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Precalculus Notes: Unit 5 – Trigonometric Identities Syllabus Objectives: 3.3 – The student will simplify trigonometric expressions and prove trigonometric identities (fundamental identities). 3.4 – The student will solve trigonometric equations with and without technology. Identity: a statement that is true for all values for which both sides are defined Example from algebra: 3 x 8 11 3 x 13 sin Reciprocal Identities csc Quotient Identities tan Recall: Unit Circle r 1, x 1 csc 1 sin sec sin cos cos , y 1 sec 1 cos cos cot cos sin cot or cos sin 1 cot 1 tan tan x 1 x y 1 sin y x, y Note: sin 2 Pythagorean Theorem: x 2 y2 1 sin Pythagorean Identity: sin 2 To derive the other Pythagorean Identities, divide the entire equation by sin 2 sin 2 sin 2 cos 2 sin 2 1 sin 2 Pythagorean Identities 1 cot 2 sin 2 csc 2 cos 2 2 sin 2 cos 2 1 1 cot 2 cos 2 cos 2 csc 2 and then by cos 2 1 cos 2 tan 2 cos 2 tan 2 1 1 : 1 sec 2 sec 2 Simplifying Trigonometric Expressions: Look for identities Change everything to sine and cosine and reduce Note that the equations in bold are the trig identities used when simplifying. All of the other steps are algebra steps. Ex1: Use basic identities to simplify the expressions. cos cot sin 2 a) cot 1 cos2 sin Page 1 of 23 cos 2 1 sin 2 1 cos 2 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities cos sin 1 cos 2 cot b) tan csc tan csc cos sin sin 2 1 sin csc x 1 csc x 1 2 csc x 1 cos 2 x cos 2 x cos 2 x sin 2 x cos 2 x 2 cot x cos 2 x 2 . cos 2 x csc 2 x 1 cos 2 x 1 cot 2 1 sin 2 x 2 sin x cos x 1 sin sec csc x 1 csc x 1 cos 2 x cos sin csc 1 cos Ex2: Simplify the expression Use algebra: sin sin cos tan sin cos sin csc 2 cot 2 csc 2 1 csc 2 x Solving Trigonometric Equations Isolate the trigonometric function. Solve for x using inverse trig functions. Note – There may be more than one solution or no solution. Ex3: Solve the equation 4 sin 2 x 4 4sin 2 x 4 sin 2 x 0 sin 2 x 1 Find values of x for which x 0 in the interval 0, 2 1 sin 1 1 and x sin x sin 1 1 : . 1 x 3 2 2 , Exploration: Consider a right triangle. Φ c a θ b Note that and sin cos csc sec Page 2 of 23 are complementary. Write the trig functions for each angle. What do you notice? b c c b cos sin sec csc a c c a tan cot cot tan b a c b Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities are equal to the cofunctions of θ, when **The trig functions of and are complementary. Cofunction Identities: sin 90 cos or sin cos 90 sin or cos sec 90 csc or sec csc 90 sec or csc tan 90 cot or cot cot 90 tan or cot cos 2 sin 2 csc 2 sec 2 tan 2 tan 2 Exploration: Consider an angle, θ, and its opposite, as shown in the coordinate grid. Compare the trig functions of each angle. z θ x −θ z y sin y , sin z y z cos csc z , csc y z y sec x , cos z z , sec x **Cosine and secant are the only EVEN f f x f x x z z x tan y , tan x y x cot x , cot y x y trig functions. All the rest are ODD f x . x Odd-Even Identities: sin sin and csc csc tan tan and cot cot cos Page 3 of 23 cos and sec sec Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities Ex4: Simplify the expression sin sin x csc x sin x csc x sin x x . x csc 1 sin x 1 sin x sin x csc x csc x Simplifying Trigonometric Expressions: Simplify using the following