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Chapter 3: Trigonometric Identities Chapter 3 Overview Two important algebraic aspects of trigonometric functions must be considered: how functions interact with each other and how they interact with their arguments ⎛π ⎞ π (the expression inside the function – e.g., ( x −1) in sin ⎜ ( x −1)⎟ ). 4 ⎝4 ⎠ First, trigonometric functions are inextricably linked because of how they were defined––via the coordinates of points on the terminal side of an angle. Since the coordinates are related by the Pythagorean theorem, there are some relationships that will be true regardless of the angle. These are referred to as identities and will act as formulas. THESE MUST BE MEMORIZED. Second, since the trigonometric function is the operation and no multiplication is present, rules on how the functions work with their arguments will be provided. Without multiplication, the associative, commutative, and distributive rules do not hold. In particular, sin ( a + b) ≠ sin a + sin b and sin2x ≠ 2sin x . New rules for the “algebra” of trigonometric operations will also be developed. 95 3-1: Quotient, Reciprocal, and Pythagorean Identities REMEMBER: sinθ = y r cosθ = x r tanθ = y x cscθ = r y secθ = r x cot θ = x y Based on these definitions, here are the basic identities. The Reciprocal Identities: cscθ = 1 sinθ secθ = 1 cosθ cotθ = 1 tanθ The Quotient Identities: tanθ = sinθ cosθ cotθ = cosθ sinθ tanθ = secθ cscθ cot θ = cscθ secθ The Pythagorean Identities*: tan 2 θ + 1 = sec2 θ sin 2 θ + cos2 θ = 1 *Recall: sin 2 θ + cos 2 θ = ( sin θ ) + ( cosθ ) 2 2 96 1+ cot 2 θ = csc2 θ The Pythagorean identities have alternative versions as well: sin 2 θ + cos2 θ = 1 1− cos2 θ = sin 2 θ 1− sin 2 θ = cos2 θ tan 2 θ +1= sec2 θ sec2 θ −1= tan 2 θ sec2 θ − tan 2 θ = 1 1+ cot 2 θ = csc2 θ csc2 θ −1= cot 2 θ csc2 θ − cot 2 θ = 1 These alternative forms are very useful because they are “difference of squares” binomials that can be factored. For example, 1− sin 2 θ = (1− sinθ )(1+ sinθ ) LEARNING OUTCOME Prove basic trigonometric identities. The proofs of the identities are algebraic in nature, meaning that the use of multiplication, addition, and common denominators will cause one side of the equation to simplify to the other. Similar to proofs in Geometry where one can work both top down and bottom up, in these proofs the two sides can simplify to the same thing. EX 1 Prove csc x tan x cos x = 1 csc x tan x cos x = 1 sin x i i cos x = sin x cos x 1 Notice that the answer is the process, not the final line; the final line was given. 97 EX 2 Prove cot xsec x + csc x = 2csc x This one can be done quickly if the substitution cot x = csc x is used. sec x This would give: csc x sec x + csc x sec x = csc x + csc x = 2csc x cot xsec x + csc x = Another, longer way, would be to turn the whole problem into sin x and cos x . cos x 1 1 + sin x cos x sin x 1 1 = + sin x sin x 2 = sin x = 2csc x cot xsec x + csc x = EX 3 Prove 1 1 + = 2csc A csc A+ cot A csc A− cot A ⎛ csc A− cot A ⎞ ⎛ csc A+ cot A ⎞ 1 1 1 1 + = +⎜ ⎜ ⎟ ⎟ csc A+ cot A csc A− cot A csc A+ cot A ⎝ csc A− cot A ⎠ ⎝ csc A+ cot A ⎠ csc A− cot A csc A− cot A+ csc A+ cot A = (csc A+ cot A)(csc A− cot A) 2csc A A− cot 2 A 2csc A = 1 = 2csc A = csc2 98 3-1 Free Response Homework Prove the following identities. 1. cos2 x + tan 2 xcos2 x = 1 2. 2cosθ = 3. 4 + ( tan σ − cot σ ) = sec2 σ + csc2 σ 4. (cosθ )(cosθ ) − (−sinθ )(sinθ ) = sec2 θ 5. (sin x )(−sin x ) − (cos x )(cos x ) = −csc2 x 6. − 7. − 8. (cos a + sin a)(cos a + sin a) − 1 = sin a 9. tanW (cotW cosW + sinW ) = secW 10. sec x sin x − = cot x sin x cos x 11. sin x 1− cos x + = 2csc x 1− cos x sin x 12. cos2 θ = 13. tan b + cot b = 14. cos x = cot xcsc x + cot 2 x 1− cos x 15. 1+ sin x = 2sec2 x + 2sec x tan x −1 1− sin x cosθ tanθ + sinθ tanθ 2 cos2 θ sin 2 x (−sinθ ) = secθ tanθ cos2 θ cosϕ = −csc ϕ cot ϕ sin 2 ϕ 2cos a cscb cosb 99 cotθ cscθ tanθ − sinθ cscθ 16. State some strategies used for approaching the previous identities. 3-1 Multiple Choice Homework 1. (a) (b) (c) (d) (e) 2. (a) (b) (c) (d) (e) 3. Where defined, sin x − sin x csc x − csc x − cos x If the ratio of sin x to cos x is 1:2, then the ratio of tan x to cot x is 1:4 1:2 1:1 2:1 4:1 If ( sec x ) ( tan x ) < 0 , which of the following must be true? I. tan x < 0 (a) (b) (c) (d) (e) 4. csc x − 1 = sin x − 1 II. csc x cot x < 0 III. x is in the third or fourth quadrant I only II only III only II and III I and II Simplify (a) sin∂ 1 ( tan∂+ cot ∂) csc∂ (b) sec∂ (c) tan∂ (d) 1 100 (e) cos∂ 5. If 90° < α <180° and 180° < β < 270° , then which of the following must be FALSE? (a) (b) (c) (d) (e) 6. (a) (b) (c) (d) (e) tan α = tan β cosα = cos β sin α = tan β tan α = sin β sec β = secα To which of the following is (sin y − cos y) cos y sec y sec y − 2sin y sec y − 2 tan y −2sin y −2 tan y 101 2 equal? 3-2: Trigonometric Identities and Factoring Hopefully the answer to #16 in the previous homework looked something like this: Techniques for Approaching Trigonometric Proofs: 1. 2. 3. Get down to two or less trig functions. These might be Sin and Cos, but be aware of how Sec and Tan, or Csc and Cot, work together. Do Algebra: a. Common Denominators b. Adding “like terms” c. Distribute and/or FOIL Squares indicate possible Pythagorean identities Another technique under the “Do Algebra” heading would be Factoring. LEARNING OUTCOME Prove trigonometric identities that involve factoring. It might be best to go back and review factoring at this point. There are three basic modes of factoring: 1. 2. 