Trigonometric Graphs Practice

1

Domain of Trigonometric Function

Problem

Find the domain of $y = \sqrt{\frac{1}{2} - \cos x}$ .

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Answer:

\left\{x \mid 2k\pi + \frac{\pi}{3} \leq x \leq 2k\pi + \frac{5\pi}{3}, k \in \mathbb{Z}\right\}

Solution

For the function to be defined, we need:

\frac{1}{2} - \cos x \geq 0

\cos x \leq \frac{1}{2}

Since $\cos\frac{\pi}{3} = \frac{1}{2}$ , we have:

\frac{\pi}{3} + 2k\pi \leq x \leq \frac{5\pi}{3} + 2k\pi, \quad k \in \mathbb{Z}

2

Even Function Parameter

Problem

Given $f(x) = \sin\left(\frac{x}{3} + \varphi\right)$ where $0 \leq \varphi < 2\pi$ is an even function, find $\varphi$ .

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Answer:

\frac{3\pi}{2}

Solution

For $f(x)$ to be even, we need $f(-x) = f(x)$ for all $x$ :

\sin\left(-\frac{x}{3} + \varphi\right) = \sin\left(\frac{x}{3} + \varphi\right)

This requires $\cos\left(\frac{\varphi}{3}\right) = 0$ , so:

\frac{\varphi}{3} = k\pi + \frac{\pi}{2}, \quad k \in \mathbb{Z}

\varphi = 3k\pi + \frac{3\pi}{2}, \quad k \in \mathbb{Z}

Since $0 \leq \varphi < 2\pi$ , we have $\varphi = \frac{3\pi}{2}$ .

3

Graph Translation and Symmetry

Problem

Translate $y = \sin x + \cos x$ left by $m$ units ( $0 < m < \pi$ ). If the resulting graph is symmetric about the y-axis, find $m$ .

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Answer:

\frac{3\pi}{4}

Solution

First, rewrite the function:

y = \sin x + \cos x = \sqrt{2}\sin\left(x + \frac{\pi}{4}\right)

After translating left by $m$ units:

y = \sqrt{2}\sin\left(x + m + \frac{\pi}{4}\right)

For y-axis symmetry (even function):

m + \frac{\pi}{4} = k\pi + \frac{\pi}{2}, \quad k \in \mathbb{Z}

m = k\pi + \frac{\pi}{4}, \quad k \in \mathbb{Z}

Since $0 < m < \pi$ , when $k = 0$ : $m = \frac{3\pi}{4}$ .

4

Period of Product Function

Problem

Find the minimum positive period of $f(x) = \sin\frac{x}{3}\cos\frac{x}{3}$ .

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Answer:

3\pi

Solution

Using the double-angle formula:

f(x) = \sin\frac{x}{3}\cos\frac{x}{3} = \frac{1}{2}\sin\frac{2x}{3}

The period of $\sin\frac{2x}{3}$ is:

T = \frac{2\pi}{\frac{2}{3}} = 3\pi

5

Period of Tangent Composition

Problem

Find the minimum positive period of $f(x) = \frac{2\tan x}{1-\tan^2 x}$ .

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Answer:

\pi

Solution

Recognize the double-angle tangent formula:

f(x) = \frac{2\tan x}{1-\tan^2 x} = \tan 2x

where $x \neq \frac{k\pi}{2}$ and $\tan x \neq \pm 1$ (i.e., $x \neq \frac{\pi}{4} + \frac{k\pi}{2}$ ).

The period of $\tan 2x$ is:

T = \frac{\pi}{2} \cdot 2 = \pi

6

Range of Trigonometric Function

Problem

Find the range of $f(x) = \sin^2 x - \cos^2 x$ for $x \in \left[\frac{\pi}{12}, \frac{2\pi}{3}\right]$ .

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Answer:

\left[-\frac{\sqrt{3}}{2}, 1\right]

Solution

Simplify the function:

f(x) = \sin^2 x - \cos^2 x = -\cos 2x

For $x \in \left[\frac{\pi}{12}, \frac{2\pi}{3}\right]$ :

2x \in \left[\frac{\pi}{6}, \frac{4\pi}{3}\right]

In this interval, $\cos 2x \in \left[-1, \frac{\sqrt{3}}{2}\right]$ .

Therefore:

f(x) = -\cos 2x \in \left[-\frac{\sqrt{3}}{2}, 1\right]

7

Range and Domain Relationship

Problem

Given $f(x) = \sin\left(\omega x + \frac{\pi}{6}\right)$ with domain $[m, n]$ ( $m < n$ ) and range $[0, 1]$ , find the range of $n - m$ .

