19 Given that $$ an A=2 $$ and $$ \sin B=-\frac{2}{3} $$, where $$ A\frac{\pi}{2} $$ and $$ B\frac{\pi}{2} $$, find the value of $$ \sin \Lambda $$ and of $$ an B $$ if (i) angles $$ \Lambda $$ and $$ B $$ are in the same quadrant, (ii) angles $$ A $$ and $$ B $$ are in different quadrants. Show working to justify your answer.

Question

19 Given that $$ 	an A=2 $$ and $$ \sin B=-\frac{2}{3} $$, where $$ A>\frac{\pi}{2} $$ and $$ B>\frac{\pi}{2} $$, find the value of $$ \sin \Lambda $$ and of $$ 	an B $$ if (i) angles $$ \Lambda $$ and $$ B $$ are in the same quadrant, (ii) angles $$ A $$ and $$ B $$ are in different quadrants. Show working to justify your answer.

UpStudy AI · Accepted Answer

To solve the problem, we need to find the values of $$ \sin \Lambda $$ and $$ \tan B $$ under the given conditions. Let's break it down step by step.

### Step 1: Analyze the given information

1. **For angle $$ A $$**:
   - $$ \tan A = 2 $$
   - Since $$ A > \frac{\pi}{2} $$, angle $$ A $$ is in the second quadrant.

2. **For angle $$ B $$**:
   - $$ \sin B = -\frac{2}{3} $$
   - Since $$ B > \frac{\pi}{2} $$, angle $$ B $$ is in the third quadrant (as sine is negative in the third and fourth quadrants).

### Step 2: Find $$ \cos A $$ and $$ \sin A $$

Using the identity $$ \tan A = \frac{\sin A}{\cos A} $$, we can express $$ \sin A $$ and $$ \cos A $$ in terms of a right triangle.

Let:
- Opposite side = 2
- Adjacent side = -1 (since $$ A $$ is in the second quadrant, cosine is negative)

Using the Pythagorean theorem:
$$
\text{Hypotenuse} = \sqrt{(-1)^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5}
$$

Thus, we have:
$$
\sin A = \frac{2}{\sqrt{5}}, \quad \cos A = -\frac{1}{\sqrt{5}}
$$

### Step 3: Find $$ \cos B $$

Using the identity $$ \sin^2 B + \cos^2 B = 1 $$:
$$
\left(-\frac{2}{3}\right)^2 + \cos^2 B = 1
$$
$$
\frac{4}{9} + \cos^2 B = 1
$$
$$
\cos^2 B = 1 - \frac{4}{9} = \frac{5}{9}
$$
$$
\cos B = -\frac{\sqrt{5}}{3} \quad (\text{since } B \text{ is in the third quadrant})
$$

### Step 4: Find $$ \tan B $$

Using the definition of tangent:
$$
\tan B = \frac{\sin B}{\cos B} = \frac{-\frac{2}{3}}{-\frac{\sqrt{5}}{3}} = \frac{2}{\sqrt{5}}
$$

### Step 5: Find $$ \sin \Lambda $$

#### Case (i): Angles $$ \Lambda $$ and $$ B $$ are in the same quadrant (third quadrant)

Since both angles are in the third quadrant, we can use the sine and cosine values we found for $$ B $$:
$$
\sin \Lambda = \sin B = -\frac{2}{3}
$$

#### Case (ii): Angles $$ A $$ and $$ B $$ are in different quadrants (A in second, B in third)

In this case, we can use the sine addition formula:
$$
\sin \Lambda = \sin(A + B) = \sin A \cos B + \cos A \sin B
$$
Substituting the values we found:
$$
\sin \Lambda = \left(\frac{2}{\sqrt{5}}\right)\left(-\frac{\sqrt{5}}{3}\right) + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{2}{3}\right)
$$
Calculating each term:
$$
= -\frac{2}{3} + \frac{2}{3} = 0
$$

### Final Results

1. For case (i): 
   - $$ \sin \Lambda = -\frac{2}{3} $$
   - $$ \tan B = \frac{2}{\sqrt{5}} $$

2. For case (ii):
   - $$ \sin \Lambda = 0 $$
   - $$ \tan B = \frac{2}{\sqrt{5}} $$

Thus, the values are:
- $$ \sin \Lambda = -\frac{2}{3} $$ (same quadrant) or $$ 0 $$ (different quadrants)
- $$ \tan B = \frac{2}{\sqrt{5}} $$ in both cases.
For case (i), when angles $$ \Lambda $$ and $$ B $$ are in the same quadrant: - $$ \sin \Lambda = -\frac{2}{3} $$ - $$ \tan B = \frac{2}{\sqrt{5}} $$ For case (ii), when angles $$ A $$ and $$ B $$ are in different quadrants: - $$ \sin \Lambda = 0 $$ - $$ \tan B = \frac{2}{\sqrt{5}} $$

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