strategies. Note that the equations in bold are the trig identities used when simplifying. All of the other steps are algebra steps. Ex5: Simplify the expression by factoring. cos3 x cos x sin 2 x cos3 x cos x sin 2 x cos x cos 2 x sin 2 x cos x 1 cos x Ex6: Simplify the expression by combining fractions. sin x 1 cos x cos x sin x sin x sin x cos x 1 cos x 1 cos x sin x sin 2 x cos 2 x cos x 1 cos x sin x 1 cos x 1 cos x sin x sin x 1 cos x cos 2 x 1 sin 2 x cos 2 x 1 csc x 1 sin x cos x sin x sin 2 x cos x cos 2 x 1 cos x sin x 1 sin x sin 2 x csc x Solving Trigonometric Equations: Solve using the following strategies. Find all solutions for each equation in the interval 0, 2 . Ex7: Solve the equation by isolating the trig function. 2cos x 1 0 cos x 1 2 x 5 3 3 These are values of x where the cosine is equal to , 1 . 2 Ex8: Solve the equation by extracting square roots. 4 sin 2 x 3 0 sin 2 x 3 4 sin x 3 2 x 2 4 5 These are values of x where the sine is equal to , , 3 3 3 3 , Ex9: Solve the equation by factoring. 2 cos 2 x cos x 1 2 cos 2 x Set equal to zero. Factor. 2cos x 1 0 1 cos x 2 5 x , 3 3 Note: It may be easier to use u-substitution with u quadratic equation that can be factored. Page 4 of 23 0 2 cos x 1 cos x 1 Set each factor equal to zero. Solve each equation. cos x 1 0 cos x 1 0 cos x x 1 x 3 , , 5 3 cos x to help students visualize the equation as a Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 3 . 2 Precalculus Notes: Unit 5 – Trigonometric Identities Ex10: Solve the equation by factoring. 2sec x sin x sec x sec x 2 sin x 1 0 Factor out GCF. Use zero product property. sec x Solve each equation. 1 cos x x 0 0 2sin x 1 0 1 sin x 2 0 x 5 6 6 , Note: It is possible for an equation to have no solution. 2 Ex11: Solve by rewriting in a single trig function. 2 sin x 2 Substitute Pyth. Identity. 2 1 cos x Simplify algebraically. 2 2 cos 2 x Factor and solve. 3cos x 3cos x Ex12: Solve using trig substitutions. Rewrite sin 3 x Rewrite sin x cos x 0 sin 3 x cos x cos x tan x tan x . sin 2 x tan x tan x sin 2 x tan x tan x 0 0 x 0, , cos sin 2 1 1 cos 2 0 1 , cos x 1 2 x 0, 5 3 3 , tan x sin 2 x sin x cos x Use the zero product property and solve. 3 2 2 cos 2 x 3cos x 1 sin x sin 2 x . Set equal to zero and factor. x sin 3 3 2 cos x 1 cos x 1 3cos x 2 tan x sin 2 x sin x cos x tan x tan x sin 2 x 1 0, sin x 1 0 x 2 , 3 2 3 2 2 , Ex13: Find the approximate solution using the calculator. 4cos x 1 Isolate the trig function. cos x 1 4 To find x, we need to find the inverse cosine of ¼. x When solving an equation in the interval 0, 2 cos 1 4 1 x 1.318 , be sure to be in Radian mode. You Try: Make the suggested trigonometric substitution and then use the Pythagorean Identities to write the resulting function as a multiple of a basic trig function. 4 x2 , x 2 cos QOD: Explain the relationship between trig functions and their cofunctions. Page 5 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities Syllabus Objective: 3.3 – The student will simplify trigonometric expressions and prove trigonometric identities. Trigonometric Identity: an equation involving trigonometric functions that is a true equation for all values of x Tips for Proving Trigonometric Identities: 1. Manipulate only one side of the equation. Start with the more complicated side. 2. Look for any identities (use all that you have learned so far). 