Solving Trigonometric Equations to Find AB

In this article, we will explore how to solve for angles A and B given the equations (sin(A - B) frac{1}{2}) and (cos(A B) frac{1}{2}). These types of problems are common in trigonometry and are useful in various fields, including engineering, physics, and mathematics.

Step 1: Solving for A - B

Given the equation (sin(A - B) frac{1}{2}), we can determine that:

[A - B npi - (-1)^n frac{pi}{6} quad text{for } n in mathbb{Z}]

This yields the following specific cases for (n):

(A - B frac{pi}{6} 2kpi) (A - B frac{5pi}{6} 2kpi)

Step 2: Solving for A B

Given the equation (cos(A B) frac{1}{2}), we can determine that:

[A B 2mpi pm frac{pi}{3} quad text{for } m in mathbb{Z}]

This yields the following specific cases for (m):

(A B frac{pi}{3} 2mpi) (A B frac{5pi}{3} 2mpi)

Step 3: Solving the System of Equations

We now have the following cases to solve:

(A - B frac{pi}{6} 2kpi) and (A B frac{pi}{3} 2mpi) (A - B frac{pi}{6} 2kpi) and (A B frac{5pi}{3} 2mpi) (A - B frac{5pi}{6} 2kpi) and (A B frac{pi}{3} 2mpi) (A - B frac{5pi}{6} 2kpi) and (A B frac{5pi}{3} 2mpi)

Case 1: (A - B frac{pi}{6}) and (A B frac{pi}{3})

Adding the equations:

[2A frac{pi}{6} frac{pi}{3} frac{pi}{6} frac{2pi}{6} frac{3pi}{6} frac{pi}{2}]

[A frac{pi}{4}]

Substituting to find B:

[A B frac{pi}{3} implies frac{pi}{4} B frac{pi}{3}]

[B frac{pi}{3} - frac{pi}{4} frac{4pi - 3pi}{12} frac{pi}{12}]

Case 2: (A - B frac{pi}{6}) and (A B frac{5pi}{3})

Adding the equations:

[2A frac{pi}{6} frac{5pi}{3} frac{pi}{6} frac{10pi}{6} frac{11pi}{6}]

[A frac{11pi}{12}]

Substituting to find B:

[A B frac{5pi}{3} implies frac{11pi}{12} B frac{5pi}{3}]

[B frac{5pi}{3} - frac{11pi}{12} frac{20pi - 11pi}{12} frac{9pi}{12} frac{3pi}{4}]

Summary of Solutions

The pairs of solutions (A, B) are as follows:

(A frac{pi}{4}, B frac{pi}{12}) (A frac{11pi}{12}, B frac{3pi}{4})

You can also consider other integer multiples of (pi) to find additional solutions, depending on the specific context or restrictions of (A) and (B).

Alternative Approach

Let's consider another approach using (sin(A - B) frac{1}{2}) and (cos(A B) frac{1}{2}):

(sin(A - B) sin30^circ frac{1}{2}) (A - B 30^circ) (cos(A B) cos60^circ frac{1}{2}) (A B 60^circ) Add the two equations:

[2A 90^circ implies A 45^circ]

Substitute to find B:

[A B 60^circ implies 45^circ B 60^circ]

[B 15^circ]

The solutions are:

(A 45^circ, B 15^circ)

Conclusion

The solutions to the given trigonometric equations can be found in multiple ways, and additional solutions may be obtained by considering other integer multiples of (pi). These techniques are fundamental in solving complex trigonometric problems and are widely applicable in various mathematical and engineering contexts.