3. Factoring by rules Factoring by guess and check Factoring by splitting the middle term I. Factoring by Rules The first technique for factoring is called monomial factoring. It is the most common and most often forgotten. Monomial factoring is the reverse of distribution, and involves pulling out factors common to each term. 102 Distribution and Monomial Factoring a ( b ± c ) = ab ± ac ab ± ac = a ( b ± c ) EX 1 Factor x 2 y + xy 2 x 2 y + xy 2 = xy ( x + y ) What would this look like in a trigonometric identity proof? EX 2 Use factoring to prove cos2 x + tan 2 xcos2 x = 1 cos2 x + tan 2 xcos2 x = cos2 x 1+ tan 2 x ( ) cos2 x i sec2 x 1 cos2 x i cos2 x 1 Factoring Rules for the Sum or Difference of Equal Powers: 1.* 2. x 2 − y 2 = ( x + y )( x − y ) x 3 + y 3 = ( x + y )( x 2 − xy + y 2 ) 3.** x 3 − y 3 = ( x − y )( x 2 + xy + y 2 ) 4.** x 4 − y 4 = ( x 2 + y 2 )( x 2 − y 2 ) *NB. x 2 + y 2 does not factor when working with real numbers. The rule for x 2 + y 2 involves imaginary numbers, which are not of interested at this point. **The trinomials in rules 4 and 5 are not factorable in the real numbers either. 103 EX 3 Simplify u3 − v 3 (u − v ) 3 u3 − v 3 3 = u − v ( ) (u − v )(u2 + uv + v 2 ) = (u − v )(u − v )(u − v ) u 2 + uv + v 2 (u − v )(u − v ) EX 4 Prove cot 4 w − csc 4 w = 1− 2 csc 2 w cot 4 w − csc4 w = (cot 2 w − csc2 w)(cot 2 w + csc2 w) = ( −1)(cot 2 w + csc2 w) = ( ) − csc 2 w −1+ csc 2 w = 1− 2 csc 2 w EX 5 Prove cos 3 x − sin 3 x = 1− cos xsin x cos x − sin x cos 3 x − sin 3 x = cos x − sin x ( cos x − sin x ) cos2 x + cos xsin x + sin2 x = cos x − sin x cos2 x + cos xsin x + sin 2 x = 1− cos xsin x ( ) 104 II. Factoring by “Guess and Check.” The second technique is trinomial factoring, and is a guess and check process. It involves finding numbers that are factors of the first and third terms of a trinomial and checking the FOILing. EX 6 Factor y 2 − 4 y − 5 If y 2 − 4 y − 5 is factorable, the first terms must multiply to y 2 . So, y2 − 4 y − 5 = ( y − )( y + ) The two numbers must multiply to the last number (–5) and add to the middle number (–4). In this case, the numbers must be –5 and +1. So, y 2 − 4 y − 5 = ( y −5)( y +1) EX 7 Factor 4x 2 −10x − 6 This problem is more complicated than the last because there is more than one possibility for the factors of the first and the third terms. 4x 2 −10x − 6 = ( 4x )( x ) or 4x 2 −10x − 6 = ( 2x )( 2x ) The second numbers might be 2 and –3, –2 and 3, 1 and –6, or –1 and 6. The guess and check process takes some time, but ultimately yields 4x 2 − 10x − 6 = ( 4x + 2 ) ( x − 3) = 2 ( 2x + 1) ( x − 3) 105 III. Factoring by splitting the middle term and/or by grouping. EX 7 Factor 4x 2 −10x − 6 (EX 4 again) In order to factor by grouping, –10 needs to be split into two terms. How? The proper split will be two factors of the product of first and last coefficients and that add to –10. 4 times –6 is –24. The factors of 24 are ±1 and ±2 and ±3 and ±4 and 24 12 8 6 The pair that adds to –10 are 2 and −12 . So, 4x 2 −10x − 6 = 4x 2 + 2x −12x − 6 = 2x ( 2x +1) − 6 ( 2x +1) = ( 2x − 6)( 2x +1) = 2 ( x − 3)( 2x +1) EX 8 Prove 3sin 2 θ − 8sinθ + 4 3sin 2 θ − 5sinθ − 2 3sinθ − 2 ÷ = 9sin 2 θ − 4 9sin 2 θ − 3sinθ − 2 3sinθ + 2 3sin 2 θ − 8sinθ + 4 3sin 2 θ − 5sinθ − 2 3sin 2 θ − 8sinθ + 4 9sin 2 θ − 3sinθ − 2 ÷ = ⋅ 9sin 2 θ − 4 9sin 2 θ − 3sinθ − 2 9sin 2 θ − 4 3sin 2 θ − 5sinθ − 2 = (3sinθ − 2)(sinθ − 2) ⋅ (3sinθ +1)(3sinθ − 2) (3sinθ − 2)(3sinθ + 2) (3sinθ +1)(sinθ − 2) = 3sinθ − 2 3sinθ + 2 106 EX 9 Prove 5sec2 θ −14tanθ + 3 5tanθ − 4 = 5sec2 θ − 8tanθ − 9 5tanθ + 2 5sec2 θ −14tanθ + 3 = 5sec2 θ − 8tanθ − 9 5( tan 2 θ +1) −14tanθ + 3 = 5( tan 2 θ +1) − 8tanθ − 9 5tan 2 θ + 5−14tanθ + 3 = 5tan 2 θ + 5− 8tanθ − 9 5tan 2 θ −14tanθ + 8 = 5tan 2 θ − 8tanθ − 4 (5tanθ − 4)( tanθ − 2) = (5tanθ + 2)( tanθ − 2) 5tanθ − 4 5tanθ + 2 107 3-2 Free Response Homework Factor and simplify the following expressions. 1. x 2 − 6xy + 8y2 2. 3x 3 y +10x 2 y2 − 8xy 3 3. 3x − 5x 2 − 2x 3 4. x 3 − x 2 y + xy2 − y 3 5. x 4 − 3x 2 + 2 6. k4 − r4 2 k2 + r2 (k + r) 7. x3 + x2 + x + 1 x4 − 1 8. Prove the following identities. 9. sin 2 x + cot 2 xsin 2 x = 1 10. sec 4 β − tan 4 β = 1+ 2 tan 2 β 11. cos2 w + 2cosw +1 cosw +1 = cos2 w − 3cosw − 4 cosw − 4 12. cos3 A− sec3 A = −sin 2 A+ tan 2 A + 3 cos A− sec A 13. tan 2 w − secw − 5 secw − 3 = tan 2 w + 3secw + 3 secw +1 14. 2sin 2 w + 5cosw +1 cosw − 3 = 6sin 2 w + 5cosw − 2 3cosw − 4 15. 3sec2 t − 8 tant +1 3tant − 2 = tant +1 sec2 t − tant − 3 108 ( ) x 3 + y3 ( x + y )3 16. 2csc2 y − 7cot y − 6 cot y − 4 = 2 6csc y − 5cot y −10 3cot y − 4 Simplify the following expressions. 17. x 2 + 5x x3 x ÷ i 2 x + 6x + 5 3x + 3 x +1 19. 54 − 2x 6 ÷ 2 4 x − 81 x + 9 21. x 2 + 3wx x 2 − 4wx + 3w 2 x + 3w i ÷ 3w − x x+w w2 − x 2 22. x 2 + 4xy + 3y 2 x+y −1 i x + 2y ÷ ( ) x 2 + 5xy + 6y 2 x 2 + 4xy + 4y 2 3 18. x 4 −16 1 2 ÷ ( x + 2) x2 + 4 20. (r (r ( 2 2 ) − rx ) + rx 2 2 ) −1 x + r r3 + x3 ÷ ÷ x − r r3 − x3 Prove the following identities. 23. cot B + 1 1+ tan3 B = cot B − tan B + sec2 B 25. (8 + cos x ) i (cos ( cos x − 2 ) ( cos 26. (sec x + 3sec x −10) i (sec x + 2sec x − 3) = sec x + 5 (sec x − 7sec x + 6) (sec x + sec x − 6) sec x − 6 27. cot 2 x − 4csc x − 11 csc x − 6 = csc x − 2 cot 2 x − 3 28. csc6 x − cot 6 x = 1+ 3csc2 xcot 2 x 3 2 2 2 2 24. ) = cos x − 2cos x + 4 ) x − 4 cos x + 4 2 2 109 2 sin 2 B − sin B + 1 1+ sin3 B = 1− sin B 1− sin 2 B x−4 ( 29. − 9 + 3sin x + sin 2 x 54 − 2sin 3 x 6 ÷ = sin 4 x − 81 sin 2 x + 9 3(sin x + 3) 30. (sin (sin 2 2 ) x − sin x cos x ) x + sin x cos x 2 2 ÷ sin x + cos x cos 3 x + sin 3 x sin 2 x + sin x cos x + cos2 x ÷ = − sin x − cos x cos 3 x − sin 3 x sin 2 x − sin x cos x + cos2 x 3-2 Multiple Choice Homework 1. (a) (b) (c) (d) (e) 2. (a) (b) (c) (d) (e) ) 1 ⎞ ⎛ 1 cot x ⎜ + = ⎝ sec x csc x ⎟⎠ cot x sec x csc x cot x sec x + csc x 2cot x sec x + csc x cot xsec x + cot x csc x sec x csc x 1 cot xsec x + cot x csc x Divide x 3 − 2x 2 + 3x − 18 by x − 3 . x2 + x + 6 x2 − x − 6 x 2 − 5x + 18 x 2 + 5x − 18 x 2 − 5x − 18 110 3. ( ) ( ) 6 tan 2 x −1 2 tan 2 x + 2 tan x +1 ÷ Find tan x + 4 tan 2 x + 4 tan x 2 tan x ( tan x −1) tan x +1 (b) 3tan x ( tan x −1) (a) (c) ( )( tan x ( tan x + 2 ) 3tan x − 3 (d) tan x +1 3tan x tan x − 1 (e) tan x + 1 ( 4. ) 12 tan 2 x −1 tan 2 x + 2 tan x +1 2 ) Simplify cos2 x − 3cos x 2cos x ÷ 2 cos x − 5cos x + 6 ( cos x − 2 )2 cos x ( cos x − 3) 2 cos x (b) 2 cos x − 2 (c) 2 cos x ( cos x − 3) (d) cos x − 2 (e) cos x − 2 (a) 5. (a) (b) 5x 3 y + 20x 2 y 2 + 20xy 3 Simplify 5xy ( x + 2) ( x + 2 y) 2 2 (c) x 2 + y 2 (d) x 2 + 4 y 2 (e) x 2 − 4xy + 4 y 2 111 3-3: Composite Argument and Even/Odd Rules As stated in the chapter overview, sin ( a + b) ≠ sin a + sin b . This is easily demonstrated with a pair of the special angles. If a = 30° and b = 60° , then 1 3 1+ 3 = . Clearly these sin (30° + 60°) = sin90° = 1 . But sin30° + sin60° = + 2 2 2 are not equal. So how does a trigonometric operation apply to its argument when the argument is a composite (i.e. ( a + b) )? The Composite Argument Identities: cos ( A− B ) = cos Acos B + sin Asin B cos ( A+ B ) = cos Acos B − sin Asin B sin ( A− B ) = sin Acos B − cos Asin B sin ( A+ B ) = sin Acos B + cos Asin B tan A− tan B 1+ tan Atan B tan A+ tan B tan ( A+ B ) = 1− tan Atan B tan ( A− B ) = 112 Some functions are called “odd” functions because they act like odd powers––they preserve negative signs: ( −x ) = − ( x ) . Others are called “even” functions because 3 3 they act like even powers and cancel negative signs: ( −x ) = ( x ) . 2 2 Odd Functions: Even Functions: sin ( − A) = − sin A cos ( − A) = cos A csc ( − A) = − csc A sec ( − A) = sec A tan ( − A) = − tan A cot ( − A) = − cot A LEARNING OUTCOMES Find exact trigonometric values for composite arguments. Solve equations involving composite argument rules. Prove identities involving composite rules. There are basically three tasks that can be completed using these rules: 1. 2. 3. Find exact values, Solve equations, and Prove other identities. 113 5 in Quad II, and (6, − 8) is on the terminal side of α , find 13 the exact value of sin (α + β ) . EX 1 Given sin β = 5 in Quad II, then by the Pythagorean identity sin 2 θ + cos2 θ = 1 , 13 12 cos β = − . 13 If sin β = If (6, − 8) is on the terminal side of α , then sin α = − cosα = 6 3 = . 10 5 sin (α + β ) = sin α cos β + cosα sin β ⎛ 4 ⎞ ⎛ −12 ⎞ ⎛ 3⎞ ⎛ 5 ⎞ = ⎜− ⎟⎜ ⎟+ ⎜ ⎟ ⎝ 5 ⎠ ⎝ 13 ⎠ ⎜⎝ 5 ⎟⎠ ⎝ 13⎠ 63 = 65 114 8 4 = − and 10 5 π π 3 EX 2 Solve for x ∈{All Reals} if cos xcos − sin xsin = . 3 3 2 π π cos xcos − sin xsin = 3 3 ⎛ π⎞ cos ⎜ x + ⎟ = 3⎠ ⎝ 3 2 3 2 π 3 = cos –1 3 2 ⎧ π ⎪ ± 2π n π ⎪ 6 x+ = ⎨ 3 ⎪ π − ± 2π n ⎪⎩ 6 x+ ⎧ π ⎪ − ± 2π n ⎪ x=⎨ 6 ⎪ − π ± 2π n ⎪ 2 ⎩ The final answer here is known as the general solution. It is a condensed way to express and infinite number of coterminal answers. ⎧ π ⎪ − ± 2π n ⎪ x=⎨ 6 ⎪ − π ± 2π n ⎪⎩ 2 ⎫ ⎧ 25π 13π π ⎪ ⎪ ... − ,− ,− , ⎪ ⎪ 6 6 6 ⎬= ⎨ 9π 5π π ⎪ ⎪ ... − , − ,− , ⎪⎭ ⎪⎩ 2 2 2 11π 23π , ... 6 6 3π 7π , ... 2 2 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪⎭ Vocabulary: 1. General Solution – all (the infinite number of) coterminal solutions 2. Particular Solution – the specific solutions in a given domain 115 Sometimes the request is for the particular solution. π π 1 EX 3 Solve for x ∈⎡⎣ 0, 2π ⎤⎦ if sin xcos − cos xsin = . 3 3 2 π π 1 sin xcos − cos xsin = 3 3 2 ⎛ π⎞ 1 sin ⎜ x − ⎟ = 3⎠ 2 ⎝ x− ⎛ 1⎞ π = sin –1 ⎜ ⎟ 3 ⎝ 2⎠ ⎧ ⎪ π ⎪ x− = ⎨ 3 ⎪ ⎪⎩ ⎧ ⎪ ⎪ x=⎨ ⎪ ⎪⎩ π ± 2π n 6 5π ± 2π n 6 π ± 2π n 2 7π ± 2π n 6 Of all the answers summarized by the general solution above, only two fall within the domain x ∈⎡⎣ 0, 2π ⎤⎦ : ⎧ π 7π ⎫ x=⎨ , ⎬ ⎩2 6 ⎭ 116 EX 4 Prove cos ( x + 30° ) − sin ( x + 60° ) = − sin x cos ( x + 30°) − sin ( x + 60°) = cos xcos30° − sin xsin30° − (sin xcos60° + cos xsin60°) 3 1 1 3 cos x − sin x − sin x − cos x 2 2 2 2 = −sin x = 117 3-3 Free Response Homework 1. Given tan α = 15 in Quad III, and (3, − 5) is on the terminal side of µ , find 8 the exact values of a) b) c) sin ( µ − α ) d) cos ( µ + α ) e) tan (α − µ ) f) sec (α − µ ) csc (α + µ ) cot (α + µ ) 5 in Quad III, and (12, − 5) is on the terminal side of ω , 4 find the exact values of 2. Given sec λ = − a) b) c) sin ( λ + ω ) d) cos ( λ − ω ) e) tan (ω + λ ) f) sec (ω + λ ) csc (ω − λ ) cot ( λ − ω ) Find the general solutions to the following equations. 1 for x ∈{All Reals} 2 3. sin3θ cos12° − cos3θ sin12° = 4. sin xsin60° − cos xcos60° = 5. sin φ cos15° = cos φ sin15° for x ∈{All Reals} 6. tan3x + tan x = 1− tan3x tan x for x ∈⎡⎣0, 2π ⎤⎦ 7. ⎛ ⎛ π⎞ π⎞ sec ⎜ x − ⎟ = 2 + 2sec ⎜ x − ⎟ for x ∈⎡⎣ −π , π ⎤⎦ 4⎠ 4⎠ ⎝ ⎝ 8. (cos Acos B − sin Asin B) + (sin Acos B + cos Asin B) 1 for x ∈⎡⎣0°, 360° ⎤⎦ 2 2 118 2 = 1 for x ∈{All Reals} Prove the following identities. 9. sin ( A+ B ) + sin ( A− B ) = tan Acot B sin ( A+ B ) − sin ( A− B ) 10. cos ( a + b) cosb + sin ( a + b) sinb = cosa 11. sin Acos A+ cos Asin A 2tan A = cos Acos A− sin Asin A 1− tan 2 A 12. sin Acos B + cos Asin B tan A+ tan B = cos Acos B − sin Asin B 1− tan Atan B 13. cot Acot B −1 = cot ( A+ B ) cot A+ cot B 3-3 Multiple Choice Homework 1. Which of the following is equivalent to sin ( A + 30°) + cos ( A + 60°) for what values of A? (a) sin A (b) cos A (c) 3sin A + cos A (d) 3sin A (e) 3 cos A 2. Which of the following is equivalent to sin (α + β ) + sin (α − β ) ? (a) sin2α (b) sin α 2 − β 2 ( (c) 2sin α sin β (d) 2sin α cos β (e) 2cosα sin β ) 119 3. (a) (b) (c) (d) (e) 4. (a) (b) (c) (d) (e) 5. 3 1 If sin A = , 90° ≤ A ≤180°, cos B = , and 270° ≤ B ≤ 360°, sin ( A + B ) = 5 3 –0.333 0.733 –0.832 0.954 –0.554 tan ( x1 − x2 ) = tan x1 − tan x2 1− tan x1 tan x2 tan x1 − tan x2 − 1+ tan x1 tan x2 tan x1 + tan x2 1− tan x1 tan x2 tan x1 + tan x2 1+ tan x1 tan x2 None of these cos 7π π 7π π cos − sin sin = 8 8 8 8 (a) –1 (b) 1 (c) 0 1 (d) 2 1 (e) − 2 120 3-4: Double Angle Rules As stated in the chapter overview sin2x ≠ 2sin x , and, as with the composite rules, this is easily demonstrated with a pair of the special angles. If a = 30° , then ⎛ 1⎞ 3 sin ( 2 i 30°) = sin60° = . But 2sin30° = 2 ⎜ ⎟ = 1 . Clearly these are not equal. 