If cos(A - B) = $\frac{\sqrt 3}{2}$ and cot(A + B) = $\frac{1}{\sqrt 3}$ , Where A - B and A + B are acute angles, then (2A - 3B) is equal to:

Question

If cos(A - B) = $\frac{\sqrt 3}{2}$  and cot(A + B) =  $\frac{1}{\sqrt 3}$ , Where A - B and A + B are acute angles, then (2A - 3B) is equal to:

Prepp Team · Accepted Answer

Solving Trigonometric Equations to Find Angle Values

The problem provides two trigonometric equations involving combinations of two angles, A and B. We are given that $ \cos(A - B) = \frac{\sqrt 3}{2} $ and $ \cot(A + B) = \frac{1}{\sqrt 3} $. We are also told that $ (A - B) $ and $ (A + B) $ are acute angles. Our goal is to find the value of the expression $ (2A - 3B) $.

Step 1: Determine the values of (A - B) and (A + B)

We use the given trigonometric equations and the fact that $ (A-B) $ and $ (A+B) $ are acute angles (between $ 0^\circ $ and $ 90^\circ $).

For the first equation, $ \cos(A - B) = \frac{\sqrt 3}{2} $. We know that $ \cos(30^\circ) = \frac{\sqrt 3}{2} $. Since $ (A-B) $ is acute, the only possible value for $ (A-B) $ is $ 30^\circ $.
So, $ A - B = 30^\circ $ (Equation 1)
For the second equation, $ \cot(A + B) = \frac{1}{\sqrt 3} $. We know that $ \cot(60^\circ) = \frac{1}{\sqrt 3} $. Since $ (A+B) $ is acute, the only possible value for $ (A+B) $ is $ 60^\circ $.
So, $ A + B = 60^\circ $ (Equation 2)

Step 2: Solve the system of linear equations for A and B

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

Equation 1: $ A - B = 30^\circ $
Equation 2: $ A + B = 60^\circ $

We can solve this system by adding the two equations:

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

$ A - B + A + B = 90^\circ $

$ 2A = 90^\circ $

Dividing by 2, we get:

$ A = \frac{90^\circ}{2} = 45^\circ $

Now substitute the value of A into either Equation 1 or Equation 2 to find B. Let's use Equation 2:

$ A + B = 60^\circ $

$ 45^\circ + B = 60^\circ $

Subtract $ 45^\circ $ from both sides:

$ B = 60^\circ - 45^\circ = 15^\circ $

So, we have $ A = 45^\circ $ and $ B = 15^\circ $.

Step 3: Calculate the value of (2A - 3B)

Now we substitute the values of A and B into the expression $ (2A - 3B) $:

$ 2A - 3B = 2(45^\circ) - 3(15^\circ) $

$ 2A - 3B = 90^\circ - 45^\circ $

$ 2A - 3B = 45^\circ $

Thus, the value of $ (2A - 3B) $ is $ 45^\circ $.

Revision Table: Key Trigonometric Values

Angle $ \theta $	$ \cos(\theta) $	$ \cot(\theta) $
$ 0^\circ $	1	Undefined
$ 30^\circ $	$ \frac{\sqrt{3}}{2} $	$ \sqrt{3} $
$ 45^\circ $	$ \frac{\sqrt{2}}{2} $	1
$ 60^\circ $	$ \frac{1}{2} $	$ \frac{1}{\sqrt{3}} $
$ 90^\circ $	0	0

Additional Information: Acute Angles and Trigonometric Functions

An acute angle is an angle that measures less than $ 90^\circ $. In the first quadrant (angles from $ 0^\circ $ to $ 90^\circ $), all basic trigonometric functions (sine, cosine, tangent, cosecant, secant, cotangent) have positive values. The problem specifies that $ (A-B) $ and $ (A+B) $ are acute angles, which simplifies finding their exact values from the given cosine and cotangent values, as we only consider the principal values in the acute range.

Solving a system of linear equations, as done in Step 2, is a common technique used in many areas of mathematics, including trigonometry problems like this one, when you have two or more equations involving the same variables.

Understanding the standard trigonometric values for common angles like $ 30^\circ, 45^\circ, $ and $ 60^\circ $ is crucial for solving many trigonometry problems efficiently.

Prepp Team · Answer

60&deg;

Prepp Team · Answer

15&deg;

Prepp Team · Answer

30&deg;

Angle \( \theta \)	\( \cos(\theta) \)	\( \cot(\theta) \)
\( 0^\circ \)	1	Undefined
\( 30^\circ \)	\( \frac{\sqrt{3}}{2} \)	\( \sqrt{3} \)
\( 45^\circ \)	\( \frac{\sqrt{2}}{2} \)	1
\( 60^\circ \)	\( \frac{1}{2} \)	\( \frac{1}{\sqrt{3}} \)
\( 90^\circ \)	0	0