### Instruction: Compute $\cos 75^\circ.$.### Response: I don't remember a formula or identity for $\cos 75^\circ$, so I'll try to express it as a sum or difference of angles that I do know, like $45^\circ$ or $30^\circ$.
I notice that $75^\circ = 45^\circ + 30^\circ$, so I can use the identity $\cos (A + B) = \cos A \cos B - \sin A \sin B$ with $A = 45^\circ$ and $B = 30^\circ$.
I know that $\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$, so I plug those values into the identity.
I get $\cos 75^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$, which simplifies to $\frac{\sqrt{6} - \sqrt{2}}{4}$.
That's the exact value of $\cos 75^\circ$, but I can also use a calculator to get an approximate decimal value if needed.