Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Discover comprehensive answers to your questions from knowledgeable professionals on our user-friendly platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.

Given:
[tex]\[
\sin (A) = \frac{4}{5}, \quad \frac{\pi}{2} \ \textless \ A \ \textless \ \pi \quad \text{and} \quad \sin (B) = \frac{-2 \sqrt{5}}{5}, \quad \pi \ \textless \ B \ \textless \ \frac{3 \pi}{2}
\][/tex]

What is the value of \(\cos (A - B)\)?

A. \(-\frac{2 \sqrt{5}}{25}\)

B. \(-\frac{\sqrt{5}}{5}\)

C. \(\frac{2 \sqrt{5}}{5}\)

D. [tex]\(\frac{11 \sqrt{5}}{25}\)[/tex]


Sagot :

To find the value of \(\cos(A - B)\), we need to use the trigonometric identities and the given information about the angles.

Given:
- \(\sin (A) = \frac{4}{5}\) and \(\frac{\pi}{2} < A < \pi\)
- \(\sin (B) = \frac{-2 \sqrt{5}}{5}\) and \(\pi < B < \frac{3 \pi}{2}\)

Let's work through this step-by-step:

### Step 1: Determine \(\cos(A)\)

Since \(\frac{\pi}{2} < A < \pi\), angle \(A\) is in the second quadrant, where sine is positive and cosine is negative. We use the Pythagorean identity:

[tex]\[ \sin^2(A) + \cos^2(A) = 1 \][/tex]

Substituting the given value of \(\sin(A)\):

[tex]\[ \left(\frac{4}{5}\right)^2 + \cos^2(A) = 1 \][/tex]
[tex]\[ \frac{16}{25} + \cos^2(A) = 1 \][/tex]
[tex]\[ \cos^2(A) = 1 - \frac{16}{25} \][/tex]
[tex]\[ \cos^2(A) = \frac{25}{25} - \frac{16}{25} \][/tex]
[tex]\[ \cos^2(A) = \frac{9}{25} \][/tex]
[tex]\[ \cos(A) = \pm \frac{3}{5} \][/tex]

Since \(A\) is in the second quadrant, \(\cos(A)\) is negative:

[tex]\[ \cos(A) = -\frac{3}{5} \][/tex]

### Step 2: Determine \(\cos(B)\)

Since \(\pi < B < \frac{3 \pi}{2}\), angle \(B\) is in the third quadrant where both sine and cosine are negative. Again, we use the Pythagorean identity:

[tex]\[ \sin^2(B) + \cos^2(B) = 1 \][/tex]

Substituting the given value of \(\sin(B)\):

[tex]\[ \left(\frac{-2 \sqrt{5}}{5}\right)^2 + \cos^2(B) = 1 \][/tex]
[tex]\[ \frac{4 \cdot 5}{25} + \cos^2(B) = 1 \][/tex]
[tex]\[ \frac{20}{25} + \cos^2(B) = 1 \][/tex]
[tex]\[ \cos^2(B) = 1 - \frac{20}{25} \][/tex]
[tex]\[ \cos^2(B) = \frac{25}{25} - \frac{20}{25} \][/tex]
[tex]\[ \cos^2(B) = \frac{5}{25} \][/tex]
[tex]\[ \cos(B) = \pm \frac{\sqrt{5}}{5} \][/tex]

Since \(B\) is in the third quadrant, \(\cos(B)\) is negative:

[tex]\[ \cos(B) = -\frac{\sqrt{5}}{5} \][/tex]

### Step 3: Calculate \(\cos(A - B)\)

Now, we use the formula for \(\cos(A - B)\):

[tex]\[ \cos(A - B) = \cos(A)\cos(B) + \sin(A)\sin(B) \][/tex]

Substitute the known values:

[tex]\[ \cos(A - B) = \left(-\frac{3}{5}\right)\left(-\frac{\sqrt{5}}{5}\right) + \left(\frac{4}{5}\right)\left(\frac{-2 \sqrt{5}}{5}\right) \][/tex]
[tex]\[ \cos(A - B) = \frac{3 \sqrt{5}}{25} + \frac{-8 \sqrt{5}}{25} \][/tex]
[tex]\[ \cos(A - B) = \frac{3 \sqrt{5} - 8 \sqrt{5}}{25} \][/tex]
[tex]\[ \cos(A - B) = \frac{-5 \sqrt{5}}{25} \][/tex]
[tex]\[ \cos(A - B) = -\frac{\sqrt{5}}{5} \][/tex]

Thus, the value of \(\cos(A - B)\) is \(-\frac{\sqrt{5}}{5}\).

So, the correct answer is:

[tex]\[ \boxed{-\frac{\sqrt{5}}{5}} \][/tex]