Cosine Function

Approach:

The cosine function is defined through the point $(\cos x, \sin x)$ on the unit circle, so it is meaningful for every real $x$, and its value is the abscissa of that point, which ranges over $[-1, 1]$.

Steps:

1. By the unit-circle definition, $\cos x$ is meaningful for every real $x$, so the domain is $\mathbb{R}$.
2. The abscissa of a point on the unit circle satisfies $-1 \leq \cos x \leq 1$, so the range is $[-1, 1]$.
3. At $x = 0$ we have $\cos x = 1$ and at $x = \pi$ we have $\cos x = -1$, so both endpoints are attained.

Answer:

Domain: $\mathbb{R}$; Range: $[-1, 1]$.

Approach:

Read the period off the graph of the cosine function and check its parity by comparing $\cos(-x)$ with $\cos x$.

Steps:

1. The addition formula gives $\cos(x + 2\pi) = \cos x$, so $2\pi$ is a period; the zeros of $\cos x$ occur only at $x = \frac{\pi}{2} + k\pi$, so no smaller positive number can be a period.
2. Computing $f(-x) = \cos(-x) = \cos x = f(x)$, the function is even.
3. Hence the period is $2\pi$ and the graph is symmetric about the $y$-axis.

Answer:

Period is $2\pi$, it is an even function.

Approach:

First fix the shape and the key points of the two curves, then compare their periods and symmetries: the sine graph passes through the origin, the cosine graph through $(0, 1)$, and the two differ only by a horizontal shift of $\frac{\pi}{2}$.

Steps:

1. The graph of $y = \sin x$ passes through the origin and is monotonically increasing on $[-\frac{\pi}{2}, \frac{\pi}{2}]$; the graph of $y = \cos x$ passes through $(0, 1)$ and is monotonically decreasing on $[0, \pi]$.
2. From $\cos x = \sin(x + \frac{\pi}{2})$ the two graphs differ only by a horizontal shift of $\frac{\pi}{2}$.
3. Both return to their starting position after a shift of $2\pi$, so the period is $2\pi$ for each of them.
4. From $\sin(-x) = -\sin x$ we know $\sin x$ is odd and its graph is symmetric about the origin; from $\cos(-x) = \cos x$ we know $\cos x$ is even and its graph is symmetric about the $y$-axis.

Answer:

Both $y = \sin x$ and $y = \cos x$ have period $2\pi$. $\sin x$ is an odd function, $\cos x$ is an even function.

Approach:

Test each option with $f(-x) = \pm f(x)$: the sine and the tangent change sign under the substitution, while the cosine and the square do not.

Steps:

1. (A) $f(-x) = \sin(-x) = -\sin x = -f(x)$, so $y = \sin x$ is an odd function.
2. (B) $f(-x) = \cos(-x) = \cos x = f(x)$, so $y = \cos x$ is an even function.
3. (C) $f(-x) = \tan(-x) = -\tan x = -f(x)$, so $y = \tan x$ is an odd function.
4. (D) $f(-x) = (-x)^2 = x^2 = f(x)$, so $y = x^2$ is an even function.

Answer:

(A), (C) are odd functions, (B), (D) are not.