Peters Presentation

Page 420 #100 

Define the inverse secant function by restricting the domain of the secant function to the intervals [0, π/2) and (π/2, π] and sketch its graph.

First, to make an inverse secant function graph, you must switch the domain and the range of the secant function.  The range of secant is (-∞, -1] , [1, ∞)  The domain of secant is x ≠ π/2 + Nπ 

You are given a restricted domain of secant, [0, π/2) and (π/2, π]. This becomes the range of inverse secant.  The domain of inverse secant is (-∞, -1] , [1, ∞) 

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To find the points on an inverse secant graph, make a yx table. Angles on the unit circle correspond to the y values, and the x values are the secants at each angle.

y

π/6

π/4

π/3

x

(2√3)/3

√2

2

2π/3 3π/4 -2

5π/6

π

-√2 -(2√3)/3 -1

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This is an inverse secant function graph.

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Y=arcsec(x) if and only if sec(y) = x – Range: [0, π/2) and (π/2, π] – Domain: [-∞ , -1] , [1, ∞)

#102. Use the results of #99-101 to evaluate the following without using a calculator.  (a) arcsec √2  (b) arcsec 1  (c) arccot (-√3)  (d) arccsc 2 

(a) arcsec√2 

What angle has a secant of √2? – Or, more simply, the cosine of 1/(√2)

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1/(√2) simplifies to (√2)/2

π/4 on the unit circle has a cosine of (√2)/2

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7π/4 also has a cosine of √2/2, and therefore a secant of √2.

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Refer back to the restrictions on the range of inverse secant: [0, π/2) and (π/2, π]

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7π/4 is greater than π and so exceeds the restriction.

(b) arcsec 1 Use the same method.  A secant of 1 means the angle also has a cosine of 1. 

0 on the unit circle has a cosine of 1.

(c) arccot (-√3) Use the same method.  A cotangent of - √3 means the angle has a tangent of 1/(- √3).  1/(- √3) simplifies to (- √3)/3 

5π/6 on the unit circle has a cotangent of (-√3)/3.

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The answer cannot be 11π/6, although it also has a cotangent of -√3.

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Referring back to problem 99, the cotangent function is given a restriction on its domain: (0, π). This means the inverse cotangent function will have the same restriction on its range, or y-values.

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11π/6 is greater than π, and so exceeds the restrictions.

(d) arccsc 2 Use the same method.  A cosecant of 2 means the angle has a sin of ½. 

π/6 has a sin of ½, and therefore a cosecant of 2.

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The answer cannot be 5π/6, although it has a cosecant of 2.

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Referring back to problem 101, the cosecant function is given a restriction of [- π/2, 0) and (0, π/2]. This means the inverse cosecant function will have the same restriction on its range.

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5π/6 is greater than π/2, and so exceeds the restrictions.