Chapter 6
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Chapter 6. Inverse Circular Functions and Trigonometric Equations. 6.1. Inverse Circular Functions. Any horizontal line will intersect the graph of a one-to-one function in at most one point. The inverse function of the one-to-one function f is defined as. Horizontal Line Test.

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Chapter 6

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Chapter 6

Chapter 6

Inverse Circular Functions and Trigonometric Equations


Chapter 6

6.1

Inverse Circular Functions


Horizontal line test

Any horizontal line will intersect the graph of a one-to-one function in at most one point.

The inverse function of the one-to-one function f is defined as

Horizontal Line Test


Inverse functions review

Inverse Functions Review

  • 1. In a one-to-one function, each x-value corresponds to only one y-value and each y-value corresponds to only one x-value.

  • 2. If a function f is one-to-one, then f has an inverse function f-1.

  • 3. The domain of f is the range of f-1and the range of f is the domain of f-1.

  • 4. The graphs of f and f-1 are reflections of each other about the line y = x.

  • 5. To find f-1(x) from f(x), follow these steps.

    • Replace f(x) with y and interchange x and y.

    • Solve for y.

    • Replace y with f-1(x).


Inverse sine function

Inverse Sine Function

  • means that x = sin y, for

  • Example: Find


Examples

sin1 2

Not possible to evaluate because there is no angle whose sine is 2.

Examples


Inverse sine function1

Inverse Sine Function


Inverse cosine function

Inverse Cosine Function

  • means that x = cos y, for

  • Example: Find


Inverse cosine function1

Inverse Cosine Function


Inverse tangent function

Inverse Tangent Function

  • means that x = tan y, for


Inverse tangent function1

Inverse Tangent Function


Other inverse functions

Other Inverse Functions


Examples1

Examples

  • Find the degree measure of  in the following.

  • a)  = arctan 1b)  = sec1 2

  • a)  must be in (90, 90), since 1 > 0,  must be in quadrant I. The alternative statement, tan  = 1, leads to  = 45.

  • b) sec  = 2, for sec1x,  is in quadrant I or II. Because 2 is positive,  is in quadrant I and sec 60 = 2.


Example

Example

  • Find y in radians if y = arctan(6.24).

    • Calculator in radian mode

    • Enter tan1(6.24) y  1.411891065

  • Find y in radians if y = arccos 2.

    • Calculator in radian mode

    • Enter cos1(2) (error message since the domain of the inverse cosine function is [1, 1].


Example evaluate the expression without using a calculator

Example: Evaluate the expression without using a calculator.

  • Let

  • The inverse tangent function yields values only in quadrants I and IV, since 3/2 is positive,  is in quadrant I.


Example evaluate the expression without using a calculator continued

Example: Evaluate the expression without using a calculator continued

  • Sketch and label the triangle.

  • The hypotenuse is


Example1

Example

  • Evaluate the expression without using a calculator.

  • Let arcsin 2/5 = B

  • Since arcsin 2/5 = B, sin B = 2/5. Sketch a triangle in quadrant I, find the length of the third side, then find tan B.


Example continued

Example continued


Chapter 6

6.2

Trigonometric Equations I


Solving a trigonometric equation

Solving a Trigonometric Equation

  • Decide whether the equation is linear or quadratic in form, so you can determine the solution method.

  • If only one trigonometric function is present, first solve the equation for that function.

  • If more than one trigonometric function is present, rearrange the equation so that one side equals 0. Then try to factor and set each factor equal to 0 to solve.


Solving a trigonometric equation continued

Solving a Trigonometric Equation continued

  • If the equation is quadratic in form, but not factorable, use the quadratic formula. Check that solutions are in the desired interval.

  • Try using identities to change the form of the equation. If may be helpful to square both sides of the equation first. If this is done, check for extraneous solutions.


Example linear method

Example: Linear Method

  • Solve 2 cos2x 1 = 0

  • Solution: First, solve for cos x on the unit circle.