2 ⎝ 2⎠ So if sin2x ≠ 2sin x , what does sin2x equal? sin2x = sin ( x + x ) = sin xcos x + cos xsin x = 2sin xcos x The Double Angle Argument Identities: sin ( 2 A) = 2 sin A cos A cos ( 2 A) = cos2 A− sin 2 A = 1− 2sin 2 A = 2cos2 A−1 tan ( 2 A) = 2tan A 1− tan 2 A or tan ( 2 A) = sin ( 2 A) cos ( 2 A) LEARNING OUTCOMES Find exact trigonometric values for double angle trigonometric functions. Solve equations involving double angle rules. Prove identities involving double angle rules. 121 The same three types of problems will be done with the double angle rules as were done with the composite rules. EX 1 Given sin β = tan ( 2β ) . If sin β = 5 in Quad II, find the exact value of sin ( 2β ) , cos ( 2β ) , and 13 5 12 in Quad II, then by the Pythagorean identity, cos β = − . 13 13 sin ( 2β ) = 2sin β cos β ⎛ 5 ⎞ ⎛ 12 ⎞ − ⎝ 13⎟⎠ ⎜⎝ 13 ⎟⎠ = 2⎜ =− 120 169 cos ( 2β ) = cos2 β − sin 2 β 2 ⎛ 12 ⎞ ⎛ 5⎞ = ⎜− ⎟ −⎜ ⎟ ⎝ 13 ⎠ ⎝ 13⎠ 2 144 25 − 169 169 119 = 169 = tan ( 2β ) = sin ( 2β ) cos ( 2β ) 120 = 169 119 169 120 =− 119 − 122 EX 2 Solve for x if 4sin xcos x = 2 . 4sin xcos x = 2 2 2sin xcos x = 2 2 sin ( 2x ) = 2 2x = sin –1 ⎧ ⎪ ⎪ 2x = ⎨ ⎪ ⎪⎩ ⎧ ⎪ ⎪ x=⎨ ⎪ ⎪⎩ EX 3 Prove tan x = 2 2 π ± 2π n 4 3π ± 2π n 4 π ±πn 8 3π ±πn 8 sin 2x 1+ cos2x It would be appropriate in this problem to work both sides of the equation toward the same goal. sin2x 1+ cos2x sin x 2sin xcos x = cos x 1+ 2cos2 x −1 tan x = ( 2sin xcos x 2cos2 x sin x = cos x = 123 ) 3-4 Free Response Homework 1. Given sin A = 1 in Quad II, and ( −7, 24) is on the terminal side of B, find 3 the exact values of a) b) c) sin ( 2 A) d) cos ( 2B ) e) tan ( 2 A) f) sec ( 2B ) csc ( 2 A) cot ( 2B ) Find the general solutions to the following equations. Use exact values wherever possible. 2. 4cosθ sinθ = 2 for θ ∈{All Reals} 3. cos2 x − sin 2 x − 1 = 2cos ( 2x ) for x ∈ ⎡⎣0, 2π ⎤⎦ 4. 2sin ( 2x − 30°) cos ( 2x − 30°) = − 5. csc2x + 6. 2tan x = 1 for x ∈⎡⎣ −2π , 2π ⎤⎦ 1− tan 2 x 7. 1− tan 2 x = 3 for x ∈{All Reals} 2tan x 3 for x ∈{All Reals} 2 1 = 6 for x ∈{All Reals} sin xcos x Prove the following identities. 1− tan 2 B 1+ tan 2 B 8. cos2B = 9. cos2β = cos4 β − sin 4 β 124 10. cot A + tan A = 2csc ( 2 A) 11. cot 2 φ −1= cos2φ cot 2 φ + cos2φ 12. ⎛ 1− tan 2 x cos2 x − sin 2 x ⎞ 2 ⎜ 2tan x + 2sin xcos x ⎟ = 4csc 2x − 4 ⎝ ⎠ 13. δ (1+ tanδ ) tan2δ = 1−2tan tan δ 2 3-4 Multiple Choice Homework 1. Which of the following is NOT equal to (a) cos x − sin x tan x 2sin x (b) sec x − cot x (c) 2cos x − sec x (d) sec x − 2sin 2 x tan x cos2x (e) All are equal to cos x 2. (a) (b) (c) (d) (e) Which expression equals cot 2x ? 2cot x 1+ cot 2 x 2 cot x 1− cot 2 x 2sin x 1− tan 2 x 2 tan x 1− tan 2 x 1 1 cot x − tan x 2 2 125 cos2x ? cos x 3. If cos x = 4 3π ≤ x ≤ 2π , then tan 2x = and 5 2 7 24 24 − 25 3 − 4 24 − 7 7 25 (a) − (b) (c) (d) (e) 4. (a) (b) (c) (d) (e) 5. (a) (b) (c) (d) (e) For what value between 0° and 360° does cos2x = 2 cos x ? 68.5° or 291.5° 68.5° only 103.9° or 256.1° 90° or 270° 111.5° or 248.5° If 3sin 2θ 1 = and 0° ≤ θ ≤180° , then θ = 1− cos2θ 2 0° 0° or 180° 80.5° 0° or 80.5° 99.5° 126 3 1 If sin A = , 90° < A <180°, cos B = , and 270° < B < 360° , the value of 5 3 tan ( A + B ) is 6. (a) (b) (c) (d) (e) 0.30 3.19 3.34 1.05 0.31 127 3-5: Half Angle Rules Just as there are Double Angle Identities, there are half angle formulas. The Half Angle Argument Identities: ⎛1 ⎞ 1 sin ⎜ A⎟ = ± (1− cos A) 2 ⎝2 ⎠ ⎛1 ⎞ 1 cos ⎜ A⎟ = ± (1+ cos A) 2 ⎝2 ⎠ ⎛1 ⎞ 1− cos A tan ⎜ A⎟ = ± 1+ cos A ⎝2 ⎠ ⎛1 ⎞ sin A tan ⎜ A⎟ = ⎝ 2 ⎠ 1+ cos A ⎛ 1 ⎞ 1− cos A tan ⎜ A⎟ = sin A ⎝2 ⎠ Note the ± on three of the formulas. This does really mean “plus OR minus,” not “plus AND minus” as usually meant in mathematics. That is, in these formulas, one OR the other is correct, but not both. Determine the sign from the 1 size and quadrant of A . 2 The same three types of problems will be done with the half angle rules as were done with the double angle and composite rules. LEARNING OUTCOMES Find exact trigonometric values for half angles. Solve equations involving half angle rules. Prove identities involving half angle rules. 128 EX 1 Given sin β = − ⎛1 ⎞ 5 and 540° < β < 630° , find the exact value of sin ⎜ β ⎟ , 13 ⎝2 ⎠ ⎛1 ⎞ ⎛1 ⎞ cos ⎜ β ⎟ , and tan ⎜ β ⎟ . ⎝2 ⎠ ⎝2 ⎠ If sin β = − 5 12 and 540° < β < 630° , then cos β = − . 13 13 1 1 Also, since 540° < β < 630° , then 270° < β < 315° . This means that β is 2 2 in QIV, and the cosine will be positive and the sine negative. ⎛1 ⎞ 1 sin ⎜ β ⎟ = ± (1− cos β ) 2 ⎝2 ⎠ ⎛1 ⎞ 1 cos ⎜ β ⎟ = ± (1+ cos β ) 2 ⎝2 ⎠ =− 1 ⎛ ⎛ 12 ⎞ ⎞ 1− − 2 ⎜⎝ ⎜⎝ 13 ⎟⎠ ⎟⎠ =+ =− 1 ⎛ 25 ⎞ 2 ⎜⎝ 13 ⎟⎠ = 1⎛ 1 ⎞ 2 ⎜⎝ 13⎟⎠ = 1 26 = −5 26 ⎛1 ⎞ sin β tan ⎜ β ⎟ = ⎝ 2 ⎠ 1+ cos β 5 − 13 = ⎛ 12 ⎞ 1+ ⎜ − ⎟ ⎝ 13 ⎠ 5 − = 13 1 13 = −5 129 1 ⎛ ⎛ 12 ⎞ ⎞ 1+ − 2 ⎜⎝ ⎜⎝ 13 ⎟⎠ ⎟⎠ EX 2 Solve for x if sin x = 3. 1− cos x The left side of the equation is not one of the formulas, but it is the reciprocal of one of the formulas. sin x = 3 1− cos x 1− cos x 1 = sin x 3 ⎛1 ⎞ 1 tan ⎜ x ⎟ = ⎝2 ⎠ 3 1 1 x = tan –1 2 3 1 π x = ±πn 2 6 π x = ± 2π n 3 ⎛1 ⎞⎛ ⎛ 1 ⎞⎞ EX 3 Prove tan ⎜ x ⎟ ⎜ 2cot x + tan ⎜ x ⎟ ⎟ = 1 ⎝2 ⎠⎝ ⎝ 2 ⎠⎠ This is an interesting problem because there are two places to use a half angle rule, and the same formula is not needed both times. ⎛1 ⎞⎛ ⎛ 1 ⎞⎞ sin x ⎛ 2cos x 1− cos x ⎞ tan ⎜ x ⎟ ⎜ 2cot x + tan ⎜ x ⎟ ⎟ = + sin x ⎟⎠ ⎝2 ⎠⎝ ⎝ 2 ⎠ ⎠ 1+ cos x ⎜⎝ sin x sin x ⎛ 1+ cos x ⎞ 1+ cos x ⎜⎝ sin x ⎟⎠ =1 = 130 3-5 Free Response Homework Given the following values and quadrant ranges, find the exact values of ⎛1 ⎞ a) sin ⎜ x ⎟ ⎝2 ⎠ ⎛1 ⎞ b) cos ⎜ x ⎟ ⎝2 ⎠ ⎛1 ⎞ c) tan ⎜ x ⎟ ⎝2 ⎠ 4 and 0° < x < 90° 5 1. cos x = 3. cos x = − 2. cos x = − 4 and 90° < x < 180° 5 24 and 180° < x < 270° 25 Given the following values and quadrant ranges, find the exact values of a) ⎛1 ⎞ csc ⎜ x ⎟ ⎝2 ⎠ 4. sin x = − 5. cos x = b) ⎛1 ⎞ sec ⎜ x ⎟ ⎝2 ⎠ c) ⎛1 ⎞ cot ⎜ x ⎟ ⎝2 ⎠ 24 and − 90° < x < 0° 25 4 and 630° < x < 720° 5 Find the general solutions to the following equations. Use exact values wherever possible. 