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Answer:

\left[\frac{\pi}{2}, \pi\right)

Solution

For $x \in [m, n]$ , we have $\omega x + \frac{\pi}{6} \in \left[\omega m + \frac{\pi}{6}, \omega n + \frac{\pi}{6}\right]$ .

For range $[0, 1]$ , the argument must cover from $0$ to $\frac{\pi}{2}$ (at minimum) but less than a full period.

Thus:

\frac{\pi}{2} \leq \left(\omega n + \frac{\pi}{6}\right) - \left(\omega m + \frac{\pi}{6}\right) < \pi

\frac{\pi}{2} \leq \omega(n - m) < \pi

For $\omega = 1$ : $n - m \in \left[\frac{\pi}{2}, \pi\right)$

8

Zero Points Problem

Problem

Given $f(x) = \sin\left(\omega x + \frac{\pi}{6}\right)$ ( $\omega > 0$ ). If $f\left(\frac{\pi}{6}\right) = 0$ and $f(x)$ has exactly one zero in $\left[\frac{\pi}{6}, \frac{5\pi}{24}\right]$ , find the minimum value of $\omega$ .

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Answer:

29

Solution

From $f\left(\frac{\pi}{6}\right) = 0$ :

\sin\left(\frac{\omega\pi}{6} + \frac{\pi}{6}\right) = 0

\frac{\omega\pi}{6} + \frac{\pi}{6} = k\pi, \quad k \in \mathbb{Z}

\omega = 6k - 1

The interval length is $\frac{5\pi}{24} - \frac{\pi}{6} = \frac{\pi}{24}$ .

For exactly one zero, the half-period must satisfy $\frac{T}{2} \geq \frac{\pi}{24}$ , so $T \geq \frac{\pi}{12}$ .

Since $T = \frac{2\pi}{\omega}$ , we need $\omega \leq 24$ .

But with constraints, solving $24 < 6k - 1 < 48$ gives $k = 5, 6, 7, 8$ .

When $k = 5$ : $\omega = 29$ (minimum).

9

Monotonicity Interval

Problem

Given $f(x) = \sin\left(\omega x + \frac{\pi}{3}\right)$ ( $\omega > 0$ ). If $f(x)$ is monotonically increasing on $\left[-\frac{2\pi}{3}, \frac{\pi}{6}\right]$ , find the range of $\omega$ .

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Answer:

(0, 1]

Solution

For $x \in \left[-\frac{2\pi}{3}, \frac{\pi}{6}\right]$ :

\omega x + \frac{\pi}{3} \in \left[-\frac{2\omega\pi}{3} + \frac{\pi}{3}, \frac{\omega\pi}{6} + \frac{\pi}{3}\right]

For $f(x)$ to be increasing, the argument must stay in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ :

\begin{cases} -\frac{2\omega\pi}{3} + \frac{\pi}{3} \geq -\frac{\pi}{2} \\ \frac{\omega\pi}{6} + \frac{\pi}{3} \leq \frac{\pi}{2} \end{cases}

From the first inequality: $\omega \leq \frac{5}{4}$

From the second inequality: $\omega \leq 1$

Combined with $\omega > 0$ : $0 < \omega \leq 1$

10

Maximum Value Problem

Problem

Given $f(x) = 2\cos^2 x(\sin x + \cos x)$ ( $\omega > 0$ ) has its graph symmetric about $x = \frac{\pi}{12}$ , and $f(x)$ has no minimum value on $\left[0, \frac{\pi}{3}\right]$ . Find $\omega$ .

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Answer:

\frac{3}{2}

Solution

Simplify $f(x)$ :

f(x) = 2\cos^2 x(\sin x + \cos x)

= (1 + \cos 2x)(\sin x + \cos x)

= \sqrt{2}\sin\left(2\omega x + \frac{\pi}{4}\right)

For symmetry about $x = \frac{\pi}{12}$ :

2\omega \cdot \frac{\pi}{12} + \frac{\pi}{4} = k\pi + \frac{\pi}{2}, \quad k \in \mathbb{Z}

\omega = 6k + \frac{3}{2}, \quad k \in \mathbb{Z}

For no minimum on $\left[0, \frac{\pi}{3}\right]$ , the minimum point $x = \frac{5\pi}{8\omega}$ must satisfy:

\frac{5\pi}{8\omega} \geq \frac{\pi}{3} \Rightarrow \omega \leq \frac{15}{8}

From $0 < 6k + \frac{3}{2} \leq \frac{15}{8}$ , we get $k = 0$ , so $\omega = \frac{3}{2}$ .

Trigonometric Graphs Practice

Move from graph reasoning into triangle application