Which of the following is not equivalent to tan ((2pi)/(7)) ? tan ((9pi)/(7)) tan ((16 pi)/(7)) tan(-(12 pi)/(7)) tan ((19 pi)/(7))

Periodicity of Tangent Function: Understand the periodicity of the tangent function. The tangent function has a period of π, which means that tan(θ) = tan(θ + kπ) for any integer k.Evaluate tan((9pi)/7): Evaluate tan((9pi)/7). Since tan(θ) = tan(θ + π), we can subtract π from (9pi)/7 to see if it is equivalent to tan((2pi)/7). tan((9pi)/7) = tan((9pi)/7 - π) = tan((9pi)/7 - (7pi)/7) = tan((2pi)/7)Evaluate tan((16pi)/7): Evaluate tan((16pi)/7). Similarly, subtract 2π from (16pi)/7 to see if it is equivalent to tan((2pi)/7). tan((16pi)/7) = tan((16pi)/7 - 2π) = tan((16pi)/7 - (14pi)/7) = tan((2pi)/7)Evaluate tan(-(12pi)/7): Evaluate tan(-(12pi)/7). We can add 2π to -(12pi)/7 to see if it is equivalent to tan((2pi)/7). tan(-(12pi)/7) = tan(-(12pi)/7 + 2π) = tan(-(12pi)/7 + (14pi)/7) = tan((2pi)/7)Evaluate tan((19pi)/7): Evaluate tan((19pi)/7). Subtract 3π from (19pi)/7 to see if it is equivalent to tan((2pi)/7). tan((19pi)/7) = tan((19pi)/7 - 3π) = tan((19pi)/7 - (21pi)/7) = tan(-(2pi)/7) Since tan(θ) = -tan(-θ), we have tan(-(2pi)/7) = -tan((2pi)/7).Identify Non-Equivalent Expression: Determine which expression is not equivalent to tan((2pi)/7). From the previous steps, we have shown that tan((9pi)/7), tan((16pi)/7), and tan(-(12pi)/7) are all equivalent to tan((2pi)/7). However, tan((19pi)/7) is equivalent to -tan((2pi)/7), which is not the same as tan((2pi)/7).

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Which of the following is not equivalent to
tan ((2pi)/(7)) ?

tan ((9pi)/(7))

tan ((16 pi)/(7))

tan(-(12 pi)/(7))

tan ((19 pi)/(7))

Which of the following is not equivalent to $\tan \frac{2 \pi}{7}$ ? $\newline$ $\tan \frac{9 \pi}{7}$ $\newline$ $\tan \frac{16 \pi}{7}$ $\newline$ $\tan \left(-\frac{12 \pi}{7}\right)$ $\newline$ $\tan \frac{19 \pi}{7}$

View step-by-step help

Home

Math Problems

Algebra 1

Multiplication with rational exponents

Full solution

Q. Which of the following is not equivalent to

\tan \frac{2 \pi}{7}

\newline

\tan \frac{9 \pi}{7}

\newline

\tan \frac{16 \pi}{7}

\newline

\tan \left(-\frac{12 \pi}{7}\right)

\newline

\tan \frac{19 \pi}{7}

Periodicity of Tangent Function: Understand the periodicity of the tangent function. The tangent function has a period of $\pi$ , which means that $\tan(\theta) = \tan(\theta + k\pi)$ for any integer $k$ .
Evaluate $\tan\left(\frac{9\pi}{7}\right):</b> Evaluate \$\tan\left(\frac{9\pi}{7}\right)$ . $\newline$ Since $\tan(\theta) = \tan(\theta + \pi)$ , we can subtract $\pi$ from $\frac{9\pi}{7}$ to see if it is equivalent to $\tan\left(\frac{2\pi}{7}\right)$ . $\newline$ $\tan\left(\frac{9\pi}{7}\right) = \tan\left(\frac{9\pi}{7} - \pi\right) = \tan\left(\frac{9\pi}{7} - \frac{7\pi}{7}\right) = \tan\left(\frac{2\pi}{7}\right)$
Evaluate $\tan\left(\frac{16\pi}{7}\right)$ : Evaluate $\tan\left(\frac{16\pi}{7}\right)$ . Similarly, subtract $2\pi$ from $\frac{16\pi}{7}$ to see if it is equivalent to $\tan\left(\frac{2\pi}{7}\right)$ . $\tan\left(\frac{16\pi}{7}\right) = \tan\left(\frac{16\pi}{7} - 2\pi\right) = \tan\left(\frac{16\pi}{7} - \frac{14\pi}{7}\right) = \tan\left(\frac{2\pi}{7}\right)$
Evaluate $\tan\left(-\frac{12\pi}{7}\right)$ : Evaluate $\tan\left(-\frac{12\pi}{7}\right)$ . We can add $2\pi$ to $-\frac{12\pi}{7}$ to see if it is equivalent to $\tan\left(\frac{2\pi}{7}\right)$ . $\tan\left(-\frac{12\pi}{7}\right) = \tan\left(-\frac{12\pi}{7} + 2\pi\right) = \tan\left(-\frac{12\pi}{7} + \frac{14\pi}{7}\right) = \tan\left(\frac{2\pi}{7}\right)$
Evaluate $\tan\left(\frac{19\pi}{7}\right)$ : Evaluate $\tan\left(\frac{19\pi}{7}\right)$ . Subtract $3\pi$ from $\frac{19\pi}{7}$ to see if it is equivalent to $\tan\left(\frac{2\pi}{7}\right)$ . $\tan\left(\frac{19\pi}{7}\right) = \tan\left(\frac{19\pi}{7} - 3\pi\right) = \tan\left(\frac{19\pi}{7} - \frac{21\pi}{7}\right) = \tan\left(-\frac{2\pi}{7}\right)$ Since $\tan(\theta) = -\tan(-\theta)$ , we have $\tan\left(-\frac{2\pi}{7}\right) = -\tan\left(\frac{2\pi}{7}\right)$ .
Identify Non-Equivalent Expression: Determine which expression is not equivalent to $\tan\left(\frac{2\pi}{7}\right)$ . From the previous steps, we have shown that $\tan\left(\frac{9\pi}{7}\right)$ , $\tan\left(\frac{16\pi}{7}\right)$ , and $\tan\left(-\frac{12\pi}{7}\right)$ are all equivalent to $\tan\left(\frac{2\pi}{7}\right)$ . However, $\tan\left(\frac{19\pi}{7}\right)$ is equivalent to $-\tan\left(\frac{2\pi}{7}\right)$ , which is not the same as $\tan\left(\frac{2\pi}{7}\right)$ .

Which of the following is not equivalent to $\tan \frac{2 \pi}{7}$ ? $\newline$ $\tan \frac{9 \pi}{7}$ $\newline$ $\tan \frac{16 \pi}{7}$ $\newline$ $\tan \left(-\frac{12 \pi}{7}\right)$ $\newline$ $\tan \frac{19 \pi}{7}$

Full solution

More problems from Multiplication with rational exponents

Related topics