must be a mistake in the equation or question as stated.
If we suppose there was a typographical error and the equation was meant to read 2sin(A+B)=1, then sin(A+B) = 1/2. Therefore, A+B might be 30 degrees because sin(30°) = 1/2. Without the proper values, we cannot solve for A and B.
Therefore, to find the values of A and B given the equations √2 cos(A−B)=1 and 2sin(A+B)=3, we need accurate values or corrections where the sine value does not exceed 1.