Answer: (a) IV (b) positive (c) [tex]\frac{\pi}{4}[/tex] (d) C (e) [tex]\sqrt{2}[/tex]
Step-by-step explanation:
a) [tex]\frac{15\pi}{4}[/tex] - [tex]\frac{8\pi}{4}[/tex] = [tex]\frac{7\pi}{4}[/tex], which is located in Quadrant IV per the Unit Circle.
b) sec is [tex]\frac{1}{cos}[/tex]. Cos is the x-coordinate. The x-coordinate inQuadrant IV is positive.
c) the reference angle is the angle from [tex]\frac{7\pi}{4}[/tex] to 2π = [tex]\frac{\pi}{4}[/tex]
d) since the angle is below the x-axis and the reference angle is[tex]\frac{\pi}{4}[/tex], then the angle is equal to [tex]-sec(\frac{\pi}{4}[/tex])
e) sec = [tex]\frac{1}{cos}[/tex] ⇒ sec [tex]\frac{7\pi}{4}[/tex] = [tex]\frac{2}{\sqrt{2} }[/tex] = [tex]\frac{2}{\sqrt{2} }[/tex][tex](\frac{\sqrt{2}} {\sqrt{2}})[/tex] = [tex]\frac{2\sqrt{2} }{2}[/tex] = [tex]\sqrt{2}[/tex]
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Answer: (a) [tex]\frac{3\pi}{2}[/tex] (b) (0, -1) (c) A (d) -1
Step-by-step explanation:
a) [tex]\frac{11\pi}{2}[/tex] - [tex]\frac{4\pi}{2}[/tex] = [tex]\frac{7\pi}{2}[/tex] - [tex]\frac{11\pi}{2}[/tex] = [tex]\frac{3\pi}{2}[/tex]
b) [tex]\frac{3\pi}{2}[/tex] is on the Unit Circle at (0, -1)
c) sin = [tex]\frac{opposite}{hypotenuse}[/tex] which equals [tex]\frac{y}{r}[/tex] on the Unit Circle.
d) sin is the y-coordinate. sin [tex](\frac{3\pi}{2})[/tex] = -1