3. Change everything to sine or cosine. 4. Use algebra (common denominators, factoring, etc) to simplify. 5. Each step should have one change only. 6. The final step should have the same expression on both sides of the equation. Note: Your goal when proving a trig identity is to make both sides look identical! For all of the following examples, prove that the identity is true. The trig identities used in the substitutions are in bold. Ex1: cos3 x 1 sin 2 x cos x Start with the right side (more complicated). cos3 x 1 sin 2 x cos x sin 2 x cos 2 x cos x cos3 x cos 2 x 1 cos 2 x 1 sin 2 x cos3 x Ex2: 1 1 sin x 1 1 sin x 2 sec 2 x Start with the left side. Combine fractions. Simplify. Trig substitution. Identity. Page 6 of 23 1 1 sin x 1 1 sin x 1 1 sin x 1 1 2 sec 2 x sin x sin x sin x 2 1 sin 2 x 2 cos 2 x 2 sec 2 x 2 sec 2 x sin 2 x cos 2 x 1 cos x sec x 1 cos 2 x 1 sin 2 x Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities Ex3: tan 2 x 1 cos2 x 1 tan 2 x tan 2 x 1 cos2 x 1 Start with the left side. Trig substitution. sec 2 x cos 2 x 1 Trig substitution. sec 2 x sin 2 x Trig substitution. 1 cos 2 x sin 2 x Multiply. sin 2 x cos 2 x Identitiy. tan 2 x Ex4: sec x tan x tan 2 x 1 sin 2 x tan 2 x sec x tan x Change to sine/cosine. 1 cos x sin x cos x Combine fractions. Multiply num/den by conjugate. 1 sin 2 x cos x 1 sin x Trig substitution. cos 2 x cos x 1 sin x Simplify. cos x 1 sin x 1 1 cos x sec x sin x cos x tan x 1 cos x sec x sin 2 x cos x 1 sin x sin x cos x tan x sin 2 x cos 2 x cos 2 x 1 1 sin 2 x cos x 1 sin x sin 2 Start with left side. sec 2 sec 2 1 Split the fraction. sec 2 sec 2 1 sec 2 Simplify. 1 Trig substitution. 1 cos 2 Identity. sin 2 Page 7 of 23 cos 2 x 1 1 1 sin x cos x 1 sin x 1 sin x cos x 1 sin x Multiply. sec 2 sec 2 cos 2 x sec 2 x cos x 1 sin x Start with the left side. Ex5: tan 2 x sin 2 1 sec 2 sin 2 1 sec cos sin 2 cos 2 1 sin 2 1 cos 2 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities Challenge: Try to prove the identity above in another way. You Try: Prove the identity. cos x sin x 2 cos x sin x 2 2 QOD: List at least 5 strategies you can use when proving trigonometric identities. Page 8 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities Syllabus Objective: 3.3 – The student will simplify trigonometric expressions and prove trigonometric identities (sum and difference identities). Recall: 36 64 100 So in general, 10 a 36 b a 64 36 64 6 8 14 b and 2 3 2 5 2 25 2 3 So in general, a b 2 a2 2 2 2 3 2 13 b2 Sum and Difference Identities sin u v sin u cos v cos u sin v cos u v cos u cos v sin u sin v tan u v tan u tan v 1 tan u tan v Note: Be careful with +/− signs! Simplifying Expressions with Sum and Differences 1. Rewrite the expression using a sum/difference identity. 2. Simplify the expression and evaluate if necessary. Ex1: Write the expression as the sine of an angle. Then give the exact value. sin sin 4 cos 12 sin cos 4 2 12 4 cos sin sin cos 12 12 sin 4 4 sin 12 sin u v 12 sin u cos v cos u sin v 1 2 6 Evaluating Trigonometric Expressions with Non-Special Angles 1. Rewrite the angle as a sum or difference of two special angles. 2. Rewrite the expression using a sum/difference identity. 