1 1 1− cos x ) = for x ∈{All Reals} ( 2 2 6. 7. − 1 1 1+ cos x ) = for x ∈ ⎡⎣0, 2π ⎤⎦ ( 2 2 8. 9. 1 3 for x ∈ ⎡⎣0, 4π ⎤⎦ 1− cos x ) = − ( 2 2 − 1 (1+ cos x ) = 0 for x ∈{All Reals} 2 131 10. sin x = −1 for x ∈{All Reals} 1+ cos x 11. 1− cos x = − 3 for x ∈ ⎡⎣ −2π , 2π ⎤⎦ sin x Prove the following identities. 12. ⎛1 ⎞ ⎛1 ⎞ tan ⎜ x ⎟ + cot ⎜ x ⎟ = 2csc x ⎝2 ⎠ ⎝2 ⎠ 13. ⎛1 ⎞ tan x tan ⎜ x ⎟ = sec x −1 ⎝2 ⎠ 14. ⎛1 ⎞ 2 tan ⎜ x ⎟ ⎝2 ⎠ = sin x 2⎛1 ⎞ 1+ tan ⎜ x ⎟ ⎝2 ⎠ 15. ⎛1 ⎞ tan ⎜ x ⎟ = csc x − cot x ⎝2 ⎠ 16. ⎛ 1 1 ⎞ sin x + cos x = 1+ sin x ⎜ 2 2 ⎟⎠ ⎝ 17. ⎛1 ⎞ ⎛1 ⎞ cos ⎜ x ⎟ + sin ⎜ x ⎟ ⎝2 ⎠ ⎝2 ⎠ = sec x + tan x ⎛1 ⎞ ⎛1 ⎞ cos ⎜ x ⎟ − sin ⎜ x ⎟ ⎝2 ⎠ ⎝2 ⎠ 18. ⎛1 ⎞ ⎛1 ⎞ cos ⎜ x ⎟ − sin ⎜ x ⎟ ⎝2 ⎠ ⎝2 ⎠ cos x = ⎛1 ⎞ ⎛ 1 ⎞ 1+ sin x cos ⎜ x ⎟ + sin ⎜ x ⎟ ⎝2 ⎠ ⎝2 ⎠ 2 132 3-5 Multiple Choice Homework 1. (a) (b) (c) (d) (e) 2. (a) (b) (c) (d) (e) 3. If 1− cosθ 3 = , then θ = sin θ 3 15° 30° 45° 60° 75° θ θ tan + cot = 2 2 1 cot θ 2 −2cscθ −2cot θ 2cot θ 2cscθ Which of the following is not equivalent to sin 40° ? (a) 1− cos 2 40° (b) 2sin 20°cos20° (c) 1− cos80° 2 1+ cos80° 2 (e) All are equivalent to sin 40° (d) 133 4. If sin θ = − (a) − 20 θ , θ lies in QIV, find cos . 29 2 3 10 10 2 5 3 10 (c) 10 2 (d) − 5 (e) None of the above (b) 5. (a) (b) (c) (d) (e) x 2 = x 1+ tan 2 2 2 tan sin x cos x tan x sin x + 1 2cos x − 1 134 3-6: Solving Equations with Identities and Algebra One of the main reasons for studying the identities in the last several sections is to use them in solving equations. These trigonometric identities will sometimes occur in the midst of a calculus question, and they will often be mixed in with factoring or other algebraic operations. Much of the carryover from Trigonometry to Calculus is in the process of solving equations. Therefore, this process will be addressed specifically. Since so many trigonometric identities are used to solve equations involving trigonometric functions, THEY MUST BE MEMORIZED. The Reciprocal Identities: cscθ = The Quotient Identities: 1 sinθ secθ = sinθ cosθ secθ tanθ = cscθ 1 cosθ cot θ = 1 tanθ cosθ sinθ cscθ cot θ = secθ tanθ = cot θ = The Pythagorean Identities: sin 2 θ + cos2 θ = 1 1− cos2 θ = sin 2 θ 1− sin 2 θ = cos2 θ tan 2 θ +1= sec2 θ sec2 θ −1= tan 2 θ sec2 θ − tan 2 θ = 1 1+ cot 2 θ = csc2 θ csc2 θ −1= cot 2 θ csc2 θ − cot 2 θ = 1 The Composite Argument Identities: cos ( A − B ) = cos A cos B + sin Asin B cos ( A+ B ) = cos Acos B − sin Asin B sin ( A− B ) = sin Acos B − cos Asin B sin ( A+ B ) = sin Acos B + cos Asin B tan A− tan B tan ( A− B ) = 1+ tan Atan B tan A+ tan B tan ( A+ B ) = 1− tan Atan B 135 Even Functions: cos ( −A ) = cos A Odd Functions: sin ( −A ) = − sin A sec ( − A) = sec A csc ( − A) = −csc A tan ( − A) = − tan A cot ( − A) = −cot A sin ( 2A ) = 2sin A cos A The Double Angle Argument Identities: cos ( 2 A) = cos2 A− sin 2 A = 1− 2sin 2 A = 2cos2 A−1 2tan A tan ( 2 A) = 1− tan 2 A 1 ⎛1 ⎞ sin ⎜ A⎟ = ± (1− cos A ) ⎝2 ⎠ 2 ⎛1 ⎞ 1 cos ⎜ A⎟ = ± (1+ cos A) 2 ⎝2 ⎠ The Half Angle Argument Identities: ⎛1 ⎞ 1− cos A tan ⎜ A⎟ = ± 1+ cos A ⎝2 ⎠ sin A = 1+ cos A 1− cos A = sin A LEARNING OUTCOME Solve equations involving the trigonometric identities. 136 EX 1 Solve 4sin 2 x = 3 exactly for x ∈⎡⎣0, 2π ⎤⎦ . This is a fairly straightforward “isolate the variable” problem. Be careful when square rooting both sides of the equation. 4sin 2 x = 3 3 sin 2 x = 4 ± 3 sin x = 2 3 2 π ± 2π n 3 2π ± 2π n 3 3 2 sin x = sin x = − ⎧ ⎪ ⎪ x=⎨ ⎪ ⎪⎩ ⎧ 4π ⎪ ± 2π n ⎪ 3 x=⎨ ⎪ 5π ± 2π n ⎪⎩ 3 Now, choose the particular answers that occur in the interval stated: x= π 2π 4π 5π , , , or 3 3 3 3 137 ) EX 2 Solve sec2 x − sec x = 2 exactly for x ∈⎡⎣0, 2π . sec2 x − sec x = 2 sec2 x − sec x − 2 = 0 (sec x − 2)(sec x + 1) = 0 sec x − 2 = 0 sec x = 2 1 cos x = 2 ⎧ π ⎪ ± 2π n ⎪ x=⎨ 3 ⎪ − π ± 2π n ⎪ 3 ⎩ So, x = sec x +1= 0 sec x = −1 cos x = −1 x = π ± 2π n π 5π , , or π 3 3 ) EX 3 Solve sec2 x − 2sec x = 3 exactly for x ∈⎡⎣0, 2π . sec2 x − 2sec x = 3 sec2 x − 2sec x − 3 = 0 (sec x − 3)(sec x +1) = 0 sec x +1= 0 sec x = −1 cos x = −1 x = π ± 2π n sec x − 3 = 0 sec x = 3 1 cos x = 3 x = ±1.231± 2π n So, x = 1.231, 5.052, or π 138 As with proofs in Geometry or with trigonometric identities, there is no algorithmic approach to solving trigonometric equations. Each problem is different and only experience can really guide the approach to a particular problem. However, there is a helpful strategy sequence. Ask these questions: 1. 2. 3. Are the arguments different? Are the trig functions different? Is there any algebra to perform? Strategies for Solving Trigonometric Equations: I. If the arguments are different, make them the same. Usually use: Composite Rules Double or Half Angle Rules II. If the arguments are the same but the trigonometric functions are different, make them the same, by using: Reciprocal or Quotient Rules Pythagorean Identities 1 tan x formulas 2 Double Angle Argument Formulas III. Do the algebra: Set equation equal to 0 and factor (Remember: Do NOT divide by a variable!!!) Find common denominators (then recheck for Pythagorean identities) IV. Get to a place where there is one trigonometric function to inverse away, thus isolating the variable. As with proofs in Geometry or with trigonometric identities, there is no shortcut to understanding. Most people need a great deal of experience with these before enlightenment occurs. 