To find the exact value of \(\cos(-75^\circ)\), we can use the cosine angle subtraction identity.
First, note that:
\[ \cos(-\theta) = \cos(\theta) \]
Thus,
\[ \cos(-75^\circ) = \cos(75^\circ) \]
Next, we can express \(75^\circ\) as \(45^\circ + 30^\circ\):
\[ \cos(75^\circ) = \cos(45^\circ + 30^\circ) \]
Using the cosine addition formula:
\[ \cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b) \]
Substituting \(a = 45^\circ\) and \(b = 30^\circ\):
\[ \cos(75^\circ) = \cos(45^\circ)\cos(30^\circ) - \sin(45^\circ)\sin(30^\circ) \]
Now, we know:
\[ \cos(45^\circ) = \frac{\sqrt{2}}{2}, \quad \cos(30^\circ) = \frac{\sqrt{3}}{2}, \quad \sin(45^\circ) = \frac{\sqrt{2}}{2}, \quad \sin(30^\circ) = \frac{1}{2} \]
Plugging these values into the formula gives:
\[ \cos(75^\circ) = \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) \]
This simplifies to:
\[ \cos(75^\circ) = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} \]
Combining the terms:
\[ \cos(75^\circ) = \frac{\sqrt{6} - \sqrt{2}}{4} \]
Therefore:
\[ \cos(-75^\circ) = \cos(75^\circ) = \frac{\sqrt{6} - \sqrt{2}}{4} \]
From the options provided, this corresponds to:
\[ \text{start fraction start root 6 end root minus start root 2 end root over 4 end fraction} \]
So the exact value of \(\cos(-75^\circ)\) is:
\[ \frac{\sqrt{6} - \sqrt{2}}{4} \]