### Instruction: Compute $\cos 45^\circ$.### Response: I recall that $\cos 45^\circ$ is one of the special values that can be found using the unit circle or a 45-45-90 right triangle.
I draw a right triangle with both acute angles equal to 45 degrees, and the hypotenuse equal to 1.
I label the legs as x and y, and use the Pythagorean theorem to find their lengths.
I get $x^2 + y^2 = 1$, and since $x = y$, I can simplify this to $2x^2 = 1$, or $x^2 = \frac{1}{2}$.
Taking the positive square root, I get $x = \frac{1}{\sqrt{2}}$.
This is also equal to y, since the triangle is isosceles.
I remember that $\cos 45^\circ$ is equal to the ratio of the adjacent side to the hypotenuse, so $\cos 45^\circ = \frac{x}{1} = x = \frac{1}{\sqrt{2}}$.
I also recall that some teachers prefer to rationalize the denominator, so I multiply the fraction by $\frac{\sqrt{2}}{\sqrt{2}}$ to get $\cos 45^\circ = \frac{\sqrt{2}}{2}$.