copyright © 2013, 2009, 2005 pearson education, inc. 1 6 inverse circular functions and...
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Inverse Circular Functions and Trigonometric Equations
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6.1 Inverse Circular Functions
6.2 Trigonometric Equations I
6.3 Trigonometric Equations II
Inverse Circular Functions and Trigonometric Equations6
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Inverse Circular Functions6.1Inverse Functions ▪ Inverse Sine Function ▪ Inverse Cosine Function ▪ Inverse Tangent Function ▪ Remaining Inverse Circular Functions ▪ Inverse Function Values
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Find y in each equation.
6.1 Example 1 Finding Inverse Sine Values (page 249)
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6.1 Example 1 Finding Inverse Sine Values (cont.)
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6.1 Example 1 Finding Inverse Sine Values (cont.)
is not in the domain of the inverse sine function, [–1, 1], so does not exist.
A graphing calculator will give an error message for this input.
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Find y in each equation.
6.1 Example 2 Finding Inverse Cosine Values (page 250)
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Find y in each equation.
6.1 Example 2 Finding Inverse Cosine Values (cont)
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6.1 Example 3 Finding Inverse Function Values (Degree-Measured Angles) (page 253)
Find the degree measure of θ in each of the following.
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6.1 Example 4 Finding Inverse Function Values With a Calculator (page 254)
(a) Find y in radians if
With the calculator in radian mode, enter as
y = 1.823476582
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6.1 Example 4(b) Finding Inverse Function Values With a Calculator (page 254)
(b) Find θ in degrees if θ = arccot(–0.2528).
A calculator gives the inverse cotangent value of a negative number as a quadrant IV angle.
The restriction on the range of arccotangent implies that the angle must be in quadrant II, so, with the calculator in degree mode, enter arccot(–0.2528) as
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6.1 Example 4(b) Finding Inverse Function Values With a Calculator (cont.)
θ = 104.1871349°
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6.1 Example 5 Finding Function Values Using Definitions of the Trigonometric Functions (page 254)
Evaluate each expression without a calculator.
Since arcsin is defined only in quadrants I and IV, and
is positive, θ is in quadrant I.
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6.1 Example 5(a) Finding Function Values Using Definitions of the Trigonometric Functions (cont.)
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6.1 Example 5(b) Finding Function Values Using Definitions of the Trigonometric Functions (page 266)
Since arccot is defined only in quadrants I and II, and
is negative, θ is in quadrant
II.
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6.1 Example 5(b) Finding Function Values Using Definitions of the Trigonometric Functions (cont.)
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6.1 Example 6(a) Finding Function Values Using Identities
(page 255) Evaluate the expression without a calculator.
Use the cosine difference identity:
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6.1 Example 6(a) Finding Function Values Using Identities
(cont.) Sketch both A and B in quadrant I. Use the Pythagorean theorem to find the missing side.
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6.1 Example 6(a) Finding Function Values Using Identities
(cont.)
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6.1 Example 6(b) Finding Function Values Using Identities
(page 255) Evaluate the expression without a calculator.
Use the double-angle sine identity:
sin(2 arccot (–5))
Let A = arccot (–5), so cot A = –5.
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6.1 Example 6(b) Finding Function Values Using Identities
(cont.) Sketch A in quadrant II. Use the Pythagorean theorem to find the missing side.
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6.1 Example 6(b) Finding Function Values Using Identities
(cont.)
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6.1 Example 7(a) Finding Function Values in Terms of u
(page 256) Write , as an algebraic expression in u.
Sketch θ in quadrant I. Use the Pythagorean theorem to find the missing side.
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6.1 Example 7(a) Finding Function Values in Terms of u
(cont.)
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6.1 Example 7(b) Finding Function Values in Terms of u
(page 256) Write , u > 0, as an algebraic expression in u.
Sketch θ in quadrant I. Use the Pythagorean theorem to find the missing side.
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6.1 Example 7(b) Finding Function Values in Terms of u
(cont.)
Use the double-angle sine identity to find sin 2θ.
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Trigonometric Equations I6.2Solving by Linear Methods ▪ Solving by Factoring ▪ Solving by Quadratic Methods ▪ Solving by Using Trigonometric Identities
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6.2 Example 1a Solving a Trigonometric Equation by Linear Methods (page 262)
is positive in quadrants I and III.
The reference angle is 30° because
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6.2 Example 1 Solving a Trigonometric Equation by Linear Methods (cont.)
Solution set: {30°, 210°}
b) for all solutions
Solution set: {30° + 180°n, where n is any integer}
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6.2 Example 2 Solving a Trigonometric Equation by Factoring (page 263)
or
Solution set: {90°, 135°, 270°, 315°}
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6.2 Example 3 Solving a Trigonometric Equation by Factoring (page 263)
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6.2 Example 3 Solving a Trigonometric Equation by Factoring (cont.)
has one solution,
has
two solutions, the angles in
quadrants III and IV with the
reference angle .729728:
3.8713 and 5.5535.
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Trigonometric Equations II6.3Equations with Half-Angles ▪ Equations with Multiple Angles
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6.3 Example 1 Solving an Equation Using a Half-Angle Identity (page 269)
(a) over the interval and (b) give all solutions.
is not in the requested domain.
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6.3 Example 1 Solving an Equation Using a Half-Angle Identity (cont.)
This is a cosine curve with period
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6.3 Example 2 Solving an Equation With a Double Angle
(page 270)
Factor.
or
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6.3 Example 3 Solving an Equation Using a Multiple Angle Identity (page 270)
From the given interval 0° ≤ θ < 360°, the interval for 2θ is 0° ≤ 2θ < 720°.
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6.3 Example 3 Solving an Equation Using a Multiple Angle Identity (cont.)
Since cosine is negative in quadrants II and III, solutions over this interval are