3. Evaluate the expression. or Ex2: Find the exact value of cos195 . 1. 195 2. 150 cos u v Page 9 of 23 45 cos u cos v cos195 sin u sin v cos 150 45 cos 150 45 cos150 cos 45 sin150 sin 45 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities 3. cos150 cos 45 3 2 sin150 sin 45 2 2 1 2 Ex3: Write as one trig function and find an exact value. 1. tan u v 2. tan 80 3. tan 135 tan u tan v 1 tan u tan v 55 2 2 6 4 2 4 6 4 2 tan 80 tan 55 1 tan 80 tan 55 tan 80 tan 55 1 tan 80 tan 55 tan 80 55 tan 135 1 Evaluating Trig Functions Given Other Trig Function(s) Ex4: Find cos u v given cos u cos u v cos u cos v sin u sin v 15 , 17 u 3 and sin v 2 4 ,0 5 v 2 . We must find cos v and sin u . Draw the appropriate right triangles in the coordinate plane. cos u 15 , 17 u 3 : 2 4 ,0 5 sin v v 2 : 15 u 5 4 17 v Use the Pythagorean Theorem to find the missing sides. 152 U 2 U 17 2 V2 8 cos u cos v Page 10 of 23 sin u sin v 52 V 3 8 . In Quadrant I, cosine is positive, so cos v 17 In Quadrant III, sine is negative, so sin u cos u v 42 15 17 3 5 8 17 4 5 45 85 32 85 77 85 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 3 . 5 Precalculus Notes: Unit 5 – Trigonometric Identities Proving Identities Ex5: Verify the identity. sin tan cos cos tan Start with the left side. sin tan cos cos Trig substitution: Split the fraction: tan sin u cos v cos u sin v sin cos cos sin cos cos sin cos cos cos cos cos Simplify: sin cos sin cos Trig substitution: tan tan You Try: Verify the cofunction identity sin sin cos tan 2 tan cos QOD: Give an example of a function for which f a b Then give an example of a function for which f a b Page 11 of 23 sin u v using the angle difference identity. f a f a f b for all real numbers a and b. f b for all real numbers a and b. Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities Syllabus Objective: 3.3 – The student will simplify trigonometric expressions and prove trigonometric identities (double angle and power-reducing identities). Ex1: Derive the double angle identities using the sum identities. 1. sin 2u sin u sin u 2. cos 2u u u cos u cos u 3. tan 2u 2 sin u cos u cos u cos u sin u sin u cos 2 u sin 2 u u u tan u sin u cos u sin u cos u u tan u tan u 1 tan u tan u tan u u 2 tan u 1 tan 2 u Double Angle Identities sin 2 2 sin cos cos 2 cos 2 sin 2 2 tan tan 2 1 tan 2 There are two other ways to write the double angle identity for cosine. Use the Pythagorean identity. cos 2 cos 2 1 cos 2 cos 2 1 2 sin 2 1 sin 2 cos 2 2 cos 2 sin 2 cos 2 sin 2 cos 2 Page 12 of 23 1 sin 2 1 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities Evaluating Double-Angle Trigonometric Functions Ex: Find the exact value of cos2u given cot u 5, 3 2 2 . u 5 u 1 u will be in Quadrant IV and forms a right triangle as labeled. Using the Pythagorean Theorem, we have c 2 cos 2 u Double Angle Identity: cos 2 u 2 3 2 2u 2 2 3 3 2 c 26 5 cos u 1 26 Note: If u is in Quadrant IV, 12 sin 2 u 2 5 cos 2u 52 2 25 26 26 26 1 26 , sin u 24 26 1 26 12 13 2 , then for 2u we have u 4 , which is in Quadrant IV. So it makes sense that cos2u is positive. 2u Solving Trigonometric Equations Ex2: Find the solutions to 4sin x cos x 1 in 0, 2 Rewrite the equation. 2 2sin x cos x 1 Trig substitution. 2 sin 2 x Isolate trig function. sin 2 x Solve for the argument. 2x 1 . sin2 x 1 2 sin 1 1 2 Because the argument is 2x, we must revisit the domain. 0, 2 0 x 2 . Therefore, 2 0 2x 2 2 2x Solve for x. Page 13 of 23 x 6 2sin x cos x 0 2x , 2x is the restriction for x. So 4 . 5 , 2x 6 13 , 2x 6 17 6 5 13 17 , , 12 12 12 12 , Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities Rewriting a Multiple Angle Trig Function to a Single Angle Ex3: Express sin 3x in terms of sin x . sin 3 x sin 2 x Rewrite argument as a sum Sum identity x sin x cos2 x sin2 x cos x sin x 1 2 sin 2 x Double angle identities Pythagorean identity 2 sin x cos x cos x sin x 2 sin 3 x 2 sin x cos 2 x sin x 2 sin 3 x 2 sin x 1 sin 2 x sin x 2 sin 3 x 2 sin x 2 sin 3 x 3sin x 4sin 3 x Simplify Verifying a Trig Identity Ex4: Verify sin 2 2 tan 1 tan 2 Start with left side. sin 2 Pythagorean identity 2 tan 1 tan 2 2 tan sec 2 sin 2 cos 1 cos 2 Rewrite in sines/cosines Simplify sin cos 2 cos 1 2sin cos 2 Double angle identity sin2 cos 2 cos 2 sin 2 1 2 sin 2 Recall: cos 2 2 cos 2 cos 2 Solving for sin 2 . and cos 2 1 , we can derive the power reducing identities. cos 2 1 2 sin 2 cos 2 2 cos 2 sin 2 1 cos 2 2 cos 2 1 cos 2 2 tan 2 sin 2 cos 2 Page 14 of 23 1 cos 2 2 1 cos 2 2 1 1 cos 2 1 cos 2 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities Power Reducing Identities sin 2 cos 2 tan 2 1 cos 2 2 1 cos 2 2 1 cos 2 1 cos 2 Ex5: Express cos5 x in terms of trig functions with no power greater than 1. cos 5 x Rewrite as a product cos 2 x cos 2 x cos x 1 cos 2 x 1 cos 2 x cos x 2 2 1 1 2 cos 2 x cos 2 2 x cos x 4 Power reducing identity Multiply 1 1 2 cos 2 x 4 Power reducing identity 1 cos 4 x cos x 2 You Try: 1. Find the solutions to 2cos x sin 2 x 2. Verify 2 cos 2 sin 2 cot tan 0 in 0, 2 . . QOD: How do you convert from a cosine function to a sine function? Explain. Page 15 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities Syllabus Objective: 3.3 – The student will simplify trigonometric expressions and prove trigonometric identities (half angle identities). Recall: sin 2 1 cos 2 2 Solving for sin u u , we have sin 2 2 u u . We have sin 2 2 2 Let 1 cos u 2 1 cos u . All of the other half-angle identities can be derived 2 in a similar manner. Half-Angle Identities sin u 2 1 cos u 2 cos u 2 1 cos u 2 tan u 2 1 cos u 1 cos u Note: The u 2 u tan 2 tan Note: There are 2 others for tangent. will be decided based upon which quadrant 1 cos u sin u sin u 1 cos u u lies in. 2 Evaluating Trig Functions Ex1: Find the exact value of cos Rewrite as a half angle cos 12 . 1 2 6 cos 12 1 cos Half angle identity 6 2 3 2 1 Evaluate 2 2 3 4 Choose sign 2 3 2 12 is in Quadrant I, where cosine is positive. Note: This can also be found using a difference identity. cos Page 16 of 23 12 cos 3 4 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities Solving a Trig Equation Ex2: Solve the equation sin x Half-angle identity sin x in 0, 2 2 . 1 cos x 2 sin x 1 cos x 2 1 cos x 1 cos 2 x 2 sin 2 x Square both sides Pythagorean identity 2 2 cos 2 x Set equal to zero 1 cos x 2 2 cos x cos x 1 0 Factor 2 cos x 1 cos x 1 2 cos x 1 0 0 cos x 1 0 1 cos x 2 2 4 x , 3 3 Zero product property cos x x 1 0 x 0, 2 4 , 3 3 You Try: 1. Find the exact value of tan 22.5 . 