Which of the following is equivalent to sin ((pi)/(7)) ? sin ((13 pi)/(7)) sin ((8pi)/(7)) sin(-(pi)/(7)) sin ((15 pi)/(7))

Properties of Sine Function: Understand the properties of the sine function. The sine function has a period of 2π, which means sin(θ) = sin(θ + 2πk) for any integer k. Also, sin(θ) = sin(π - θ) for any angle θ.Evaluate sin((13pi)/(7)): Evaluate sin((13pi)/(7)). Since 13π/7 = π + 6π/7, we can use the property sin(θ) = sin(θ + 2πk) with k = -1 to find an equivalent expression. sin((13pi)/(7)) = sin((13pi)/(7) - 2π) = sin((13pi)/(7) - (14π/7)) = sin(-π/7). This is not equivalent to sin(π/7) because sin(θ) is not generally equal to sin(-θ).Evaluate sin((8pi)/(7)): Evaluate sin((8pi)/(7)). Since 8π/7 = π + π/7, we can use the property sin(θ) = sin(π - θ). sin((8pi)/(7)) = sin(π - π/7) = sin(6π/7). This is not equivalent to sin(π/7) because 6π/7 is not the same angle as π/7.Evaluate sin(-(pi)/(7)): Evaluate sin(-(pi)/(7)). Using the odd function property of sine, which is sin(-θ) = -sin(θ), we get: sin(-(pi)/(7)) = -sin((pi)/(7)). This is not equivalent to sin(π/7) because of the negative sign.Evaluate sin((15pi)/(7)): Evaluate sin((15pi)/(7)). Since 15π/7 = 2π + π/7, we can use the property sin(θ) = sin(θ + 2πk) with k = -1 to find an equivalent expression. sin((15pi)/(7)) = sin((15pi)/(7) - 2π) = sin((15pi)/(7) - (14π/7)) = sin(π/7). This is equivalent to sin(π/7) because we have subtracted a full period of the sine function.

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Math Problems

Algebra 1

Multiplication with rational exponents

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Q. Which of the following is equivalent to

\sin \frac{\pi}{7}

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\sin \frac{13 \pi}{7}

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\sin \frac{8 \pi}{7}

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\sin \left(-\frac{\pi}{7}\right)

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\sin \frac{15 \pi}{7}

Properties of Sine Function: Understand the properties of the sine function. The sine function has a period of $2\pi$ , which means $\sin(\theta) = \sin(\theta + 2\pi k)$ for any integer $k$ . Also, $\sin(\theta) = \sin(\pi - \theta)$ for any angle $\theta$ .
Evaluate $\sin\left(\frac{13\pi}{7}\right)$ : Evaluate $\sin\left(\frac{13\pi}{7}\right)$ . $\newline$ Since $\frac{13\pi}{7} = \pi + \frac{6\pi}{7}$ , we can use the property $\sin(\theta) = \sin(\theta + 2\pi k)$ with $k = -1$ to find an equivalent expression. $\newline$ $\sin\left(\frac{13\pi}{7}\right) = \sin\left(\frac{13\pi}{7} - 2\pi\right) = \sin\left(\frac{13\pi}{7} - \frac{14\pi}{7}\right) = \sin\left(-\frac{\pi}{7}\right)$ . $\newline$ This is not equivalent to $\sin\left(\frac{\pi}{7}\right)$ because $\sin(\theta)$ is not generally equal to $\sin(-\theta)$ .
Evaluate $\sin\left(\frac{8\pi}{7}\right)$ : Evaluate $\sin\left(\frac{8\pi}{7}\right)$ . $\newline$ Since $\frac{8\pi}{7} = \pi + \frac{\pi}{7}$ , we can use the property $\sin(\theta) = \sin(\pi - \theta)$ . $\newline$ $\sin\left(\frac{8\pi}{7}\right) = \sin\left(\pi - \frac{\pi}{7}\right) = \sin\left(\frac{6\pi}{7}\right)$ . $\newline$ This is not equivalent to $\sin\left(\frac{\pi}{7}\right)$ because $\frac{6\pi}{7}$ is not the same angle as $\frac{\pi}{7}$ .
Evaluate $\sin\left(-\frac{\pi}{7}\right)$ : Evaluate $\sin\left(-\frac{\pi}{7}\right)$ . Using the odd function property of sine, which is $\sin(-\theta) = -\sin(\theta)$ , we get: $\sin\left(-\frac{\pi}{7}\right) = -\sin\left(\frac{\pi}{7}\right)$ . This is not equivalent to $\sin\left(\frac{\pi}{7}\right)$ because of the negative sign.
Evaluate $\sin\left(\frac{15\pi}{7}\right)$ : Evaluate $\sin\left(\frac{15\pi}{7}\right)$ . $\newline$ Since $\frac{15\pi}{7} = 2\pi + \frac{\pi}{7}$ , we can use the property $\sin(\theta) = \sin(\theta + 2\pi k)$ with $k = -1$ to find an equivalent expression. $\newline$ $\sin\left(\frac{15\pi}{7}\right) = \sin\left(\frac{15\pi}{7} - 2\pi\right) = \sin\left(\frac{15\pi}{7} - \frac{14\pi}{7}\right) = \sin\left(\frac{\pi}{7}\right)$ . $\newline$ This is equivalent to $\sin\left(\frac{\pi}{7}\right)$ because we have subtracted a full period of the sine function.