Find the value of
`(i) sin 22^(@) 30'` (ii) cos `22^(@) 30'` (iii) tan `22^(@) 30'`

Answer

To find the values of \( \sin 22^\circ 30' \), \( \cos 22^\circ 30' \), and \( \tan 22^\circ 30' \), we can follow these steps:

### Step 1: Convert 22 degrees 30 minutes to decimal degrees
30 minutes is equal to \( \frac{30}{60} = 0.5 \) degrees. Therefore:
\[ 22^\circ 30' = 22.5^\circ \]

### Step 2: Calculate \( \sin 22^\circ 30' \)
We will use the identity:
\[ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \]
Here, \( \theta = 22.5^\circ \) so \( 2\theta = 45^\circ \). Thus:
\[ \sin^2 22.5^\circ = \frac{1 - \cos 45^\circ}{2} \]
Since \( \cos 45^\circ = \frac{1}{\sqrt{2}} \):
\[ \sin^2 22.5^\circ = \frac{1 - \frac{1}{\sqrt{2}}}{2} \]
\[ = \frac{\sqrt{2} - 1}{2\sqrt{2}} \]
So:
\[ \sin 22.5^\circ = \sqrt{\frac{\sqrt{2} - 1}{2\sqrt{2}}} \]

### Step 3: Calculate \( \cos 22^\circ 30' \)
We will use the identity:
\[ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \]
Using \( \theta = 22.5^\circ \):
\[ \cos^2 22.5^\circ = \frac{1 + \cos 45^\circ}{2} \]
Substituting \( \cos 45^\circ = \frac{1}{\sqrt{2}} \):
\[ \cos^2 22.5^\circ = \frac{1 + \frac{1}{\sqrt{2}}}{2} \]
\[ = \frac{\sqrt{2} + 1}{2\sqrt{2}} \]
Thus:
\[ \cos 22.5^\circ = \sqrt{\frac{\sqrt{2} + 1}{2\sqrt{2}}} \]

### Step 4: Calculate \( \tan 22^\circ 30' \)
Using the identity:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
We have:
\[ \tan 22.5^\circ = \frac{\sin 22.5^\circ}{\cos 22.5^\circ} \]
Substituting the values we found:
\[ \tan 22.5^\circ = \frac{\sqrt{\frac{\sqrt{2} - 1}{2\sqrt{2}}}}{\sqrt{\frac{\sqrt{2} + 1}{2\sqrt{2}}}} \]
This simplifies to:
\[ \tan 22.5^\circ = \sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}} \]

### Final Results
1. \( \sin 22^\circ 30' = \sqrt{\frac{\sqrt{2} - 1}{2\sqrt{2}}} \)
2. \( \cos 22^\circ 30' = \sqrt{\frac{\sqrt{2} + 1}{2\sqrt{2}}} \)
3. \( \tan 22^\circ 30' = \sqrt{\frac{\sqrt{2} - 1}{\sqrt{2} + 1}} \)