2. Solve the equation sin 2 x 2 cos x 0 in 0, 2 . QOD: Explain why two of the half-angle identities do not have +/− signs. Page 17 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities Syllabus Objective: 3.5 – The student will solve application problems involving triangles (Law of Sines). Deriving the Law of Sines: Consider the two triangles. C C b A b a h A c B a h B c h h and sin B . b a b sin A and h a sin B In the acute triangle, sin A Solve for h. h In the obtuse triangle, sin sin A a sin C The same type of argument can be used to show that c Substitute. a sin B b sin A can be rewritten as B sin B h . a sin B b sin B b sin A . a Law of Sines: The ratio of the sine of an angle to the length of its opposite side is the same for all three angles of any triangle. C a b sin A a sin B b sin C or c a sin A b sin B c sin C A B c Solving a Triangle: finding all of the missing sides and angles Note: The Law of Sines can be used to solve triangles given AAS and ASA. Ex1: Solve the triangle ABC given that B 15 , C 52 , and b 9. C a 9 A sin B b c sin C c B We are given AAS, so we will use the Law of Sines. sin15 9 sin 52 c Solve for c: c sin15 Find m A using the triangle sum. m A 180 sin B b Solve for a: a sin15 sin A a sin15 9 sin113 a ABC : m A 113 , m B Page 18 of 23 15 , m C 52 , a 15 52 32, b 9, c 9 sin 52 9 sin 52 sin15 c 27.4 113 9 sin113 c 9 sin113 sin15 27.4 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 32 Precalculus Notes: Unit 5 – Trigonometric Identities Ambiguous Case: When given SSA, there could be 2 triangles, 1 triangle, or no triangles that can be created with the given information. Ex1: Solve the triangle ABC (if possible) when m C sin C c Given SSA, use Law of Sines. sin A a 54 , a sin 54 7 7. 10, c sin A 10 Solve for A. 10sin 54 7sin A sin A 10sin 54 7 A sin 10sin 54 7 1 sin 1 1.1557 Solution: There is no possible triangle with the given information because there is no angle with a sine greater than 1. Ex2: Solve the triangle ABC (if possible) when m C sin C c Given SSA, use Law of Sines. sin B b 31 , b sin 31 29 46, c 29 . sin B 46 Solve for B. 46sin 31 29sin B 46sin 31 29 sin B B sin 1 46sin 31 29 54.8 Note that the calculator only gives the acute angle measure for B. There does exist an obtuse angle B with the same sine. m B 180 54.8 125.2 This is also an appropriate measure of an angle in a triangle, so there are 2 triangles that can be formed with the given information. Triangle 1 Triangle 2 m A 180 sin C c a sin 31 sin A a 54.8 31 sin 31 29 29 sin 94.2 Triangle 1: m A 94.2 , m B 94.2 m A 180 sin 94.2 a sin C c a 56.2 54.8 , m C Triangle 2: m A 23.8 , m B 125.2 , m C Page 19 of 23 sin A a a sin 31 31 , a 31 , a 125.2 31 56.2, b 22.7, b sin 31 29 29 sin 23.8 46, c 46, c 23.8 sin 23.8 a a 22.7 29 29 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities Application Problems 1. Draw a picture! 2. Use the Law of Sines to solve for what is asked in the problem. Ex3: The angle of elevation to a mountain is 3.5 . After driving 13 miles, the angle of elevation is 9 . Approximate the height of the mountain. z 3.5° θ 13 h 9° First, find θ: (not drawn to scale) 180 9 171 . Therefore, the third angle in the small triangle = 5.5°. Using the law of sines, we know that 13 sin 5.5 z sin171 Now we can use right triangle trig to find h: sin 3.5 You Try: Solve the triangle sin 5.5 z h 21.22 ABC (if possible) when m B 13sin171 h z 21.22 . 