Evalute: `"sin"15^(@)`
`cos15^(@), "tan"15^(@)`

Answer

To evaluate \( \sin 15^\circ \), \( \cos 15^\circ \), and \( \tan 15^\circ \), we will use the angle subtraction formulas for sine and cosine.

### Step 1: Evaluate \( \sin 15^\circ \)

We can express \( 15^\circ \) as \( 45^\circ - 30^\circ \). Using the sine subtraction formula:

\[ \sin(a - b) = \sin a \cos b - \cos a \sin b \]

Let \( a = 45^\circ \) and \( b = 30^\circ \):

\[ \sin 15^\circ = \sin(45^\circ - 30^\circ) = \sin 45^\circ \cos 30^\circ - \cos 45^\circ \sin 30^\circ \]

Substituting the known values:

\[ \sin 45^\circ = \frac{1}{\sqrt{2}}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \cos 45^\circ = \frac{1}{\sqrt{2}}, \quad \sin 30^\circ = \frac{1}{2} \]

Now substituting these values into the equation:

\[ \sin 15^\circ = \left(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\right) \]

Calculating this gives:

\[ \sin 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}} \]

### Step 2: Evaluate \( \cos 15^\circ \)

Using the cosine subtraction formula:

\[ \cos(a - b) = \cos a \cos b + \sin a \sin b \]

Again, let \( a = 45^\circ \) and \( b = 30^\circ \):

\[ \cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos 30^\circ + \sin 45^\circ \sin 30^\circ \]

Substituting the known values:

\[ \cos 15^\circ = \left(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}\right) + \left(\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\right) \]

Calculating this gives:

\[ \cos 15^\circ = \frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} + 1}{2\sqrt{2}} \]

### Step 3: Evaluate \( \tan 15^\circ \)

Using the definition of tangent:

\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]

Thus,

\[ \tan 15^\circ = \frac{\sin 15^\circ}{\cos 15^\circ} = \frac{\frac{\sqrt{3} - 1}{2\sqrt{2}}}{\frac{\sqrt{3} + 1}{2\sqrt{2}}} \]

The \( 2\sqrt{2} \) cancels out:

\[ \tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \]

### Final Results

- \( \sin 15^\circ = \frac{\sqrt{3} - 1}{2\sqrt{2}} \)
- \( \cos 15^\circ = \frac{\sqrt{3} + 1}{2\sqrt{2}} \)
- \( \tan 15^\circ = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \)