holt algebra 2 14-6 solving trigonometric equations 14-6 solving trigonometric equations holt...
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Holt Algebra 2
14-6 Solving Trigonometric Equations14-6 Solving Trigonometric Equations
Holt Algebra 2
Warm UpWarm Up
Lesson PresentationLesson Presentation
Lesson QuizLesson Quiz
Holt Algebra 2
14-6 Solving Trigonometric Equations
Warm Up
Solve.
1. x2 + 3x – 4 = 0
2. 3x2 + 7x = 6
Evaluate each inverse trigonometric
function.
3. Tan-1 1
4. Sin-1
x = 1 or – 4
45°
– 60
Holt Algebra 2
14-6 Solving Trigonometric Equations
Solve equations involving trigonometric functions.
Objective
Holt Algebra 2
14-6 Solving Trigonometric Equations
Unlike trigonometric identities, most trigonometric equations are true only for certain values of the variable, called solutions. To solve trigonometric equations, apply the same methods used for solving algebraic equations.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Example 1: Solving Trigonometric Equations with Infinitely Many Solutions
Find all the solutions of sinθ = sinθ +
Method 1 Use algebra.
Solve for θ over the principal value of sine, –90° ≤ θ ≤ 90°.
sinθ sinθ =
sinθ = sinθ +
sinθ =
Subtract sinθ from both sides.
Combine like terms.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Example 1 Continued
sinθ =
θ = 30°
θ = sin-1
Multiply by 2.
Apply the inverse sineθ.
Find θ when sinθ =
Find all real number value of θ, where n is an integer.
θ = 30° + 360°n Use the period of the sine function.
Use reference angles to find other values of θ.
θ = 150° + 360°n
Holt Algebra 2
14-6 Solving Trigonometric Equations
Example 1 Continued
1
–1
–90 90
Method 2 Use a graph.
Graph y = sinθ and
y = sinθ + in the same viewing window for –90° ≤ θ ≤ 90°.
Use the intersect feature of your graphing calculator to find the points of intersection.
The graphs intersect at θ = 30°. Thus, θ = 30° + 1360°n, where n is an integer.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Check It Out! Example 1
Find all of the solutions of 2cosθ + = 0.
Method 1 Use algebra.
Solve for θ over the principal value of sine, 0 ≤ θ ≤ .
2cosθ = Subtract from both sides.
cosθ = Divide both sides by 2.
θ = cos-1 –
θ = 150°
Apply the inverse cosineθ.
Find θ when cosine θ = .
Holt Algebra 2
14-6 Solving Trigonometric Equations
Check It Out! Example 1 Continued
Use reference angles to find other values of θ.
θ = 150° + 360°n, 210° +360°n.
Method 2 Use a graph.
Graph y = 2cosθ and
y = in the same
viewing window for
–360° ≤ θ ≤ 360°.
–360 360
2
–2
The graphs intersect at θ = 150°. Thus, θ = 150° + 360°n, where n is an integer.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Some trigonometric equations can be solved by applying the same methods used for quadratic equations.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Example 2A: Solving Trigonometric Equations in Quadratic Form
Solve each equation for the given domain.
4tan2θ – 7 tanθ + 3 = 0 for 0° ≤ θ ≤ 360°.
4tan2θ – 7 tanθ + 3 = 0 Factor the quadratic expression by comparing it with 4x2 – 7x + 3 = 0.
(tanθ – 1)(4tanθ – 3) = 0 Apply the Zero Product Property.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Example 2A Continued
tanθ = 1 or tan θ =
θ = tan-1(1) θ = tan-1
= 45° or 225° ≈ 36.9° or 216.9°Use a calculator.
Find all angles for 0°≤ θ ≤360°.
Apply the inverse tangent.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Example 2B: Solving Trigonometric Equations in Quadratic Form
2cos2θ – cosθ = 1 for 0 ≤ θ ≤ .
2cos2θ – cosθ – 1 = 0
(2cosθ + 1) (cosθ – 1) = 0
cosθ = or cosθ = 1
Subtract 1 from both sides.
Factor the quadratic expression by comparing it with 2x2 – x + 1 = 0.
Apply the Zero Product Property.
Find both angles for 0 ≤ θ ≤ .
θ = or θ = 0
Holt Algebra 2
14-6 Solving Trigonometric Equations
Check It Out! Example 2a
Solve each equation for 0 ≤ θ ≤ 2.
cos2 θ + 2cosθ = 3
cos2 θ + 2cosθ – 3 = 0
(cosθ – 1)(cosθ + 3) = 0
cosθ = 1 or cosθ = –3
cosθ = – 3 has no solution because –3 ≤ cosθ ≤ 1.
cosθ = 2 or 0
Subtract 3 from both sides.
Factor the quadratic expression by comparing it to x2 +2x – 3 = 0.
Apply the Zero Product Property.
The only solution will come from cosθ = 1.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Check It Out! Example 2b Solve each equation for 0 ≤ θ ≤ 2.
sin2θ + 5 sinθ – 2 = 0
The equation is in quadratic form but can not be easily factored. Use the quadratic formula.
sinθ =
Holt Algebra 2
14-6 Solving Trigonometric Equations
Check It Out! Example 2b Continued
Apply the inverse sine.
Use a calculator. Find both angles.
