Solving Trigonometric Equations for Angles A and B

We are given the following trigonometric equations:

• \(sin\left(\frac{2A+B}{2}\right) = \frac{\sqrt{3}}{2}\)
• \(cos\left(\frac{2A-B}{2}\right) = \frac{\sqrt{3}}{2}\)

We are also given the condition \(0^{\circ}<\frac{2A+B}{2}<90^{\circ}\).

Let's analyze each equation separately to find the values of the angles \(\frac{2A+B}{2}\) and \(\frac{2A-B}{2}\).

Finding the Value of the First Angle

From the first equation, \(sin\left(\frac{2A+B}{2}\right) = \frac{\sqrt{3}}{2}\). We know that in the range \(0^{\circ}\) to \(90^{\circ}\), the angle whose sine is \(\frac{\sqrt{3}}{2}\) is \(60^{\circ}\). The given condition \(0^{\circ}<\frac{2A+B}{2}<90^{\circ}\) confirms that this is the correct angle.

So, we have:

\[ \frac{2A+B}{2} = 60^{\circ} \]

Multiplying both sides by 2 gives us our first linear equation:

\[ 2A + B = 120^{\circ} \quad (1) \]

Finding the Value of the Second Angle

From the second equation, \(cos\left(\frac{2A-B}{2}\right) = \frac{\sqrt{3}}{2}\). We know that the angle whose cosine is \(\frac{\sqrt{3}}{2}\) is \(30^{\circ}\) (or \(330^{\circ}\), \(-30^{\circ}\), etc.). Without a specific range for this angle, we usually take the principal value or a value that fits common trigonometric scenarios. Let's assume the simplest positive acute angle.

So, we have:

\[ \frac{2A-B}{2} = 30^{\circ} \]

Multiplying both sides by 2 gives us our second linear equation:

\[ 2A - B = 60^{\circ} \quad (2) \]

Solving the System of Linear Equations

Now we have a system of two linear equations with two variables A and B:

• \(2A + B = 120^{\circ}\) (1)
• \(2A - B = 60^{\circ}\) (2)

We can solve this system by adding the two equations:

Adding (1) and (2):

\[ (2A + B) + (2A - B) = 120^{\circ} + 60^{\circ} \]

\[ 4A = 180^{\circ} \]

Now, divide by 4 to find the value of A:

\[ A = \frac{180^{\circ}}{4} = 45^{\circ} \]

Now substitute the value of A (\(45^{\circ}\)) into equation (1) to find the value of B:

\[ 2(45^{\circ}) + B = 120^{\circ} \]

\[ 90^{\circ} + B = 120^{\circ} \]

Subtract \(90^{\circ}\) from both sides:

\[ B = 120^{\circ} - 90^{\circ} = 30^{\circ} \]

So, we have found that \(A = 45^{\circ}\) and \(B = 30^{\circ}\).

Calculating sin[3(A - B)]

We need to find the value of \(sin[3(A - B)]\). First, let's find the value of \(A - B\).

\[ A - B = 45^{\circ} - 30^{\circ} = 15^{\circ} \]

Now, let's find the value of \(3(A - B)\).

\[ 3(A - B) = 3(15^{\circ}) = 45^{\circ} \]

Finally, we need to calculate \(sin[3(A - B)]\), which is \(sin(45^{\circ})\).

\[ sin(45^{\circ}) = \frac{1}{\sqrt{2}} \]

Thus, the value of \(sin[3(A - B)]\) is \(\frac{1}{\sqrt{2}}\).

Summary of Steps

• Used the given sine and cosine equations to determine the values of the angles \(\frac{2A+B}{2}\) and \(\frac{2A-B}{2}\).
• Formed a system of linear equations for A and B.
• Solved the system to find the values of A and B.
• Calculated \(A - B\) and then \(3(A - B)\).
• Evaluated the sine of the resulting angle.

Final Answer Derivation

Given: \(sin\left(\frac{2A+B}{2}\right) = \frac{\sqrt{3}}{2}\) with \(0^{\circ}<\frac{2A+B}{2}<90^{\circ}\) \(cos\left(\frac{2A-B}{2}\right) = \frac{\sqrt{3}}{2}\)

From the first equation and range, \(\frac{2A+B}{2} = 60^{\circ}\), so \(2A + B = 120^{\circ}\).

From the second equation, \(\frac{2A-B}{2} = 30^{\circ}\), so \(2A - B = 60^{\circ}\).

Adding the two linear equations: \((2A + B) + (2A - B) = 120^{\circ} + 60^{\circ}\) \(4A = 180^{\circ}\) \(A = 45^{\circ}\)

Substituting A into the first equation: \(2(45^{\circ}) + B = 120^{\circ}\) \(90^{\circ} + B = 120^{\circ}\) \(B = 30^{\circ}\)

Now find \(A - B\): \(A - B = 45^{\circ} - 30^{\circ} = 15^{\circ}\)

Now find \(3(A - B)\): \(3(A - B) = 3(15^{\circ}) = 45^{\circ}\)

Finally, calculate \(sin[3(A - B)]\): \(sin(45^{\circ}) = \frac{1}{\sqrt{2}}\)

Revision Table: Key Trigonometric Values

Additional Information: Solving Systems of Equations

A system of linear equations involves two or more linear equations with the same set of variables. In this problem, we had variables A and B.

The system was:

• \(2A + B = 120^{\circ}\)
• \(2A - B = 60^{\circ}\)

We used the elimination method to solve this system. Here's how it works:

1. Align the equations with like terms in columns (A terms, B terms, constant terms).
2. Look for variables with coefficients that are opposites or the same. In our case, the B terms (\(+B\) and \(-B\)) are opposites.
3. Add or subtract the equations to eliminate one variable. Adding the equations eliminated B, leaving us with an equation only in terms of A.
4. Solve the resulting single-variable equation for that variable (we found A = \(45^{\circ}\)).
5. Substitute the value found back into either of the original equations to solve for the other variable (we substituted A into the first equation to find B = \(30^{\circ}\)).
6. Check the solution by substituting both values into the other original equation. \(2(45^{\circ}) - 30^{\circ} = 90^{\circ} - 30^{\circ} = 60^{\circ}\), which matches the second equation.

This method is effective when variables have coefficients that are easy to eliminate by adding or subtracting.