Find `sin^(-1)(sintheta).cos^(-1)(costheta),tan^(-1)(tantheta),cot^(-1)(cottheta)` for `theta in ((5pi)/2,3pi)`

Answer

To solve the problem, we need to evaluate the following expressions for \( \theta \) in the interval \( \left(\frac{5\pi}{2}, 3\pi\right) \):

1. \( \sin^{-1}(\sin \theta) \)
2. \( \cos^{-1}(\cos \theta) \)
3. \( \tan^{-1}(\tan \theta) \)
4. \( \cot^{-1}(\cot \theta) \)

### Step 1: Determine the Range of \( \theta \)

Given \( \theta \) is in the interval \( \left(\frac{5\pi}{2}, 3\pi\right) \):
- Convert \( \frac{5\pi}{2} \) to degrees:
\[ \frac{5\pi}{2} = \frac{5 \times 180}{2} = 450^\circ \]
- Convert \( 3\pi \) to degrees:
\[ 3\pi = 3 \times 180 = 540^\circ \]
Thus, \( \theta \) is in the interval \( (450^\circ, 540^\circ) \).

### Step 2: Evaluate \( \sin^{-1}(\sin \theta) \)

In the interval \( (450^\circ, 540^\circ) \):
- The sine function is periodic with a period of \( 360^\circ \).
- We can express \( \theta \) as:
\[ \theta = 450^\circ + x \quad \text{where } 0 < x < 90^\circ \]
- Therefore, \( \sin \theta = \sin(450^\circ + x) = \sin(90^\circ + x) = \cos x \).
- Since \( \sin^{-1}(\sin \theta) \) will yield the principal value:
\[ \sin^{-1}(\sin \theta) = 3\pi - \theta \]

### Step 3: Evaluate \( \cos^{-1}(\cos \theta) \)

In the interval \( (450^\circ, 540^\circ) \):
- The cosine function is also periodic with a period of \( 360^\circ \).
- We can express \( \theta \) as:
\[ \theta = 450^\circ + x \quad \text{where } 0 < x < 90^\circ \]
- Therefore, \( \cos \theta = \cos(450^\circ + x) = \cos(90^\circ + x) = -\sin x \).
- Since \( \cos^{-1}(\cos \theta) \) will yield:
\[ \cos^{-1}(\cos \theta) = \theta - 2\pi \]

### Step 4: Evaluate \( \tan^{-1}(\tan \theta) \)

In the interval \( (450^\circ, 540^\circ) \):
- The tangent function is periodic with a period of \( 180^\circ \).
- We can express \( \theta \) as:
\[ \theta = 450^\circ + x \quad \text{where } 0 < x < 90^\circ \]
- Therefore, \( \tan \theta = \tan(450^\circ + x) = \tan(x) \).
- Thus, we have:
\[ \tan^{-1}(\tan \theta) = \theta - \pi \]

### Step 5: Evaluate \( \cot^{-1}(\cot \theta) \)

In the interval \( (450^\circ, 540^\circ) \):
- The cotangent function is periodic with a period of \( 180^\circ \).
- We can express \( \theta \) as:
\[ \theta = 450^\circ + x \quad \text{where } 0 < x < 90^\circ \]
- Therefore, \( \cot \theta = \cot(450^\circ + x) = -\tan(x) \).
- Thus, we have:
\[ \cot^{-1}(\cot \theta) = \theta - 2\pi \]

### Final Answers

1. \( \sin^{-1}(\sin \theta) = 3\pi - \theta \)
2. \( \cos^{-1}(\cos \theta) = \theta - 2\pi \)
3. \( \tan^{-1}(\tan \theta) = \theta - \pi \)
4. \( \cot^{-1}(\cot \theta) = \theta - 2\pi \)