139 ) EX 4 Solve cos3θ cos12° = sin3θ sin12° exactly for θ ∈⎡⎣0°, 360° . The arguments in this equation are different from one another, so they must be made the same. Since there are no triple angle rules, there must be something else. Four trigonometric functions in these pairs look like a composite rule. Do some minor rearranging to see it more clearly. cos3θ cos12° = sin3θ sin12° cos3θ cos12° − sin3θ sin12° = 0 cos (3θ + 12°) = 0 ⎪⎧ 90° ± 360n° ⎩⎪ −90° ± 360n° 3θ + 12° = ⎨ ⎧⎪ 78° ± 360n° ⎪⎩ −102° ± 360n° 3θ = ⎨ ⎧⎪ 26° ±120n° ⎩⎪ −34° ±120n° θ =⎨ θ = 26°, 86°, 146°, 206°, 266° or 326° ) EX 5 Solve sin 2 x + cos x − 1 = 0 exactly for x ∈⎡⎣0°, 360° . This cannot be factored like EX 2 because the trigonometric functions are different. But the Pythagorean identities can be used to make them the same. sin 2 x + cos x − 1 = 0 1− cos2 x + cos x − 1 = 0 −cos2 x + cos x = 0 cos x ( −cos x + 1) = 0 cos x = 0 x = ±90 ± 360n° −cos x + 1 = 0 cos x = 1 x = 0° ± 360n° x = 0°, 90° or 270° 140 ) The interval will not always be a standard x ∈⎡⎣0, 2π . EX 6 Solve cos2 x − sin 2 x = ) 1 exactly for x ∈⎡⎣ −π , π . 2 cos2 x − sin 2 x = cos2x = 1 2 1 2 π ± 2π n 4 π x = ± ±πn 8 2x = ± x= π 7π π 7π ,− , − , or 8 8 8 8 141 3-6 Free Response Homework Solve the following equations in the given interval. Use exact values when possible. 3 in θ ∈⎡⎣0°, 360° ⎤⎦ 2 1. cos2 2θ − sin 2 2θ = 2. 3− 3sin x − 2cos2 x = 0 in x ∈⎡⎣ −2π , 2π ⎤⎦ 3. sin3θ cos12° − cos3θ sin12° = 4. ⎛ ⎛ π⎞ π⎞ sec ⎜ x − ⎟ = 2 + 2sec ⎜ x − ⎟ in x ∈(0, 2π ⎤⎦ 4⎠ 4⎠ ⎝ ⎝ 5. ⎛1 ⎞ 2cos2 ⎜ x ⎟ − 2 = 3cos x in x ∈( −π , π ) ⎝2 ⎠ 6. cos4θ − cos2θ = 0 in θ ∈⎡⎣0°, 360° ⎤⎦ 7. 4sin xcos x = 3 in x ∈⎡⎣0, 2π ⎤⎦ 8. sin 2 x − cos2 x = 9. cosθ cos20° − sinθ sin 20° = 10. 2cos2 x = 3 + 2sin 2 x in x ∈⎡⎣ −180°, 0° ⎤⎦ 11. tan 4 x − 4 tan 2 x + 3 = 0 in x ∈⎢ 12. 3 3 sin 2 3t − 2sin t cos t − 2 = 0 in t ∈⎡⎣0°, 360° 2 2 1 in θ ∈⎡⎣0°, 360° ⎤⎦ 2 3 in x ∈⎡⎣0, 2π ⎤⎦ 2 1 in θ ∈⎡⎣0°, 360° ⎤⎦ 2 ⎡ −π ⎣ 4 , π⎞ 4 ⎟⎠ 142 ) 13. 3− 3sin x − 2cos2 x = 0 in x ∈{All Reals} 14. 2cot 2 x − 5csc x + 5 = 0 on x ∈⎡⎣ −2π , 2π 15. tan 4 x − sec4 x = −3 in x ∈⎡⎣ −45°, 0° ⎤⎦ 16. tan 2 x − sec2 x − sec x − 2 = 0 in x ∈⎡⎣ −2π , 2π ⎤⎦ 17. sin 2 A − 2sin (3A + π ) = −cos2 A in A ∈ π , 3π 18. 2sin ( 2x − 30°) cos ( 2x − 30°) = − 19. ⎡ π π⎞ sec2 ( x − 4) + sin (3x ) = 1+ tan 2 ( x − 4) in x ∈⎢ − , ⎟ ⎣ 2 2⎠ 20. cos2xsin x + sin2xcos x − 2sin3x = 21. csc2x + 22. 1− cos x = 3 in x ∈ 0, 2π sin x 23. 1− tan 2 x cos2 x − sin 2 x + = 1 in x ∈(0, 2π ) 2tan x 2sin xcos x ) ( ) 3 in x ∈⎡⎣0°, 720° ⎤⎦ 2 1 in x ∈⎡⎣0, 2π 2 ) 1 = 6 in x ∈⎡⎣0, π ⎤⎦ sin xcos x ( ) 24. 1 2 in x ∈⎡⎣0, 2π ⎤⎦ 1− cos x ) = ( 2 2 25. 4tan x + 3 = tan x in x ∈(0, 2π ⎤⎦ 26. ⎛π ⎞ π π sin3acos + cos3asin + tan 2 a = sec2 a in a ∈⎜ , π ⎟ 2 2 ⎝2 ⎠ 143 27. (cos 28. 1 sin x tan x 2 − sin 2 1 x = 0 on x ∈⎡ −2π , 2π ) ⎣ 2 2 29. 2csc2 θ + 5cotθ = 0 in θ ∈⎡⎣ –180°, 180° ⎤⎦ 30. sin 2 θ − cos2 θ = 31. tan3x + tan x = tan3x tan x − 1 in x ∈⎡⎣0, 2π ⎤⎦ 32. 1− cos x = −1 in x ∈{All Reals} sin x 33. sinθ = sin2θ in θ ∈( –90°, 270°) 34. 2tan 2 θ − 3secθ = 0 in θ ∈⎡⎣ –180°, 180° ⎤⎦ 35. ⎛ ⎛ π⎞ π⎞ sec ⎜ 3x − ⎟ = 2 + 2sec ⎜ 3x − ⎟ in x ∈(0, 2π ⎤⎦ 4⎠ 4⎠ ⎝ ⎝ 36. cos2x − sin4x = 0 in x ∈( −π , π ) 37. ⎛1 ⎞ 2cos2 ⎜ x ⎟ − 2 = 2cos x in x ∈( −π , π ) ⎝2 ⎠ 2 ) x − sin 2 x − ( 2sin xcos x ) = 0 in x ∈⎡⎣ −2π , 2π 2 2 1 in θ ∈( –90°, 270°) 2 144 ) 3-6 Multiple Choice Homework 1. (a) (b) (c) (d) (e) 2. (a) (b) (c) (d) (e) If 2 sin 2x = 3cos 2x and 0 ≤ 2x ≤ π , then x = 2 0.25 0.52 0.49 0.39 0.63 If 4 sin x + 3 = 0 on 0 ≤ x ≤ 2π , then x = 5.435 0.848 3.990 or 5.435 0.848 or 5.435 –0.848 3. Solve the equation sin15x + cos15x = 0 . What is the sum of the three smallest positive solutions? (a) (b) (c) (d) (e) π 20 π 3 7π 20 21π 10 21π 4 145 4. (a) (b) (c) (d) (e) For what positive value(s) x ≤ 180° of does tan 2x = 2 cot 2x ? 54.7° 25° and 155° 27.4° and 117.4° 27.4°, 62.6°, 117.4°, and 152.6° None of the above 5. For all positive angles less than 360°, if csc ( 2x + 30° ) = cos ( 3y − 15° ) , the sum of x and y is (a) (b) (c) (d) (e) 185° 5° 30° 215° 95° 146 Trigonometric Identities Practice Group Test––Page 1 Find EXACT values (no decimals). Show all work. For problems #1-6, (3, − 4) is on the terminal side of A and 270° ≤ A ≤ 360°, csc B = 90° ≤ B ≤ 180°, and tanC = 7 and −180° ≤ C ≤ −90° . Find the EXACT value of: 24 1. sin ( A − B ) 2. cos2 A 3. ⎛1 ⎞ tan ⎜ C ⎟ ⎝2 ⎠ 4. csc ( A + B ) 5. tan2B 6. ⎛1 ⎞ cos ⎜ A⎟ ⎝2 ⎠ 147 13 and 5 7. 1 1 1 1 Prove: tan x + cot x = csc xsec x 2 2 2 2 9. Solve EXACTLY: sin 2 2x − 2sin xcos x = 2 for x ∈⎡⎣ −π , π 10. Solve EXACTLY: −cos2 x = sin 2 x − 2sin (3x + π ) for x ∈⎡⎣0, 2π 8. 148 1+ tan 2 x = sec2x Prove: 1− tan 2 x ) ) Trigonometric Identities Practice Group Test––Page 2 Find EXACT values (no decimals). Show all work. For problems #1-6, (3, − 4) is on the terminal side of A and 270° ≤ A ≤ 360°, csc B = 90° ≤ B ≤ 180°, and tanC = 7 and −180° ≤ C ≤ −90° . Find the EXACT value of: 24 1. cos ( A − B ) 2. sin2 A 3. ⎛1 ⎞ cot ⎜ C ⎟ ⎝2 ⎠ 4. sec ( A + B ) 5. cot 2B 6. ⎛1 ⎞ sin ⎜ A⎟ ⎝2 ⎠ 149 13 and 5 7. ⎛1 ⎞ 2 tan ⎜ x ⎟ ⎝2 ⎠ Prove: = sin x 2⎛1 ⎞ 1+ tan ⎜ x ⎟ ⎝2 ⎠ 9. Solve EXACTLY: sin 2 2x − 2sin xcos x = 2 for x ∈⎡⎣ −3π , − π 10. Solve EXACTLY: −cos2 x = sin 2 x − 2sin (3x + π ) for x ∈⎡⎣ −2π , 0 8. 150 Prove: cos2β = ) ) 1− tan 2 β 1+ tan 2 β Trigonometric Identities Practice Group Test––Page 3 Find EXACT values (no decimals). Show all work. For problems #1-6, (3, − 4) is on the terminal side of A and 270° ≤ A ≤ 360°, csc B = 90° ≤ B ≤ 180°, and tanC = 7 and −180° ≤ C ≤ −90° . Find the EXACT value of: 24 1. sin ( A + B ) 2. sec2 A 3. ⎛1 ⎞ tan ⎜ B⎟ ⎝2 ⎠ 4. csc ( A − B ) 5. tan2 A 6. ⎛1 ⎞ cos ⎜ C ⎟ ⎝2 ⎠ 151 13 and 5 7. 1 1 Prove: tan x + cot x = 2csc x 2 2 9. Solve EXACTLY: sin 2 2x − 2sin xcos x = 2 for x ∈⎡⎣ −2π , 0 10. Solve EXACTLY: −cos2 x = sin 2 x − 2sin (3x + π ) for x ∈⎡⎣ −3π , − π 8. 