copyright © 2013, 2009, 2005 pearson education, inc. 1 6 inverse circular functions and...

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 1

6

Inverse Circular Functions and Trigonometric Equations



6.1 Inverse Circular Functions

6.2 Trigonometric Equations I

6.3 Trigonometric Equations II

Inverse Circular Functions and Trigonometric Equations6


Inverse Circular Functions6.1Inverse Functions ▪ Inverse Sine Function ▪ Inverse Cosine Function ▪ Inverse Tangent Function ▪ Remaining Inverse Circular Functions ▪ Inverse Function Values


Find y in each equation.

6.1 Example 1 Finding Inverse Sine Values (page 249)


6.1 Example 1 Finding Inverse Sine Values (cont.)


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.



6.1 Example 2 Finding Inverse Cosine Values (page 250)



6.1 Example 2 Finding Inverse Cosine Values (cont)


6.1 Example 3 Finding Inverse Function Values (Degree-Measured Angles) (page 253)

Find the degree measure of θ in each of the following.


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


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


6.1 Example 4(b) Finding Inverse Function Values With a Calculator (cont.)

θ = 104.1871349°


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.


6.1 Example 5(a) Finding Function Values Using Definitions of the Trigonometric Functions (cont.)


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.


6.1 Example 5(b) Finding Function Values Using Definitions of the Trigonometric Functions (cont.)


6.1 Example 6(a) Finding Function Values Using Identities

(page 255) Evaluate the expression without a calculator.

Use the cosine difference identity:



(cont.) Sketch both A and B in quadrant I. Use the Pythagorean theorem to find the missing side.



(cont.)


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.



(cont.) Sketch A in quadrant II. Use the Pythagorean theorem to find the missing side.



(cont.)


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.


6.1 Example 7(a) Finding Function Values in Terms of u

(cont.)


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.


6.1 Example 7(b) Finding Function Values in Terms of u

(cont.)

Use the double-angle sine identity to find sin 2θ.


Trigonometric Equations I6.2Solving by Linear Methods ▪ Solving by Factoring ▪ Solving by Quadratic Methods ▪ Solving by Using Trigonometric Identities

Copyright © 2013, 2009, 2005 Pearson Education, Inc. 286-28

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


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}


6.2 Example 2 Solving a Trigonometric Equation by Factoring (page 263)

or

Solution set: {90°, 135°, 270°, 315°}


6.2 Example 3 Solving a Trigonometric Equation by Factoring (page 263)


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.


Trigonometric Equations II6.3Equations with Half-Angles ▪ Equations with Multiple Angles


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.


6.3 Example 1 Solving an Equation Using a Half-Angle Identity (cont.)

This is a cosine curve with period


6.3 Example 2 Solving an Equation With a Double Angle

(page 270)

Factor.

or


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°.


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

copyright © 2013, 2009, 2005 pearson education, inc. 1 6 inverse circular functions and...

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