Holt Algebra 2
14-6 Solving Trigonometric Equations
You can often write trigonometric equations involving more than one function as equations of only one function by using trigonometric identities.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Example 3A: Solving Trigonometric Equations with Trigonometric Identities
Use trigonometric identities to solve each equation.
tan2θ + sec2θ = 3 for 0 ≤ θ ≤ 2π.
tan2θ + (1 + tan2θ) – 3 = 0
2tan2θ – 2 = 0
tan2θ – 1 = 0
(tanθ – 1)(tanθ + 1) = 0
tanθ = 1 or tanθ = – 1
Substitute 1 + tan2θ for sec2θ by the Pythagorean identity.
Simplify. Divide by 2.
Factor.
Apply the Zero Product Property.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Example 3A Continued
Check Use the intersect feature of your graphing calculator. A graph supports your answer.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Example 3B: Solving Trigonometric Equations with Trigonometric Identities
Use trigonometric identities to solve each equation.
cos2θ = 1 + sin2θ for 0° ≤ θ ≤ 360°
(1 – sin2θ) – 1– sin2θ = 0
–2sin2θ = 0
sin2θ = 0
sinθ = 0
θ = 0° or 180° or 360°
Substitute 1 – sin2θ for cos2θ by the Pythagorean identity.
Simplify.
Divide both sides by – 2.
Take the square root of both sides.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Example 3B Continued
cos2θ = 1+sin2θ for 0° ≤ θ ≤ 360°
Check Use the intersect feature of your graphing calculator. A graph supports your answer.
θ = 0° or 180° or 360°
Holt Algebra 2
14-6 Solving Trigonometric Equations
Check It Out! Example 3a
Use trigonometric identities to solve each equation for the given domain.
4sin2θ + 4cosθ = 5
4(1 - cos2θ) + 4cosθ – 5 = 0
4cos2θ – 4cosθ + 1 = 0
(2cos2θ – 1)2 = 0
Substitute 1 – cos2θ for sin2θ by the Pythagorean identity.
Simplify.
Factor.
Take the square root of both sides and simplify.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Check It Out! Example 3b
Use trigonometric identities to solve each equation for the given domain.
sin2θ = – cosθ for 0 ≤ θ < 2
2cosθsinθ + cosθ = 0
cosθ(2sinθ + 1) = 0
Substitute 2cosθsinθ for sin2θ by the double-angle identity.
Factor.
Apply the Zero Product Property.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Example 4: Problem-Solving Application
On what days does the sun rise at 4 A.M. on Cadillac Mountain? The time of the sunrise can be modeled by
Holt Algebra 2
14-6 Solving Trigonometric Equations
List the important information:
• The function model is
t(m) = 1.665 (m + 3) + 5.485.
• Sunrise is at 4 A.M., which is represented by t = 4.
• m represents the number of months after January 1.
11 Understand the Problem
The answer will be specific dates in the year.
Holt Algebra 2
14-6 Solving Trigonometric Equations
22 Make a Plan
Substitute 4 for t in the model. Then solve the equation for m by using algebra.
Solve33
4 = 1.665sin (m + 3) + 5.485
sin-1(–0.8918) = (m + 3)
Substitute 4 for t.
Isolate the sine term.
Apply the inverse sine θ.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Sine is negative in Quadrants lll and lV. Compute both values.
Qlll: π + sin-1(0.8918) QlV: 2π + sin-1(0.8918)
Holt Algebra 2
14-6 Solving Trigonometric Equations
Using an average of 30 days per month, the date m = 5.10 corresponds to June 4(5 months and 3 days after January 1) and m = 6.90 corresponds to July 28 (6 months and 27 days after January 1).
Holt Algebra 2
14-6 Solving Trigonometric Equations
Enter
y = 1.665sin (x + 3) + 5.485 and y = 4.
Graph the functions on the same viewing window, and find the points of intersection. The graphs intersect at early June and late July.
Look Back44
Check your answer by using a graphing calculator.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Check It Out! Example 4
The number of hours h of sunlight in a day at Cadillac Mountain can be modeled by h(d) = 3.31sin (d – 85.25) + 12.22, where d is the number of days after January 1. When are there 12 hours of sunlight.
11 Understand the Problem
The answer will be specific dates in the year.
Holt Algebra 2
14-6 Solving Trigonometric Equations
List the important information:
• The number of hours of sunlight in the day, which is represented by h = 12.
• d represents the number of days after January 1.
• The function model is
(d – 85.25) + 12.22.h(d) = 3.31sin
11 Understand the Problem
The answer will be specific dates in the year.
Holt Algebra 2
14-6 Solving Trigonometric Equations
22 Make a Plan
Substitute 12 for h in the model. Then solve the equation for d by using algebra.
Solve33
Substitute 12 for h.
Isolate the sine term.
12 = 3.31sin (d – 85.25) + 12.22
Apply the inverse sine θ.
Holt Algebra 2
14-6 Solving Trigonometric Equations
Sine is negative in Quadrants lll and lV. Compute both values.
Qlll:
81.4 ≈ d
Holt Algebra 2
14-6 Solving Trigonometric Equations
Look Back44
Check your answer by using a graphing calculator. Enter
Graph the functions on the same viewing window, and find the points of intersection. The graphs intersect in late March and late September.
y = 3.31sin (d – 85.25) + 12.22