### Instruction: Compute $\cos 15^\circ.$.### Response: One way to approach this problem is to use the cosine addition formula, which says that $\cos(a+b) = \cos a \cos b - \sin a \sin b.$
If I can find two angles $a$ and $b$ that add up to $15^\circ$ and have easy values for their sines and cosines, then I can plug them into the formula and simplify.
A good choice might be $a = 45^\circ$ and $b = -30^\circ$, since these are special angles that are multiples of $15^\circ$ and have well-known trigonometric ratios.
Using these values, I get that $\cos 15^\circ = \cos(45^\circ - 30^\circ) = \cos 45^\circ \cos (-30^\circ) - \sin 45^\circ \sin (-30^\circ).$
Now I need to recall the values of $\cos 45^\circ$, $\cos (-30^\circ)$, $\sin 45^\circ$, and $\sin (-30^\circ)$.
I can use the unit circle, a 45-45-90 triangle, or a 30-60-90 triangle to help me remember them.
From the unit circle, I know that $\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}$, since these are the coordinates of the point on the circle that makes an angle of $45^\circ$ with the positive x-axis.
Similarly, I know that $\cos (-30^\circ) = \frac{\sqrt{3}}{2}$ and $\sin (-30^\circ) = -\frac{1}{2}$, since these are the coordinates of the point on the circle that makes an angle of $-30^\circ$ with the positive x-axis.
Plugging these values into the formula, I get that $\cos 15^\circ = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \left(-\frac{1}{2}\right) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4}.$
This is a simplified exact expression for $\cos 15^\circ$, but I can also use a calculator to find a decimal approximation if needed.
To do that, I need to use the square root function and the fraction function on the calculator, and then round the result to the desired number of decimal places.
For example, if I want to round to two decimal places, I get that $\cos 15^\circ \approx 0.97.$