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Precalculus Spring 2017 Exam 5 Summary (Section 6.6 through 8.2) Section 6.6 Find the exact value using inverse sine, inverse cosine, or inverse tangent functions. State the domain and range of each inverse function. Evaluate composition functions using the inverse properties. Evaluate composition functions using right triangle trigonometry. Section 6.7 Solve a right triangle using right triangle trigonometry, the sum of the angles of a triangle, and the Pythagorean Theorem. Draw a navigational bearing. Solve a right triangle problem that uses navigation. Find the frequency of a simple harmonic motion. Section 8.1 Find the missing sides and angles of a triangle using the Law of Sines Find the area of a triangle using A = 1 ab sin C 2 Section 8.2 Find the missing sides and the cosine of the angles using the Law of Cosines Find the area of a triangle using Heron’s Area Formula Section 7.1 Simplify Trigonometric Identities State all the reciprocal, quotient, Pythagorean, co-function, and negative fundamental identities. Section 7.2 Verify Trigonometric Identities Section 7.3 Solve Trigonometric Equations Section 7.4 State the Sum and Difference Formulas Find exact values using sum and difference formulas Rewrite an expression using sum and difference formulas Prove an identity using sum and difference formulas Evaluate sum and difference formulas given two values and the quadrants they lie in. Section 7.5 State the Double Angle, Half-Angle Formulas Simplify using Multiple angle formulas Evaluate multiple angle formulas Find exact values using multiple angle formulas Solve trigonometric equations using double angle formulas and half angle formulas Use the power reducing formulas 6.6 Inverse Trigonometric Functions Inverse Sine Function y = arcsin x ( y = sin −1 x ) if and only if sin y = x , Where −1 ≤ x ≤ 1 and − π 2 ≤y≤ π 2 . This interval is chosen because it is one-to-one and takes on a full range of values. Evaluating the Inverse Sine Function 1. Find the exact value, if possible. 1 2 a) sin −1 d) sin −1 − b) arcsin(5) 1 2 c) arcsin (-1) 3 2 e) arcsin Inverse Cosine Function y = arccos x ( y = cos −1 x ) if and only if cos y = x , Where − 1 ≤ x ≤ 1 and 0 ≤ y ≤ π . This interval is chosen because it is one-to-one and takes on a full range of values. Evaluating the Inverse Cosine Function 2. Find the exact value. a) 3 cos −1 − 2 b) arccos(1) 1 2 e) arcos(-3) d) cos −1 c) cos −1 − 1 2 Inverse Tangent Function y = arctan x ( y = tan −1 x ) if and only if tan y = x , Where −∞ ≤ x ≤ ∞ and − π 2 ≤ y≤ π 2 . This interval is chosen because it is one-to-one and takes on a full range of values. Evaluate the Inverse Tangent Function 3. Find the exact value. a) 3 tan −1 − 1 c) tan −1 3 b) arctan(1) 1 3 d) tan −1 (0) e) arctan Inverse Properties of Trigonometric Functions If −1 ≤ x ≤ 1 and − π 2 ≤y≤ π 2 , then sin(arcsinx) = x and arccsin(siny)=y. If − 1 ≤ x ≤ 1 and 0 ≤ y ≤ π , then cos(arccosx)=x and arccos(cosy)=y. If x is a real number and − π 2 ≤ y≤ π 2 , then tan(arctanx)=x and arctan(tany)=y. The inverse properties do not apply for arbitrary values of x and y. It is only valid within the interval of the domain. Evaluate Composition Functions Using Inverse Properties 4. Find the exact value. a) arccos(cos − c) cos(cos −1 3π ) 11π ) 6 b) tan[arctan(14)] 3π d) sin −1 sin 4 5. Use right triangles to evaluate the composition of functions. 2 5 a) sin arccos − 3 b) tan arcsin 5 6. Write an algebraic expression that is equivalent to cos(arctan 3x). 6.7 Applications and Models Right Triangle Applications Solve the right triangle. (Find all the missing sides and angles using right triangle trigonometry) *Remember lower case letters represent side lengths and upper case letters represent angle measures. 