1.3miles 98 , b 10, c 3. QOD: Explain why SSA is the ambiguous case when solving triangles. Page 20 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities Syllabus Objective: 3.5 – The student will solve application problems involving triangles (Law of Cosines). Law of Cosines: For any triangle, ABC C a b c 2 b 2 a2 a 2 a 2 b2 b 2 2ab cos C c 2 2ac cos B c2 2bc cos A Note: In a right triangle, c 2 a2 A b2 B c c2 2ab cos 90 a2 b2 2ab 0 c2 a2 b 2 (Pythagorean Theorem) The Law of Cosines can be used to solve triangles when given SAS or SSS. Ex1: Solve the triangle ABC when m A Note: The given information is SAS. Use a 2 a2 42 2 152 b2 a2 2 42 15 cos 49 49 , b c2 15 . 42, & c 2bc cos A . 1162.36 a 34.09 Now that we have a matching pair of a side and angle, we can use the Law of Sines. a sin A 42 42sin 49 sin B sin B 34.09 Now find the two possibilities for m C using the triangle sum: C b sin B 180 34.09 sin 49 49 68.4 62.6 or C 180 68.4 or B B 49 111.6 49 , m B 111.6 , m C 19.4 , a Ex2: Solve the triangle ABC when a Note: The given information is SSS. Use a 2 312 52 2 282 2 52 28 cos A 34.09, b cos A c2 111.6 19.4 . 42, c 15 28 . 31, b 52, & c b2 68.4 19.4 Since c is the shortest side, it must be opposite the smallest angle. So m C m A 180 2bc cos A . (Or any of them!) 2527 2912 A 29.8 Now that we have a matching pair of a side and angle, we can use the Law of Sines. a sin A b sin B 31 sin 29.8 52 sin B sin B 52sin 29.8 31 B 56.5 or B Now find the two possibilities for m C using the triangle sum: C 180 29.8 56.5 93.7 or C 180 29.8 123.5 180 56.5 26.7 Since c is the shortest side, it must be opposite the smallest angle. So m C m A Page 21 of 23 29.8 , m B 123.5 , m C 26.7 , a 31, b 52, c 26.7 . 28 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 123.5 Precalculus Notes: Unit 5 – Trigonometric Identities Application Problems 1. Draw a picture! 2. Use the Law of Cosines to solve for what is asked in the problem. Ex3: A plane takes off and travels 60 miles, then turns 15° and travels for 80 miles. How far is the plane from the airport? 15° 165° 60 80 c (not drawn to scale) Using the picture, we can find the angle in the triangle to give us SAS. Use the Law of Cosines: c 2 c 2 60 2 80 2 a2 b2 2 60 80 cos165 2 ab cos C c 2 19272.89 c 138.8 miles Area of a Triangle Recall: A 1 bh 2 h c sin c h h c sin ; A 1 bc sin 2 h θ b b Formula for the Area of a Triangle Given SAS: 1 ab sin C 2 A Ex: Find the area of triangle ABC shown. B 120° 50 32 C A A 1 32 50 sin120 2 1 ac sin B . 2 We will use A A 16 50 3 2 A 400 3 sq units Heron’s Area Formula Semi-Perimeter: s a b c 2 Area of a Triangle Given SSS: A Page 22 of 23 s s a s b s c Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 Precalculus Notes: Unit 5 – Trigonometric Identities Ex4: Find the area of the triangle with side lengths 5 m, 6 m, and 9 m. Semiperimeter: s A s s a A s b a b c 2 s 5 6 9 2 s 10 s c 10 10 5 10 6 10 9 A 10 5 4 1 A 200 A 10 2 m 2 You Try: Two ships leave port with a 19° angle between their planned routes. If they are traveling at 23 mph and 31 mph, how far apart are they in 3 hours? QOD: Can there be an ambiguous case when using the Law of Cosines? Explain why or why not. Page 23 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4