152 Prove: (1+ tan ∂ ) ( tan 2∂ ) = ) ) 2 tan ∂ 1− tan ∂ Trigonometric Identities Practice Group Test––Page 4 Find EXACT values (no decimals). Show all work. For problems #1-6, (3, − 4) is on the terminal side of A and 270° ≤ A ≤ 360°, csc B = 90° ≤ B ≤ 180°, and tanC = 7 and −180° ≤ C ≤ −90° . Find the EXACT value of: 24 1. cos ( A + B ) 2. csc2 A 3. ⎛1 ⎞ cot ⎜ B⎟ ⎝2 ⎠ 4. sec ( A − B ) 5. cot 2 A 6. ⎛1 ⎞ sin ⎜ C ⎟ ⎝2 ⎠ 153 13 and 5 7. Prove: sin A + B + sin A − B 8. Prove: cos ( a + b) cosb + sin(a + b)sin b = cos a ( ) ( ) = tan Acot B sin ( A + B ) − sin ( A − B ) ) 9. Solve EXACTLY: sin 2 2x − 2sin xcos x = 2 for x ∈⎡⎣0, 2π 10. Solve EXACTLY: −cos2 x = sin 2 x − 2sin (3x + π ) for x ∈⎡⎣ −π , π 154 ) Trigonometric Identities Homework Answer Key 3-1 Free Response Homework 1. 2. 3. 4. cos2 x + tan 2 xcos2 x sin 2 x = cos2 x + cos2 x cos2 x = cos2 x + sin 2 x =1 cosθ tanθ + sinθ tanθ cosθ tanθ sinθ = + tanθ tanθ sinθ = cosθ + tanθ sinθ = cosθ + sinθ cosθ cosθ = cosθ + sinθ sinθ = cosθ + cosθ = 2cosθ 4 + ( tan σ − cot σ ) 5. (cosθ )(cosθ ) − (−sinθ )(sinθ ) cos2 θ cos2 θ + sin 2 θ = cos2 θ 1 = cos2 θ = sec2 θ (sin x )(−sin x ) − (cos x )(cos x ) sin 2 x −sin 2 x − cos2 x = sin 2 x − sin 2 x + cos2 x = sin 2 x −1 = 2 sin x = −csc2 x ( 6. 2 = 4 + tan 2 σ − 2 tan σ cot σ + cot 2 σ = 4 + tan 2 σ − 2 (1) + cot 2 σ = 2 + tan 2 σ + cot 2 σ = 2 + sec2 σ − 1+ csc2 σ − 1 = sec2 σ + csc2 σ 7. 155 − (−sinθ ) cos2 θ sinθ 1 = cosθ cosθ = tanθ secθ cosϕ sin 2 ϕ 1 cosϕ =− sin ϕ sin ϕ = −csc ϕ cot ϕ − ) 8. (cos a + sin a)(cos a + sin a) − 1 11. 2cos a cos a + 2sin acos a + sin 2 a −1 = 2cos a 2 2 cos a + sin a + 2sin acos a −1 = 2cos a 1+ 2sin acos a −1 = 2cos a 2sin acos a = 2cos a = sin a 2 9. 10. sin x 1− cos x + 1− cos x sin x = (1− cos x ) (1− cos x ) sin x sin x + sin x (1− cos x ) (1− cos x ) sin x sin 2 x + 1− 2cos x + cos2 x = sin x (1− cos x ) = = tanW (cotW cosW + sinW ) = cosW + tanW sinW sinW = cosW + sinW cosW cos2 W sin 2 W = + cosW cosW 1 = cosW = secW = 1+ 1− 2cos x sin x (1− cos x ) 2 − 2cos x sin x (1− cos x ) 2 (1− cos x ) sin x (1− cos x ) 2 sin x = 2csc x = 12. sec x sin x − sin x cos x cos x sec x sin x sin x = − cos x sin x cos x sin x 1− sin 2 x = cos xsin x cos2 x = cos xsin x cos x = sin x = cot x 13. 156 cot θ cscθ tanθ − sinθ csc θ cscθ − sinθ = cscθ sinθ = 1− cscθ = 1− sin 2 θ = cos2 θ tanb + cot b sin b cosb = + cosb sin b sin b sin b cosb cosb = + sin b cosb sin b cosb 1 = sin bcosb cscb = cosb 14. cot xcsc x + cot 2 x cos x 1 cos2 x = + sin x sin x sin 2 x cos x + cos2 x = 1− cos2 x cos x 1+ cos x = 1− cos x 1+ cos x ( ( = 15. ) )( 3-2 Free Response Homework ) cos x 1− cos x 2sec2 x + 2sec x tan x − 1 2 1 sin x = −1 2 +2 cos x cos x cos x 2 + 2sin x − cos2 x = cos2 x 2 + 2sin x − 1− sin 2 x = 1− sin 2 x 2 + 2sin x − 1− sin 2 x = 1− sin x 1+ sin x ( ( = 16. ( )( 1+ 2sin x + sin 2 x (1− sin x )(1+ sin x ) = (1+ sin x )(1+ sin x ) (1− sin x )(1+ sin x ) = 1+ sin x 1− sin x ( x − 4 y )( x − 2 y ) 2. xy ( 3x − 2 y ) ( x + 4 y ) 3. −x ( 2x − 1) ( x + 3) 4. (x 5. ( x − 1)( x + 1)( x 6. k−r k+r 7. 1 x −1 ) ) ) 8. 9. 2 ) + y2 ( x − y) 2 −2 ) x 2 − xy + y 2 ( x + y) 2 sin 2 x + cot 2 xsin 2 x = sin 2 x (1+ cot 2 x ) = sin 2 x (csc2 x ) =1 Answers vary 10. 3-1 Multiple Choice Homework 1. 2. 3. 4. 5. 6. 1. sec4 β − tan 4 β = sec2 β − tan 2 β sec2 β + tan 2 β ( )( = sec2 β + tan 2 β = 1+ tan 2 β + tan 2 β = 1+ 2tan 2 β D A C B A B 156 ) 11. 12. cos2 w + 2cosw +1 cos2 w − 3cosw − 4 ( cosw +1)( cosw +1) = ( cosw +1)( cosw − 4 ) cosw +1 = cosw − 4 = ( ) 6 (1− cos w ) + 5cosw − 2 2 2 − 2cos2 w + 5cosw +1 = 6 − 6cos2 w + 5cosw − 2 −2cos2 w + 5cosw + 3 cos3 A− sec3 A = −6cos2 w + 5cosw + 4 cos A− sec A 2 cos A− sec A) cos2 A+ cos Asec A+ sec2 A = 2cos w − 5cosw − 3 ( = 6cos2 w − 5cosw − 4 cos A− sec A ( 2cosw +1)( cosw − 3) = cos2 A+1+ sec2 A = ( 2cosw +1)( 3cosw − 4 ) = 1− sin 2 A +1+ 1+ tan 2 A cosw − 3 = = −sin 2 A+ tan 2 A+ 3 3cosw − 4 ( ( 13. 2sin 2 w + 5cosw +1 6sin 2 w + 5cosw − 2 2 1− cos2 w + 5cosw +1 14. ) ) ( ) tan 2 w − secw − 5 tan 2 w + 3secw + 3 sec 2 w − 1) − sec w − 5 ( = (sec2 w − 1) + 3sec w + 3 15. ( ( sec 2 w − sec w − 6 = sec 2 w + 3sec w + 2 (sec w + 2 )(sec w − 3) = (sec w + 2 )(sec w + 1) = 3sec2 t − 8 tant +1 sec2 t − tant − 3 3 tan 2 t +1 − 8 tant +1 = tan 2 t +1 − tant − 3 ) ) 3tan 2 t + 3− 8 tant +1 tan 2 t +1− tant − 3 3tan 2 t − 8 tant + 4 = tan 2 t − tant − 2 ( 3tant − 2 )( tant − 2 ) = ( tant +1)( tant − 2 ) 3tant − 2 = tant +1 = sec w − 3 sec w + 1 156 16. 2csc2 y − 7cot y − 6 6csc2 y − 5cot y −10 = ( 6 ( cot 1+ tan3 B − tan B + sec2 B 1+ tan B ) 1− tan B + tan 2 B ( = − tan B + 1+ tan 2 B 23. ( ) y +1) − 5cot y −10 2 cot 2 y +1 − 7cot y − 6 2 2cot 2 y + 2 − 7cot y − 6 6cot 2 y + 6 − 5cot y −10 2cot 2 y − 7cot y − 4 = 6cot 2 y − 5cot y − 4 ( 2cot y +1)( cot y − 4 ) = ( 2cot y +1)( 3cot y − 4 ) cot y − 4 = 3cot y − 4 18. = (1+ tan B ) = cot B cot B cot B + 1 cot B 1+ sin3 B 1− sin 2 B 1+ sin B ) 1− sin B + sin 2 B ( = (1+ sin B)(1− sin B) 24. ( 3 x ( x +1) = x−2 x+2 (8 + cos x ) i (cos ( cos x − 2 ) ( cos 19. − ( 9 + 3x + x 3( x + 3) 20. − 21. 22. ) 25. ) sin 2 B − sin B + 1 1− sin B 3 2 ) = 1+ tan B = 17. ( ) 2 2 ) x − 2cos x + 4 ) x − 4 cos x + 4 ( 2 + cos x )( cos2 x − 2cos x + 4 ) ( cos x − 2 )( cos x − 2 ) = i ( cos x − 2 ) (cos2 x − 2cos x + 4 ) = ( 2 + cos x )( cos x − 2 ) r 2 + rx + x 2 r 2 − rx + x 2 = cos2 x − 4 x 26. sec2 x + 3sec x −10 ( (sec 1 = = 156 2 ) i (sec x + 2sec x − 3) x − 7sec x + 6 ) (sec x + sec x − 6 ) 2 2 (sec x + 5 )(sec x − 2 ) i (sec x −1)(sec x + 3) (sec x − 6 )(sec x −1) (sec x + 3)(sec x − 2 ) sec x + 5 sec x − 6 27. cot 2 x − 4csc x − 11 cot 2 x − 3 csc 2 x − 1 − 4csc x − 11 = csc 2 x − 1− 3 csc 2 x − 4csc x − 12 = csc 2 x − 4 (csc x + 2)(csc x − 6) = (csc x + 2)(csc x − 2) ( = 28. ) csc x − 6 csc x − 2 csc6 x − cot 6 x ( ) ( ) = ( csc x − cot x ) ( csc 3 = csc 2 x − cot 2 x 2 2 3 4 x + csc 2 x cot 2 x + cot 4 x = csc 4 x + csc 2 x cot 2 x + cot 4 x ( ) ( ) ) = csc 2 x 1+ cot 2 x + csc 2 x cot 2 x + cot 2 x csc 2 x − 1 = csc 2 x − cot 2 x + 3csc 2 x cot 2 x = 1+ 3csc 2 x cot 2 x 29. 