Using the value of `cos 315 ^(@) ,` find the value of `sin 157 ""(1)/(2) ^(@and cos 157 (1)/(2) ""^(@).`

Answer

To solve the problem of finding the values of \( \sin 157.5^\circ \) and \( \cos 157.5^\circ \) using the value of \( \cos 315^\circ \), we can follow these steps:

### Step 1: Identify the Quadrant
Since \( 157.5^\circ \) is between \( 90^\circ \) and \( 180^\circ \), it lies in the second quadrant where sine is positive and cosine is negative.

**Hint:** Remember the signs of trigonometric functions in different quadrants.

### Step 2: Use the Half-Angle Formulas
We can express \( \sin 157.5^\circ \) and \( \cos 157.5^\circ \) using the half-angle formulas:
- \( \sin \theta = \sqrt{\frac{1 - \cos 2\theta}{2}} \)
- \( \cos \theta = \sqrt{\frac{1 + \cos 2\theta}{2}} \)

Here, \( \theta = 157.5^\circ \) corresponds to \( 2\theta = 315^\circ \).

**Hint:** Recall the half-angle identities for sine and cosine.

### Step 3: Find \( \cos 315^\circ \)
To find \( \cos 315^\circ \), we can use the cosine of a related angle:
\[ \cos 315^\circ = \cos(360^\circ - 45^\circ) = \cos 45^\circ = \frac{1}{\sqrt{2}} \]

**Hint:** Use the property \( \cos(360^\circ - \theta) = \cos \theta \).

### Step 4: Calculate \( \sin 157.5^\circ \)
Using the half-angle formula for sine:
\[ \sin 157.5^\circ = \sqrt{\frac{1 - \cos 315^\circ}{2}} = \sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{2}} \]
Now simplify:
\[ = \sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{2}} = \sqrt{\frac{\frac{\sqrt{2} - 1}{\sqrt{2}}}{2}} = \sqrt{\frac{\sqrt{2} - 1}{2\sqrt{2}}} \]
This can be simplified further:
\[ = \frac{1}{\sqrt{2}} \sqrt{\sqrt{2} - 1} \]

**Hint:** Simplifying square roots often involves rationalizing the denominator.

### Step 5: Calculate \( \cos 157.5^\circ \)
Using the half-angle formula for cosine:
\[ \cos 157.5^\circ = -\sqrt{\frac{1 + \cos 315^\circ}{2}} = -\sqrt{\frac{1 + \frac{1}{\sqrt{2}}}{2}} \]
Now simplify:
\[ = -\sqrt{\frac{\frac{\sqrt{2} + 1}{\sqrt{2}}}{2}} = -\sqrt{\frac{\sqrt{2} + 1}{2\sqrt{2}}} \]
This can be simplified further:
\[ = -\frac{1}{\sqrt{2}} \sqrt{\sqrt{2} + 1} \]

**Hint:** Remember that cosine is negative in the second quadrant.

### Final Results
Thus, we have:
\[ \sin 157.5^\circ = \frac{1}{\sqrt{2}} \sqrt{\sqrt{2} - 1} \]
\[ \cos 157.5^\circ = -\frac{1}{\sqrt{2}} \sqrt{\sqrt{2} + 1} \]