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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

Chapter 6

Inverse Circular Functions and Trigonometric Equations


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 function in at most one point.

  • 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 function in at most one point.

  • means that x = sin y, for

  • Example: Find


Examples

sin function in at most one point.1 2

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

Examples


Inverse sine function1
Inverse Sine Function function in at most one point.


Inverse cosine function
Inverse Cosine Function function in at most one point.

  • means that x = cos y, for

  • Example: Find


Inverse cosine function1
Inverse Cosine Function function in at most one point.


Inverse tangent function
Inverse Tangent Function function in at most one point.

  • means that x = tan y, for


Inverse tangent function1
Inverse Tangent Function function in at most one point.


Other inverse functions
Other Inverse Functions function in at most one point.


Examples1
Examples function in at most one point.

  • Find the degree measure of  in the following.

  • a)  = arctan 1 b)  = 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 function in at most one point.

  • 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 continued

  • 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.



6.2 continued

Trigonometric Equations I


Solving a trigonometric equation
Solving a Trigonometric Equation continued

  • 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 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 continued

  • Solve 2 cos2x 1 = 0

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


Example factoring
Example: Factoring continued

  • 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 continued2 sin2x + sin x 1 = 0

Set each factor equal to 0.

continued

Example: Factoring



Example squaring
Example: Squaring continued

  • Solve cos x + 1 = sin x [0, 2]

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


6.3 continued

Trigonometric Equations II


Example using a half angle identity
Example: Using a Half-Angle Identity continued

  • 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 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 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 continued

  • 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 continued

  • Over the interval


Example multiple angle identity
Example: Multiple-Angle Identity continued

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


Example multiple angle identity continued
Example: Multiple-Angle Identity continued 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 continued

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

  • Tangent and secant are related so use the identity



Example multiple angle continued1
Example: Multiple Angle continued 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}.


6.4 continued

Equations Involving Inverse Trigonometric Functions


Solving for x in terms of y using inverse function

Example continued: 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 Functions continued

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 continued

The solution set is { }.

Solving an Inverse Trigonometric Equation Using an Identity continued


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