Question

HELP WITH THESE QUESTIONS PART 2.

Answer 1

Answer: (a) IV (b) positive (c)
`(\pi)/(4)` (d) C (e)
`√(2)`

Explanation:

a)
`(15\pi)/(4)` -
`(8\pi)/(4)` =
`(7\pi)/(4)`, which is located in Quadrant IV per the Unit Circle.

b) sec is
`(1)/(cos)`. Cos is the x-coordinate. The x-coordinate inQuadrant IV is positive.

c) the reference angle is the angle from
`(7\pi)/(4)` to 2π =
`(\pi)/(4)`

d) since the angle is below the x-axis and the reference angle is
`(\pi)/(4)`, then the angle is equal to
`-sec((\pi)/(4)`)

e) sec =
`(1)/(cos)` ⇒ sec
`(7\pi)/(4)` =
`(2)/(√(2) )` =
`(2)/(√(2) )`
`(\frac{√(2)} {√(2)})` =
`(2√(2) )/(2)` =
`√(2)`

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Answer: (a)
`(3\pi)/(2)` (b) (0, -1) (c) A (d) -1

Explanation:

a)
`(11\pi)/(2)` -
`(4\pi)/(2)` =
`(7\pi)/(2)` -
`(11\pi)/(2)` =
`(3\pi)/(2)`

b)
`(3\pi)/(2)` is on the Unit Circle at (0, -1)

c) sin =
`(opposite)/(hypotenuse)` which equals
`(y)/(r)` on the Unit Circle.

d) sin is the y-coordinate. sin
`((3\pi)/(2))` = -1