Power Trigonometric Functions

Problem-solving approach: Analyze the basic properties of the sine squared function and derive using trigonometric function properties.

Detailed steps:

1. Domain: Since the domain of $\sin x$ is $\mathbb{R}$, the domain of $\sin^2 x$ is also $\mathbb{R}$

2. Range: Since $\sin x \in [-1, 1]$, we have $\sin^2 x \in [0, 1]$

3. Period: Since $\sin^2 x = \frac{1 - \cos 2x}{2}$, and $\cos 2x$ has period $\pi$, the period of $\sin^2 x$ is $\pi$

Answer:

• Domain: $\mathbb{R}$
• Range: $[0, 1]$
• Period: $\pi$

Problem-solving approach: Use the definition of even and odd functions and the properties of cosine functions for analysis.

Detailed steps:

1. Let $f(x) = \cos^3 x$

2. Calculate $f(-x) = \cos^3(-x) = [\cos(-x)]^3 = (\cos x)^3 = \cos^3 x = f(x)$

3. Since $f(-x) = f(x)$, $y = \cos^3 x$ is an even function

Answer: $y = \cos^3 x$ is an even function because $f(-x) = f(x)$

Problem-solving approach: Use trigonometric identities and completing the square method to solve.

Detailed steps:

1. Use the identity: $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x$

2. Since $\sin^2 x + \cos^2 x = 1$: $\sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x$

3. Use double angle formula: $\sin 2x = 2\sin x \cos x$, so: $\sin^2 x \cos^2 x = \frac{\sin^2 2x}{4}$

4. Therefore: $\sin^4 x + \cos^4 x = 1 - \frac{\sin^2 2x}{2}$

5. Since $\sin^2 2x \in [0, 1]$, we have $\frac{\sin^2 2x}{2} \in [0, \frac{1}{2}]$

6. Therefore $1 - \frac{\sin^2 2x}{2} \in [\frac{1}{2}, 1]$

Answer: Minimum value is $\frac{1}{2}$

Problem-solving approach: Discuss by cases: use double angle formula when $n$ is even, use periodic function definition when $n$ is odd.

Detailed steps:

Case 1: $n$ is even

Let $n = 2k$ ($k$ is a positive integer), then $f(x) = \sin^{2k} x = (\sin^2 x)^k$

Since $\sin^2 x = \frac{1 - \cos 2x}{2}$, and $\cos 2x$ has period $\pi$, $\sin^2 x$ has period $\pi$

Therefore $(\sin^2 x)^k$ also has period $\pi$

Case 2: $n$ is odd

Let $f(x) = \sin^n x$ ($n$ is odd)

Since $\sin(x + 2\pi) = \sin x$: $f(x + 2\pi) = \sin^n(x + 2\pi) = [\sin(x + 2\pi)]^n = (\sin x)^n = \sin^n x = f(x)$

Therefore $2\pi$ is a period of $f(x)$

Suppose there exists a smaller positive period $T < 2\pi$, then $\sin^n(x + T) = \sin^n x$ holds for all $x$

Since $n$ is odd, this means $\sin(x + T) = \sin x$, so $T$ must be a period of $\sin x$

The minimum positive period of $\sin x$ is $2\pi$, so $T \geq 2\pi$, contradiction

Answer:

• When $n$ is even, the function $y = \sin^n x$ has minimum positive period $\pi$
• When $n$ is odd, the function $y = \sin^n x$ has minimum positive period $2\pi$

Problem-solving approach: Use the trigonometric identity $\sin^2 x + \cos^2 x = 1$ to solve.

Detailed steps:

1. Since $\sin^2 x + \cos^2 x = 1$ holds for all $x$

2. So $f(x) = 1$ holds for all $x$

3. Therefore the range of $f(x)$ is $\{1\}$, i.e., $[1, 1]$

Answer: (D) $[1, 1]$

Problem-solving approach: Analyze the periodicity of cosine function and properties of power functions.

Detailed steps:

1. Since $\cos x$ has period $2\pi$

2. For the power function $\cos^3 x$, the period remains unchanged

3. Therefore $\cos^3 x$ has period $2\pi$

Answer: (B) $2\pi$