Which Of The Following Statements Shows How To Calculate The Reference Angle For $\theta=\frac{11 \pi}{6}$?A. $\pi-\theta$ B. $\theta-\pi$ C. $2 \pi-\theta$

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Understanding Reference Angles

In trigonometry, reference angles are used to simplify the calculation of trigonometric functions for angles that are not in the first quadrant. A reference angle is the acute angle between the terminal side of an angle and the x-axis. It is used to find the values of trigonometric functions for angles in all four quadrants.

Calculating Reference Angles for Angles Greater Than 2Ï€2\pi

When an angle is greater than 2Ï€2\pi, we can calculate its reference angle by subtracting multiples of 2Ï€2\pi from the angle. This is because the unit circle repeats every 2Ï€2\pi radians. Therefore, we can subtract multiples of 2Ï€2\pi from the angle to bring it within the range of 00 to 2Ï€2\pi.

Calculating Reference Angles for Angles in Quadrants II, III, and IV

To calculate the reference angle for an angle in quadrant II, III, or IV, we can use the following formulas:

  • For angles in quadrant II: π−θ\pi - \theta
  • For angles in quadrant III: Ï€+θ\pi + \theta
  • For angles in quadrant IV: θ\theta

Calculating the Reference Angle for θ=11π6\theta=\frac{11 \pi}{6}

To calculate the reference angle for θ=11π6\theta=\frac{11 \pi}{6}, we need to subtract multiples of 2π2\pi from the angle to bring it within the range of 00 to 2π2\pi. We can do this by subtracting 2π2\pi from the angle.

θ−2π=11π6−2π=11π6−12π6=−π6\theta - 2\pi = \frac{11 \pi}{6} - 2\pi = \frac{11 \pi}{6} - \frac{12 \pi}{6} = -\frac{\pi}{6}

However, since we are looking for the reference angle, we need to find the positive acute angle between the terminal side of the angle and the x-axis. Therefore, we can add π\pi to the angle to get the reference angle.

π−(−π6)=π+π6=7π6\pi - \left(-\frac{\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6}

However, the correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

Understanding Reference Angles

In trigonometry, reference angles are used to simplify the calculation of trigonometric functions for angles that are not in the first quadrant. A reference angle is the acute angle between the terminal side of an angle and the x-axis. It is used to find the values of trigonometric functions for angles in all four quadrants.

Calculating Reference Angles for Angles Greater Than 2Ï€2\pi

When an angle is greater than 2Ï€2\pi, we can calculate its reference angle by subtracting multiples of 2Ï€2\pi from the angle. This is because the unit circle repeats every 2Ï€2\pi radians. Therefore, we can subtract multiples of 2Ï€2\pi from the angle to bring it within the range of 00 to 2Ï€2\pi.

Calculating Reference Angles for Angles in Quadrants II, III, and IV

To calculate the reference angle for an angle in quadrant II, III, or IV, we can use the following formulas:

  • For angles in quadrant II: π−θ\pi - \theta
  • For angles in quadrant III: Ï€+θ\pi + \theta
  • For angles in quadrant IV: θ\theta

Calculating the Reference Angle for θ=11π6\theta=\frac{11 \pi}{6}

To calculate the reference angle for θ=11π6\theta=\frac{11 \pi}{6}, we need to subtract multiples of 2π2\pi from the angle to bring it within the range of 00 to 2π2\pi. We can do this by subtracting 2π2\pi from the angle.

θ−2π=11π6−2π=11π6−12π6=−π6\theta - 2\pi = \frac{11 \pi}{6} - 2\pi = \frac{11 \pi}{6} - \frac{12 \pi}{6} = -\frac{\pi}{6}

However, since we are looking for the reference angle, we need to find the positive acute angle between the terminal side of the angle and the x-axis. Therefore, we can add π\pi to the angle to get the reference angle.

π−(−π6)=π+π6=7π6\pi - \left(-\frac{\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6}

However, the correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is π−(11π6−2π)=π−(11π6−12π6)=π−(11π6−2π)=π−(11π6−12π6)=π−(−π6)=π+π6=7π6\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi + \frac{\pi}{6} = \frac{7\pi}{6} is incorrect.

The correct answer is $\pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{11\pi}{6} - 2\pi\right) = \pi - \left(\frac{11\pi}{6} - \frac{12\pi}{6}\right) = \pi - \left(\frac{-\pi}{6}\right) = \pi +