1. A c B b=14 a C = 32.9 ° 2. A safety regulation states that the maximum safe rescue height for a fire department’s ladder when at an angle of elevation of 82° is 115 feet. How long is the ladder? 3. The wave pool (Tsunami Bay) at Magic Waters water park in Rockford, IL is 100 feet long and 60 feet wide. The bottom of the pond is slanted so the depth is 6 inches at the shallow end and 5 feet at the deep end. Find the angle of depression of the bottom of the pool. Trigonometry and Bearings - North is a bearing of 0 degrees. - Compass bearings and courses are given with three digits traveling clockwise from north. (Ex. 070 ° or 112 ° ) 4. Draw a earing of 012 ° . 5. Draw a bearing of 280 ° . 6. Draw a bearing of S47° E. 7. Draw a bearing of N 31° W. 8. A ship leaves port at noon and has a bearing of S 29° W. The ship sails at 20 knots. How many nautical miles west will the ship have traveled by 6:00 P.M.? 9. A ship leaves port at noon on a course of 035° at a speed of 20 knots. At 2PM the ship changes course to 305° at a speed of 25 knots. Find the ships distance from the port of departure at 3PM when it is due north of its starting point. Harmonic Motion A point that moves on a coordinate line is said to be in simple harmonic motion if its distance d from the origin at time t is given by either d = a sin ω t or d = a cos ωt , where a and ω (the angle) are real numbers such that ω > 0 . The motion has amplitude |a|, period 2π ω , and frequency ω . 2π 10. Write the equation for the simple harmonic motion of a ball attached to the bottom of a spring. Suppose that 10 centimeters is the maximum distance the ball moves vertically upward or downward from its equilibrium position and that the time it takes to move from its maximum displacement above zero to its maximum placement below zero is 4 seconds. 11. Given the equation for simple harmonic motion d = 1 sin16π t , find: 4 a) The maximum displacement. b) The frequency. c) The value of d when t = 3. 8.1 Law of Sines Oblique Triangles: Triangles with no right angles. Law of Sines Use to solve AAS, ASA, or SSA (ambiguous case) oblique triangles. If ABC is a triangle with sides a, b, and c, then sin A sin B sin C a b c or the reciprocal form = = = = a b c sin A sin B sin C Solving a Triangle Using the Law of Sines 1. Solve ∆ABC if <A = 45 ° , <B = 30 ° , a = 14. 3. Solve ∆ABC if <B = 120 ° , b = 15, c = 11. 2. Solve ∆ABC if <B = 84 ° , a = 18, b = 9. 4. Solve ∆QRS if <S =70 ° , r = 26, s = 25. Area of an Oblique Triangle K = Area of ∆ABC 1 K = bh 2 1 K = ab sin C 2 1 K = bc sin A 2 1 K = ac sin B 2 By right triangle trigonometry we know sin C = h , so h = a sinC a 5. Two sides of a triangle have lengths 12 cm and 8 cm. The angle between them is 60° . Find the area of the triangle. 6. Find the area of a triangular lot having two sides of lengths 9 meters and 15 meters and an included angle of 150°. 8.2 Law of Cosines The law of cosines can be used to find missing information of a triangle given SAS or SSS. Finding the largest angle first is helpful because we can determine if using the Law of Sines would make solving for the other angles easier. c2 = a 2 + b2 − 2ab cos C 1. A triangle has side lengths 6, 9, 14. Find the cosine of all the angles. 2. The lengths of a triangle are 10 cm and 6 cm. The angle between them is 120 ° . Find the remaining side and the cosine of the remaining angles. Heron’s Area Formula Area = s ( s − a )( s − b)( s − c ) s = (a + b + c) / 2 where, 3. Find the area of a triangle with side lengths 3, 5, and 4 meters. 