Find the exact value of each of the following under the given conditions below. $$ \tan \alpha=-\frac{4}{3}, \frac{\pi}{2}

Question

Find the exact value of each of the following under the given conditions below. $$ \tan \alpha=-\frac{4}{3}, \frac{\pi}{2}<\alpha<\pi ; \cos \beta=\frac{\sqrt{3}}{2}, 0<\beta<\frac{\pi}{2} $$ $$ \begin{array}{llll}\text { (a) } \sin (\alpha+\beta) & \text { (b) } \cos (\alpha+\beta) & \text { (c) } \sin (\alpha-\beta) & \text { (d) } \tan (\alpha-\beta)\end{array} $$

UpStudy AI · Accepted Answer

To find the exact values of the trigonometric functions under the given conditions, we will follow these steps:

1. **Determine the values of $$\sin \alpha$$ and $$\cos \alpha$$** using the given $$\tan \alpha = -\frac{4}{3}$$ and the fact that $$\alpha$$ is in the second quadrant.
2. **Determine the values of $$\sin \beta$$ and $$\cos \beta$$** using the given $$\cos \beta = \frac{\sqrt{3}}{2}$$ and the fact that $$\beta$$ is in the first quadrant.
3. **Use the angle addition and subtraction formulas** to find $$\sin(\alpha + \beta)$$, $$\cos(\alpha + \beta)$$, $$\sin(\alpha - \beta)$$, and $$\tan(\alpha - \beta)$$.

### Step 1: Finding $$\sin \alpha$$ and $$\cos \alpha$$

Given:
$$
\tan \alpha = -\frac{4}{3}
$$
This means:
$$
\sin \alpha = -4k \quad \text{and} \quad \cos \alpha = 3k
$$
for some $$k > 0$$. Since $$\alpha$$ is in the second quadrant, $$\sin \alpha$$ is positive and $$\cos \alpha$$ is negative. Thus, we have:
$$
\sin \alpha = 4k \quad \text{and} \quad \cos \alpha = -3k
$$

Using the Pythagorean identity:
$$
\sin^2 \alpha + \cos^2 \alpha = 1
$$
Substituting the values:
$$
(4k)^2 + (-3k)^2 = 1 \implies 16k^2 + 9k^2 = 1 \implies 25k^2 = 1 \implies k^2 = \frac{1}{25} \implies k = \frac{1}{5}
$$

Now substituting back to find $$\sin \alpha$$ and $$\cos \alpha$$:
$$
\sin \alpha = 4k = 4 \cdot \frac{1}{5} = \frac{4}{5}
$$
$$
\cos \alpha = -3k = -3 \cdot \frac{1}{5} = -\frac{3}{5}
$$

### Step 2: Finding $$\sin \beta$$

Given:
$$
\cos \beta = \frac{\sqrt{3}}{2}
$$
Since $$\beta$$ is in the first quadrant, we can find $$\sin \beta$$ using the Pythagorean identity:
$$
\sin^2 \beta + \cos^2 \beta = 1
$$
Substituting the value of $$\cos \beta$$:
$$
\sin^2 \beta + \left(\frac{\sqrt{3}}{2}\right)^2 = 1 \implies \sin^2 \beta + \frac{3}{4} = 1 \implies \sin^2 \beta = 1 - \frac{3}{4} = \frac{1}{4} \implies \sin \beta = \frac{1}{2}
$$

### Step 3: Calculating the required values

Now we can calculate each of the required values:

#### (a) $$\sin(\alpha + \beta)$$
Using the formula:
$$
\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta
$$
Substituting the values:
$$
\sin(\alpha + \beta) = \left(\frac{4}{5}\right) \left(\frac{\sqrt{3}}{2}\right) + \left(-\frac{3}{5}\right) \left(\frac{1}{2}\right)
$$
Calculating:
$$
= \frac{4\sqrt{3}}{10} - \frac{3}{10} = \frac{4\sqrt{3} - 3}{10}
$$

#### (b) $$\cos(\alpha + \beta)$$
Using the formula:
$$
\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta
$$
Substituting the values:
$$
\cos(\alpha + \beta) = \left(-\frac{3}{5}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{4}{5}\right) \left(\frac{1}{2}\right)
$$
Calculating:
$$
= -\frac{3\sqrt{3}}{10} - \frac{4}{10} = -\frac{3\sqrt{3} + 4}{10}
$$

#### (c) $$\sin(\alpha - \beta)$$
Using the formula:
$$
\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta
$$
Substituting the values:
$$
\sin(\alpha - \beta) = \left(\frac{4}{5}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(-\frac{3}{5}\right) \left(\frac{1}{2}\right)
$$
Calculating:
$$
= \frac{4\sqrt{3}}{10} + \frac{3}{10} = \frac{4\sqrt{3} + 3}{10}
$$

#### (d) $$\tan(\alpha - \beta)$$
Using the formula:
$$
\tan(\alpha - \beta) = \frac{\sin(\alpha - \beta)}{\cos(\alpha + \beta)}
$$
Substituting the values:
$$
\tan(\alpha - \beta) = \frac{\frac{4\sqrt{3} + 3}{10}}{-\frac{3\sqrt{3} + 4}{10}} = \frac{4\sqrt{3} + 3}{-(3\sqrt{3} + 4)} = -\frac{4\sqrt{3} + 3}{3\sqrt{3} + 4}
$$

### Final Answers
Thus, the exact values are:
$$
\begin{array}{llll}
\text{(a) } \sin(\alpha + \beta = \frac{4\sqrt{3} - 3}{10} \
\text{(b) } \cos(\alpha + \beta) = -\frac{3\sqrt{3} + 4}{10} \
\text{(c) } \sin(\alpha - \beta) = \frac{4\sqrt{3} + 3}{10} \
\
\[ \begin{array}{llll} \text{(a) } \sin(\alpha + \beta) = \frac{4\sqrt{3} - 3}{10} \ \text{(b) } \cos(\alpha + \beta) = -\frac{3\sqrt{3} + 4}{10} \ \text{(c) } \sin(\alpha - \beta) = \frac{4\sqrt{3} + 3}{10} \ \text{(d) } \tan(\alpha - \beta) = -\frac{4\sqrt{3} + 3}{3\sqrt{3} + 4} \end{array} $$

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Trigonometry Questions