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The angle of \(15^\circ\) can be rewritten as the subtraction of \(45^\circ\) and \(30^\circ\), so the first step is to rewrite the angle in this way: \(15^\circ = 45^\circ - 30^\circ\).

Now apply the difference formulas of sine and cosine for the angle \(15^\circ\).These are given as:For sine: \(\sin(\alpha - \beta) = \sin\alpha \cos\beta - \cos\alpha \sin\beta\)For cosine: \(\cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta\)Substitute \(\alpha = 45^\circ\) and \(\beta = 30^\circ\) into these formulas to get:\(\sin(15^\circ) = \sin(45^\circ)\cos(30^\circ) - \cos(45^\circ)\sin(30^\circ)\)\(\cos(15^\circ) = \cos(45^\circ)\cos(30^\circ) + \sin(45^\circ)\sin(30^\circ)\)

Use the known sine and cosine values for \(45^\circ\) and \(30^\circ\) to evaluate the formula that was calculated in Step 2.The values for the sine and cosine of 45 degrees are both \(\frac{\sqrt{2}}{2}\), the cosine of 30 degrees is \(\frac{\sqrt{3}}{2}\), and the sine of 30 degrees is \(\frac{1}{2}\).So,\(\sin(15^\circ) = \frac{\sqrt{2}}{2} * \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} * \frac{1}{2} = \frac{\sqrt{6} - \sqrt{2}}{4}\)\(\cos(15^\circ) = \frac{\sqrt{2}}{2} * \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} * \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}\)

Now that the sine and cosine values have been found, the tangent of 15 degrees can be found using the formula: \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\).Then, you can calculate \(\tan(15^\circ)\) using the sine and cosine values which were calculated above.\(\tan(15^\circ) = \frac{\sin(15^\circ)}{\cos(15^\circ)}\)Substitute the above-calculated values of \(\sin(15^\circ)\) and \(\cos(15^\circ)\) into the formula and simplify. Remember that: \(\tan(15^\circ) = \frac{\frac{\sqrt{6} - \sqrt{2}}{4}}{\frac{\sqrt{6} + \sqrt{2}}{4}} = \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} + \sqrt{2}}\)Finally, the exact value of \(\tan(15^\circ)\) is 2 - \(\sqrt{3}\) by multiplying the numerator and denominator by the conjugate of the denominator.