4. Find the area of a triangle with side lengths 5, 6, and 7 feet. 7.1 Using Fundamental Trigonometric Identities Fundamental Trigonometric Identities Reciprocal Identities: Quotient Identities: 1 cscθ 1 cos θ = sec θ 1 tan θ = cot θ 1 sin θ 1 secθ = cosθ 1 cot θ = tan θ cscθ = sin θ = sin θ cos θ cos θ cot θ = sin θ tan θ = Negative Identities: Pythagorean Identities: sin( −θ ) = − sin θ sin 2 θ + cos2 θ = 1 cos( −θ ) = cos θ 1 + tan 2 θ = sec2 θ tan( −θ ) = − tan θ cot( −θ ) = − cot θ 1 + cot 2 θ = csc 2θ Cofunction Identities: sin θ = cos(90° − θ ) tan θ = cot(90° − θ ) csc θ = sec(90° − θ ) cos θ = sin(90° − θ ) cot θ = tan(90° − θ ) sec θ = csc(90° − θ ) Trigonometric Identities – A relationship that is true for all values of the variable for which each side of the equation is defined. (We use them to simplify expressions and prove other identities.) Simplifying a Trigonometric Expression 1. sin A tan A + sin( π 2 − A) 2. csc 2 x(1 − cos 2 x) Factoring Trigonometric Expressions (This will later be used with solving.) 3. cos3 y + cos y sin 2 y 4. cs c 2 θ − 1 5. 5 cos 2 θ + 2 cos θ − 3 Adding Trigonometric Expressions 6. sin θ cos θ + 1 + cos θ sin θ Rewriting a Trigonometric Expression 7. Rewrite 1 so that it is not in fractional form. 1 + sin x Trigonometric Substitution 8. Use substitution x = 2 tan θ , 0 < θ < π 2 , to write 4 + x2 as a trigonometric function of θ . Rewriting a Logarithm 9. Rewrite ln csc θ + ln tan θ as a single logarithm and simplify the result. 7.2 Verifying Trigonometric Identities Guidelines for Verifying Trigonometric Identities 1. Work with one side of the equation at a time. It is often better to work with the more complicated side first. 2. Look for opportunities to factor an expression, add fractions, square a binomial, or create a monomial denominator. 3. Look for opportunities to use the fundamental identities. Note which functions are in the final expression you want. Sines and cosines pair up well, as do secants and tangents, and cosecants and cotangents. 4. If the preceding guidelines do not help, try converting all terms to sines and cosines. 5. Always try something. Even paths that lead to dead ends provide insights. Verifying a Trigonometric Identity 1. 2sec2 x = 1 1 + 1 − sin x 1 + sin x 3. sec x + tan x = cos x 1 − sin x 2. (tan 2 x + 1)(cos 2 x − 1) = − tan 2 x 4. tan 4 x = tan 2 x sec 2 x − tan 2 x 7.3 Solving Trigonometric Equation The goal of solving a trigonometric equation is to get the trigonometric function alone by using algebraic methods that we use to get x alone in an equation. Once the trigonometric function is alone we can use the inverse trigonometric functions to solve for the angle measure. All trigonometric functions are periodic and so there will be infinitely many solutions to the equation. Another solution to the equation can be found by adding the period of the function to your solutions that lie within this period. Answers should be written in radians in terms of π . Solve for the missing angle. 1. tan x − 2 = − tan x 3. sec x sin x = 2 sin x 2. 4 cos 2 x − 3 = 0 4. cos 2 x − 3sin x = 3 5. 2 sec 2 x + tan x = 5 2 7. tan ( 2 x ) = 3 6. θ 2 cos − 1 = 0 2 7.4 Sum and Difference Formulas Sum and Difference Formulas sin(u + v ) = sin u cos v + cos u sin v sin(u − v) = sin u cos v − cos u sin v cos(u + v) = cos u cos v − sin u sin v cos(u − v) = cos u cos v + sin u sin v tan u + tan v tan(u + v) = 1 − tan u tan v tan u − tan v tan(u − v ) = 1 + tan u tan v Evaluating a Trigonometric Function 1. Evaluate cos 285° . 2. Find the exact value of sin 5π . 12 3. Prove that the given equation is an identity. π sin − x = cos x . 