Example factoring

Example: Factoring

  • Solve 2 cos x + sec x = 0

  • Solution

Since neither factor of the equation can equal zero, the equation has no solution.


Example factoring1

Solve 2 sin2x + sin x 1 = 0

Set each factor equal to 0.

continued

Example: Factoring


Example factoring continued

Example: Factoring continued


Example squaring

Example: Squaring

  • Solve cos x + 1 = sin x [0, 2]

Check the solutions in the original equation. The only solutions are /2 and .


Chapter 6

6.3

Trigonometric Equations II


Example using a half angle identity

Example: Using a Half-Angle Identity

  • a) over the interval [0, 2), and

  • b) give all solutions

  • Solution:

    Write the interval as the inequality


Example using a half angle identity continued

Example: Using a Half-Angle Identity continued

  • The corresponding interval for x/2 is

  • Solve

  • Sine values that corresponds to 1/2 are


Example using a half angle identity continued1

Example: Using a Half-Angle Identity continued

  • b) Sine function with a period of 4, all solutions are given by the expressions

    where n is any integer.


Example double angle

Example: Double Angle

  • Solve cos 2x = cos x over the interval [0, 2).

  • First, change cos 2x to a trigonometric function of x. Use the identity


Example double angle continued

Example: Double Angle continued

  • Over the interval


Example multiple angle identity

Example: Multiple-Angle Identity

  • Solve over the interval [0, 360).


Example multiple angle identity continued

Example: Multiple-Angle Identity continued

  • List all solutions in the interval.

  • The final two solutions were found by adding 360 to 60 and 120, respectively, giving the solution set


Example multiple angle

Example: Multiple Angle

  • Solve tan 3x + sec 3x = 2 over the interval [0, 2).

  • Tangent and secant are related so use the identity


Example multiple angle continued

Example: Multiple Angle continued


Example multiple angle continued1

Example: Multiple Angle continued

  • Use a calculator and the fact that cosine is positive in quadrants I and IV,

  • Since both sides of the equation were squared, each proposed solution must be checked. The solution set is {.2145, 2.3089, 4.4033}.


Chapter 6

6.4

Equations Involving Inverse Trigonometric Functions


Solving for x in terms of y using inverse function

Example: y = 3 cos 2x for x.

Solution: We want 2x alone on one side of the equation so we can solve for 2x, and then for x.

Solving for x in Terms of y Using Inverse Function


Solving an equation involving an inverse trigonometric function

Solving an Equation Involving an Inverse Trigonometric Function

Example: Solve 2 arcsin

Solution: First solve for arcsin x, and then for x.

The solution set is {1}.


Solving an equation involving inverse trigonometric functions

Solving an Equation Involving Inverse Trigonometric Functions

Example: Solve

Solution: Let Then sin and for u in

quadrant I, the equation becomes


Solving an equation involving inverse trigonometric functions continued

Solving an Equation Involving Inverse Trigonometric Functions continued

Sketch a triangle and label it using the facts that u is in quadrant I and

Since x = cos u, x = and the solution set is { }.


Solving an inverse trigonometric equation using an identity

Solving an Inverse Trigonometric Equation Using an Identity

Example: Solve

Solution: Isolate one inverse function on one side of the equation.

(1)


Solving an inverse trigonometric equation using an identity continued

Solving an Inverse Trigonometric Equation Using an Identity continued

Let u = arccos x, so 0  u  by definition.

(2)

Substitute this result into equation (2) to get

(3)


Solving an inverse trigonometric equation using an identity continued1

Solving an Inverse Trigonometric Equation Using an Identity continued

From equation (1) and by the definition of the arcsine function,

Since we must have

Thus x > 0. From this triangle we find that


Solving an inverse trigonometric equation using an identity continued2

Now substituting into equation (3) using

The solution set is { }.

Solving an Inverse Trigonometric Equation Using an Identity continued


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