If ` cot theta=(4)/(3)`, then find the value of `sin theta,cos theta` and `cos ectheta` in first quadrant.
If `sin theta+cos ec theta=2` then find the value of `sin^(8 )theta+cos ec^(8)theta`.

Answer

To solve the problem step by step, we will break it down into two parts as outlined in the question.

### Part 1: Finding `sin(theta)`, `cos(theta)`, and `cos(ectheta)`

1. **Given Information**: We know that \( \cot \theta = \frac{4}{3} \).
- Since \( \cot \theta = \frac{\cos \theta}{\sin \theta} \), we can express this as:
\[ \frac{\cos \theta}{\sin \theta} = \frac{4}{3} \]

2. **Assigning Values**: Let \( \cos \theta = 4k \) and \( \sin \theta = 3k \) for some scalar \( k \).

3. **Finding the Hypotenuse**: Using the Pythagorean theorem:
\[ h^2 = (\sin \theta)^2 + (\cos \theta)^2 \]
Substituting the values we have:
\[ h^2 = (3k)^2 + (4k)^2 = 9k^2 + 16k^2 = 25k^2 \]
Therefore, the hypotenuse \( h \) is:
\[ h = 5k \]

4. **Calculating `sin(theta)`**:
\[ \sin \theta = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{3k}{5k} = \frac{3}{5} \]

5. **Calculating `cos(theta)`**:
\[ \cos \theta = \frac{\text{base}}{\text{hypotenuse}} = \frac{4k}{5k} = \frac{4}{5} \]

6. **Calculating `cos(ectheta)`**:
\[ \cos(ectheta) = \frac{1}{\sin \theta} = \frac{1}{\frac{3}{5}} = \frac{5}{3} \]

### Summary of Part 1:
- \( \sin \theta = \frac{3}{5} \)
- \( \cos \theta = \frac{4}{5} \)
- \( \cos(ectheta) = \frac{5}{3} \)

### Part 2: Finding `sin^8(theta) + cos^8(theta)`

1. **Given Information**: We have \( \sin \theta + \cos(ectheta) = 2 \).
- Substituting the values we found:
\[ \frac{3}{5} + \frac{5}{3} = 2 \]

2. **Simplifying the Equation**: To find \( \sin^8 \theta + \cos^8 \theta \), we can use the identity:
\[ a^8 + b^8 = (a^4 + b^4)(a^4 - b^4) \]
where \( a = \sin \theta \) and \( b = \cos \theta \).

3. **Finding \( \sin^4 \theta \) and \( \cos^4 \theta \)**:
\[ \sin^2 \theta + \cos^2 \theta = 1 \quad \text{(Pythagorean identity)} \]
\[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta = 1 - 2\left(\frac{3}{5} \cdot \frac{4}{5}\right) = 1 - 2 \cdot \frac{12}{25} = 1 - \frac{24}{25} = \frac{1}{25} \]

4. **Finding \( \sin^8 \theta + \cos^8 \theta \)**:
\[ \sin^8 \theta + \cos^8 \theta = (\sin^4 \theta + \cos^4 \theta)^2 - 2(\sin^4 \theta \cos^4 \theta) \]
We already found \( \sin^4 \theta + \cos^4 \theta = \frac{1}{25} \).
Now we need \( \sin^4 \theta \cos^4 \theta = \left(\frac{3}{5} \cdot \frac{4}{5}\right)^4 = \left(\frac{12}{25}\right)^2 = \frac{144}{625} \).

5. **Final Calculation**:
\[ \sin^8 \theta + \cos^8 \theta = \left(\frac{1}{25}\right)^2 - 2\left(\frac{144}{625}\right) = \frac{1}{625} - \frac{288}{625} = -\frac{287}{625} \]

### Summary of Part 2:
- \( \sin^8 \theta + \cos^8 \theta = -\frac{287}{625} \)

### Final Answers:
- \( \sin \theta = \frac{3}{5} \)
- \( \cos \theta = \frac{4}{5} \)
- \( \cos(ectheta) = \frac{5}{3} \)
- \( \sin^8 \theta + \cos^8 \theta = -\frac{287}{625} \)