Question

Q 2 PLEASE HELP ME FIGURE THIS OUT

Answer 1

Answer: IV, positive, $\frac{\pi} {6}$, - sec $\frac{\pi} {6}$, $\frac{2\sqrt{3}}{3}$

Step-by-step explanation:

a) Look at the Unit Circle to see that $\frac{11\pi} {6}$ = 330°, which is located in Quadrant IV.

b) The coordinate (cos θ, sin θ) for $\frac{11\pi} {6}$ is: $(\frac{\sqrt{3}} {2},\frac{-1}{2})$

sec = $\frac{1}{cos}$ = $\frac{2}{\sqrt{3}}$ which is positive

c) Since the given angle is in Quadrant IV, which is closest to the x-axis at 360° = 2π, the reference angle can be found by subtracting the given angle $\frac{11\pi} {6}$ from 2π: $\frac{12\pi} {6}$ - $\frac{11\pi} {6}$ = $\frac{\pi} {6}$

d) the reference angle is below the x-axis so the given angle is equal to the negative of the reference angle: - sec $\frac{\pi} {6}$.

e) sec $\frac{11\pi} {6}$ = $\frac{2}{\sqrt{3}}$ = $\frac{2}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}$ = $\frac{2\sqrt{3}}{3}$

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Answer: $\frac{18\pi}{11}$, IV, $\frac{4\pi} {11}$

Step-by-step explanation:

2π is one rotation.  2π = $\frac{22\pi}{11}$

$\frac{-26\pi}{11}$ + $\frac{22\pi}{11}$ = $\frac{-4\pi}{11}$

$\frac{-4\pi}{11}$ + $\frac{22\pi}{11}$ = $\frac{18\pi}{11}$

Convert the radians into degrees to see which Quadrant it is in by setting up the proportion and cross multiplying:

$\frac{\pi}{180}$= $\frac{18\pi}{11x}$

π(11x) = (180)18π

x = $\frac{180(18\pi}{11\pi}$

x = 295°     which lies in Quadrant IV

Since the given angle is in Quadrant IV, which is closest to the x-axis at 360° = 2π, the reference angle can be found by subtracting the angle of least nonegative value$\frac{18\pi} {11}$ from 2π: $\frac{22\pi} {11}$ - $\frac{18\pi} {11}$ = $\frac{4\pi} {11}$

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Answer: $\frac{5\pi}{3}$, IV, $\frac{4\pi} {11}$, $\frac{\pi} {3}$

Step-by-step explanation:

2π is one rotation.  2π = $\frac{6\pi}{3}$

$\frac{-13\pi}{3}$ + $\frac{6\pi}{3}$ = $\frac{-7\pi}{3}$

$\frac{-7\pi}{3}$ + $\frac{6\pi}{3}$ = $\frac{-\pi}{3}$

$\frac{-\pi}{3}$ + $\frac{6\pi}{3}$ = $\frac{5\pi}{3}$

This is on the Unit Circle at 300°, which is located in Quadrant IV

Since the given angle is in Quadrant IV, which is closest to the x-axis at 360° = 2π, the reference angle can be found by subtracting the angle of least nonegative value$\frac{5\pi} {3}$ from 2π: $\frac{6\pi} {3}$ - $\frac{5\pi} {3}$ = $\frac{\pi} {3}$