2 4. Simplify. tan(θ + 3π ) 5. Find all solutions of cos( x + 6. Suppose that sin u = 7.Verify that π π ) + cos( x − ) = −1, 0 ≤ x < 2π 4 4 3 15 π and sin v = , where 0 < u < < v < π . Find sin(u + v ) and cos(u − v ) . 5 17 2 sin( x + h ) − sin x sinh 1 − cosh = (cos x) − (sin x ) , where h ≠ 0 h h h 8.Write cos(arctan1 + arccos x ) as an algebraic expression. 7.5 Multiple Angle Formula Double Angle Formulas Half Angle Formulas sin 2u = 2sin u cos u 2 2 cos 2u = cos u − sin u = 1 − 2sin 2 u sin u 1 − cos u =± 2 2 cos u 1 + cos u =± 2 2 = 2 cos2 u − 1 tan 2u = 2 tan u 2 1 − tan u tan u 1 − cos u =± 2 1 + cos u = sin u 1 + cos u = 1 − cos u sin u Power Reducing Formulas 1 − cos 2u 2 1 + cos 2u cos 2 u = 2 1 − cos 2u tan 2 u = 1 + cos 2u sin 2 u = 1. Use the double angle formula to rewrite the equation. y = 6 cos 2 x − 3 2. If sin u = 3 , 5 π 2 < u < π , find sin 2u, cos 2u, tan 2u,sin u 2 3. Derive the triple angle formula. sin 3x Solving Trigonometric Equations 4. 2 cos x + sin 2 x = 0, 0 ≤ x < 2π 5. cos 2 x = 1 − sin x for 0 ≤ x < 2π . 6. 3cos 2 x + cos x = 2 for 0 ≤ x < 2π . 2 2 7. 2 − 2sin x = 2cos Power Reducing Formulas 8. Rewrite sin 4 x as a sum of first powers of the cosines of multiple angles. x , 0 ≤ x < 2π 2 Trig Identities Practice Simplify. 1. 3. 1 1 − sec t − tan t sec t + tan t 2. cos 2 y + cos 2 y sin 2 y sec x cot x − cot x cos x cos2 x − 1 4. cos2 x tan 2 x Verify. sin x 1 − cos x 5. (csc θ + cot θ )(csc θ − cot θ ) = 1 6. csc x + cot x = 7. sin x + cot x cos x = csc x sec 2 x − 1 = sin 2 x 8. 2 sec x Things to Memorize for This Exam Basic Sine Curve y= sinx (In Degrees, radians, or decimal radians) Basic Cosine Curve y=cosx (In Degrees, radians, or decimal radians) amplitude = |a| Period = To find Phase Shift for sine and cosine: 2π . b bx – c = 0 to bx – c =2 π To find two consecutive vertical asymptotes for tangent: bx − c = − π 2 and bx − c = π 2 To find two consecutive vertical asymptotes for cotangent: bx − c = 0 and bx − c = π For cosecant graphs; vertical asymptotes are where sinx is zero. For secant graphs; vertical asymptotes are where cosx is zero. Inverse Sine Function y = arcsin x ( y = sin −1 x ) if and only if sin y = x , Where −1 ≤ x ≤ 1 and − π 2 ≤y≤ π 2 . Inverse Cosine Function y = arccos x ( y = cos −1 x ) if and only if cos y = x , Where −1 ≤ x ≤ 1 and 0 ≤ y ≤ π . Inverse Tangent Function y = arctan x ( y = tan −1 x ) if and only if tan y = x , Where −∞ ≤ x ≤ ∞ and − π 2 ≤ y≤ π 2 . Inverse Properties of Trigonometric Functions If −1 ≤ x ≤ 1 and − π 2 ≤y≤ π 2 , then sin(arcsinx) = x and arccsin(siny)=y. If −1 ≤ x ≤ 1 and 0 ≤ y ≤ π , then cos(arccosx)=x and arccos(cosy)=y. If x is a real number and − π 2 ≤ y≤ π 2 , then tan(arctanx)=x and arctan(tany)=y. Fundamental Trigonometric Identities Reciprocal Identities: 1 sin θ 1 secθ = cosθ 1 cot θ = tan θ cscθ = Quotient Identities: 1 cscθ 1 cos θ = sec θ 1 tan θ = cot θ sin θ = sin θ cos θ cos θ cot θ = sin θ tan θ = Pythagorean Identities: Co-function Identities: sin 2 θ + cos2 θ = 1 sin θ = cos(90° − θ ) cos θ = sin(90° − θ ) 1 + tan 2 θ = sec2 θ tan θ = cot(90° − θ ) cot θ = tan(90° − θ ) 1 + cot 2 θ = csc 2θ csc θ = sec(90° − θ ) sec θ = csc(90° − θ ) Sum and Difference Formulas sin(u + v ) = sin u cos v + cos u sin v sin(u − v) = sin u cos v − cos u sin v cos(u + v) = cos u cos v − sin u sin v cos(u − v) = cos u cos v + sin u sin v tan u + tan v tan(u + v) = 1 − tan u tan v tan u − tan v tan(u − v ) = 1 + tan u tan v Double Angle Formulas Half Angle Formulas sin 2u = 2sin u cos u 2 2 cos 2u = cos u − sin u = 1 − 2sin 2 u sin u 1 − cos u =± 2 2 cos u 1 + cos u =± 2 2 = 2 cos2 u − 1 * You also need to have memorized everything that we needed to memorize for previous exams!