Sine Function

Approach:

The sine function $y = \sin x$ is defined through the point $(\cos x, \sin x)$ on the unit circle, so it is meaningful for every real $x$; the ordinate of that point ranges exactly over $[-1, 1]$.

Steps:

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

Answer:

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

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:

First show by the addition formula that $2\pi$ is a period, then use the distribution of the zeros to rule out any smaller positive period.

Steps:

1. From the addition formula $\sin(x + 2\pi) = \sin x$, so $2\pi$ is a period of $y = \sin x$.
2. Looking at the zeros, $\sin x = 0$ holds exactly when $x = k\pi$, so any positive period must be an integer multiple of $\pi$.
3. But $\sin(x + \pi) = -\sin x \neq \sin x$, so $\pi$ is not a period and the smallest positive multiple can only be $2\pi$.

Answer:

The minimum positive period is $2\pi$.

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.