54 − 2sin 3 x 6 ÷ 2 4 sin x − 81 sin x + 9 2 27 − sin 3 x sin 2 x + 9 = i 6 sin 2 x − 9 sin 2 x + 9 ( ( ) ( ) )( ) 2 ( 3− sin x )( 9 + 3sin x + sin x ) (sin x + 9 ) = i 6 (sin x − 3)(sin x + 3)(sin x + 9 ) − ( 9 + 3sin x + sin x ) = 2 2 2 2 3(sin x + 3) 156 30. (sin (sin 2 2 ) x − sin x cos x ) x + sin x cos x 2 2 ÷ sin x + cos x cos 3 x + sin 3 x ÷ sin x − cos x cos 3 x − sin 3 x sin x − cos x ) ( cos x − sin x )( cos2 x + sin x cos x + sin 2 x ) ( = i 2 i 2 sin x + cos x ( ) sin x (sin x − cos x ) ( cos x + sin x )( cos2 x − sin x cos x + sin2 x ) sin 2 x (sin x + cos x ) =− 2 cos2 x + sin x cos x + sin 2 x cos2 x − sin x cos x + sin 2 x 3-2 Multiple Choice 1. 2. 3. 4. 5. D A E C B 3-3 Free Response Homework 1a. b. c. d. e. f. 2a. 5 34 −99 17 34 −5 3 34 3 −17 34 5 99 5 b. c. d. e. f. 156 −16 65 −33 65 16 63 −65 63 65 56 33 56 3. ⎪⎧14° ± 120n θ=⎨ ⎪⎩54° ± 120n 4. x = {60°, 180°} 5. φ = 15° ± 180n 6. ⎧ π 5π 9π 13π 3π 7π 11π 15π ⎫ x=⎨ , , , ,− , − , − , − ⎬ 16 16 16 16 ⎭ ⎩16 16 16 16 7. ⎧11π x=⎨ ⎩ 12 , − 12. 5π ⎫ ⎬ 12 ⎭ = tan ( A + B ) 8. A and B = {All Reals} 9. sin ( A+ B ) + sin ( A− B ) = 13. sin ( A+ B ) − sin ( A− B ) 2sin Acos B 2cos Asin B = tan Acot B = cos ( a + b) cosb + sin ( a + b) sinb = cot ( A + B ) = (cos acosb − sin asinb) cosb + (sin acosb + cos asinb) sinb = cos acos2 b − sin asinbcosb + sin acosbsinb + cos asin 2 b = cos acos2 b + cos asin 2 b = cos a cos2 b + sin 2 b = cos a 11. tan Atan B (cot Acot B − 1) tan Atan B (cot A + cot B ) 1− tan Atan B tan B + tan A 1 = tan ( A + B ) = ( tan A + tan B 1− tan Atan B cot Acot B − 1 cot A + cot B = sin Acos B + cos Asin B + sin Acos B − cos Asin B = sin Acos B + cos Asin B − (sin Acos B − cos Asin B ) 10. sin Acos B + cos Asin B cos Acos B − sin Asin B sin ( A + B ) = cos ( A + B ) ) 3-3 Multiple Choice Homework 1. 2. 3. 4. 5. sin Acos A + cos Asin A cos Acos A − sin Asin A sin ( A + A) = cos ( A + A) B C D E A 3-4 Free Response Homework = tan ( A + A) 1a. tan A + tan A 1− tan Atan A 2 tan A = 1− tan 2 A = b. c. d. 156 4 2 9 −527 625 −8 7 2 625 − 527 − −9 4 2 527 336 9. 10. 2. ⎧π ⎪⎪ ± π n θ = ⎨8 ⎪ 3π ± π n ⎪⎩ 8 3. ⎧ π 3π ⎫ x=⎨ , ⎬ ⎩2 2 ⎭ 4. ⎪⎧0° ± 90n x=⎨ ⎩⎪75° ± 90n cot A + tan A cos A sin A = + sin A cos A cos2 A + sin 2 A = sin Acos A 1 = sin Acos A 2 = 2sin Acos A = 2csc ( 2 A) 11. cos2φ cot 2 φ + cos2φ = cos2φ cot 2 φ +1 e. f. 5. 6. 7. 8. ( )( = cos 2β ⎧π ⎪⎪ ± π n x = ⎨12 ⎪ 5π ± π n ⎪⎩ 12 ( ) = (cos φ − sin φ )(csc φ ) 2 2 2 = cos2 φ csc2 φ −1 cos2 φ = −1 sin 2 φ = cot 2 φ −1 ⎧ π 5π 9π 13π ⎫ ⎪⎪ 8 , 8 , 8 , 8 , ⎪⎪ x=⎨ ⎬ ⎪− 3π , − 7π , − 11π , − 15π ⎪ ⎪⎩ 8 8 8 8 ⎪⎭ x= cos4 β − sin 4 β = cos2 β − sin 2 β cos2 β + sin 2 β 12. π π ± n 12 2 ⎛ 1− tan 2 x cos2 x − sin 2 x ⎞ + ⎜ 2sin xcos x ⎟⎠ ⎝ 2 tan x ⎛ cos2x ⎞ = ⎜ cot 2x + sin 2x ⎟⎠ ⎝ 1− tan 2 B 1+ tan 2 B 1− tan 2 B = sec2 B 1 tan 2 B = − sec2 B sec2 B = cos2 B − sin 2 B = cos2B = ( 2cot 2x ) 2 = 4cot 2 2x = 4 csc2 2x − 1 ( = 4csc2 2x − 4 156 ) 2 2 ) 13. (1+ tanδ ) tan 2δ 2 tan δ = (1+ tan δ ) 1− tan 2 δ 2 tan δ (1+ tan δ ) = (1− tanδ )(1+ tanδ ) = 2 tan δ 1− tan δ 3-4 Multiple Choice Homework 1. 2. 3. 4. 5. 6. D E D E C B 3a. ⎛1 ⎞ 7 sin ⎜ x ⎟ = ⎝2 ⎠ 5 2 b. ⎛1 ⎞ −1 cos ⎜ x ⎟ = ⎝2 ⎠ 5 2 c. ⎛1 ⎞ tan ⎜ x ⎟ = −7 ⎝2 ⎠ 4a. ⎛1 ⎞ 5 csc ⎜ x ⎟ = − 3 ⎝2 ⎠ b. ⎛1 ⎞ 5 sec ⎜ x ⎟ = ⎝2 ⎠ 4 c. ⎛1 ⎞ 4 cot ⎜ x ⎟ = − 3 ⎝2 ⎠ 5a. 3-5 Free Response Homework 1a. b. c. 2a. b. c. ⎛1 ⎞ 1 sin ⎜ x ⎟ = ⎝2 ⎠ 10 ⎛1 ⎞ 3 cos ⎜ x ⎟ = ⎝2 ⎠ 10 ⎛1 ⎞ 1 tan ⎜ x ⎟ = ⎝2 ⎠ 3 ⎛1 ⎞ 3 sin ⎜ x ⎟ = ⎝2 ⎠ 10 ⎛1 ⎞ 1 cos ⎜ x ⎟ = ⎝2 ⎠ 10 ⎛1 ⎞ tan ⎜ x ⎟ = 3 ⎝2 ⎠ 156 ⎛1 ⎞ csc ⎜ x ⎟ = − 10 ⎝2 ⎠ b. ⎛1 ⎞ 10 sec ⎜ x ⎟ = 3 ⎝2 ⎠ c. ⎛1 ⎞ cot ⎜ x ⎟ = −3 ⎝2 ⎠ 6. ⎧π ⎪⎪ ± 4π n x = ⎨3 ⎪ 5π ± 4π n ⎪⎩ 3 7. x=⎨ 8. x=⎨ 9. x = ±π ± 4π n ⎧ 8π ⎩ 3 ⎧ 2π ⎩ 3 , 10π ⎫ ⎬ 3 ⎭ , 10π ⎫ ⎬ 3 ⎭ 10. 11. 12. 13. x=− π ± 2π n 2 ⎧ 2π x = ⎨− ⎩ 3 , 14. 4π ⎫ ⎬ 3 ⎭ ⎛1 ⎞ 2tan ⎜ x ⎟ ⎝2 ⎠ ⎛1 ⎞ 1+ tan 2 ⎜ x ⎟ ⎝2 ⎠ ⎛1 ⎞ 2 tan ⎜ x ⎟ ⎝2 ⎠ = ⎛1 ⎞ sec2 ⎜ x ⎟ ⎝2 ⎠ ⎛1 ⎞ ⎛1 ⎞ tan ⎜ x ⎟ + cot ⎜ x ⎟ ⎝2 ⎠ ⎝2 ⎠ 1− cos x 1+ cos x = + sin x sin x 2 = sin x = 2csc x ⎛1 ⎞ 2sin ⎜ x ⎟ ⎛1 ⎞ ⎝2 ⎠ = cos2 ⎜ x ⎟ ⎛1 ⎞ ⎝2 ⎠ cos ⎜ x ⎟ ⎝2 ⎠ ⎛1 ⎞ ⎛1 ⎞ = 2sin ⎜ x ⎟ cos ⎜ x ⎟ ⎝2 ⎠ ⎝2 ⎠ ⎛1 ⎞ tan x tan ⎜ x ⎟ ⎝2 ⎠ ⎛1 ⎞ = sin 2 ⎜ x ⎟ ⎝2 ⎠ = sin x sin x (1− cos x ) = cos x sin x 1− cos x = cos x = sec x − 1 15. 16. csc x − cot x 1 cos x = − sin x sin x 1− cos x = sin x ⎛1 ⎞ = tan ⎜ x ⎟ ⎝2 ⎠ ⎛ 1 1 ⎞ sin x + cos x ⎜⎝ 2 2 ⎟⎠ 2 1 1 1 1 = sin 2 x + 2sin xcos x + cos2 x 2 2 2 2 ⎛1 ⎞ 1 1 = sin 2 x + cos2 x + sin2 ⎜ x ⎟ 2 2 ⎝2 ⎠ = 1+ sin x 156 17. ⎛1 ⎞ ⎛1 ⎞ cos ⎜ x ⎟ + sin ⎜ x ⎟ ⎝2 ⎠ ⎝2 ⎠ ⎛1 ⎞ ⎛1 ⎞ cos ⎜ x ⎟ − sin ⎜ x ⎟ ⎝2 ⎠ ⎝2 ⎠ ⎛ ⎛1 ⎞ ⎛ 1 ⎞⎞ ⎜ cos ⎜ x ⎟ + sin ⎜ x ⎟ ⎟ ⎝2 ⎠ ⎝ 2 ⎠⎠ ⎝ = ⎛ ⎛1 ⎞ ⎛ 1 ⎞⎞ ⎜ cos ⎜ x ⎟ − sin ⎜ x ⎟ ⎟ ⎝2 ⎠ ⎝ 2 ⎠⎠ ⎝ 18. ⎛ ⎛1 ⎞ ⎛ 1 ⎞⎞ ⎜ cos ⎜ x ⎟ + sin ⎜ x ⎟ ⎟ ⎝2 ⎠ ⎝ 2 ⎠⎠ ⎝ ⎛ ⎛1 ⎞ ⎛ 1 ⎞⎞ ⎜ cos ⎜ x ⎟ + sin ⎜ x ⎟ ⎟ ⎝2 ⎠ ⎝ 2 ⎠⎠ ⎝ ⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞ cos2 ⎜ x ⎟ + 2sin ⎜ x ⎟ cos ⎜ x ⎟ + sin 2 ⎜ x ⎟ ⎝2 ⎠ ⎝2 ⎠ ⎝2 ⎠ ⎝2 ⎠ = ⎛1 ⎞ ⎛1 ⎞ cos2 ⎜ x ⎟ − sin 2 ⎜ x ⎟ ⎝2 ⎠ ⎝2 ⎠ = ⎛1 ⎞ ⎛1 ⎞ cos ⎜ x ⎟ − sin ⎜ x ⎟ ⎝2 ⎠ ⎝2 ⎠ ⎛1 ⎞ ⎛1 ⎞ cos ⎜ x ⎟ + sin ⎜ x ⎟ ⎝2 ⎠ ⎝2 ⎠ ⎛ ⎛1 ⎞ ⎛ 1 ⎞⎞ ⎜ cos ⎜ x ⎟ − sin ⎜ x ⎟ ⎟ ⎝2 ⎠ ⎝ 2 ⎠⎠ ⎝ ⎛ ⎛1 ⎞ ⎛ 1 ⎞⎞ ⎜ cos ⎜ x ⎟ + sin ⎜ x ⎟ ⎟ ⎝2 ⎠ ⎝ 2 ⎠⎠ ⎝ ⎛ ⎛1 ⎞ ⎛ 1 ⎞⎞ cos x + sin ⎜ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 x ⎟⎠ ⎟ ⎝ ⎠ ⎛ ⎛1 ⎞ ⎛ 1 ⎞⎞ cos x + sin ⎜ ⎜⎝ 2 ⎟⎠ ⎜⎝ 2 x ⎟⎠ ⎟ ⎝ ⎠ ⎛1 ⎞ ⎛1 ⎞ cos2 ⎜ x ⎟ − sin 2 ⎜ x ⎟ ⎝2 ⎠ ⎝2 ⎠ = ⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞ cos2 ⎜ x ⎟ − 2sin ⎜ x ⎟ cos ⎜ x ⎟ + sin 2 ⎜ x ⎟ ⎝2 ⎠ ⎝2 ⎠ ⎝2 ⎠ ⎝2 ⎠ ⎛1 ⎞ 1+ sin2 ⎜ x ⎟ ⎝2 ⎠ = ⎛1 ⎞ cos2 ⎜ x ⎟ ⎝2 ⎠ 1+ sin x = cos x 1 sin x = + cos x cos x = sec x + tan x ⎛1 ⎞ cos2 ⎜ x ⎟ ⎝2 ⎠ = ⎛1 ⎞ 1+ sin2 ⎜ x ⎟ ⎝2 ⎠ cos x = 1+ sin x 3-5 Multiple Choice Homework 1. 2. 3. 4. 5. D E D A C 3-6 Free Response Homework 1. 156 ⎧7.5°, 82.5°, 97.5°, ⎫ ⎪ ⎪ θ ∈⎨172.5°, 187.5°, 262.5°,⎬ ⎪277.5°, 352.5° ⎪ ⎩ ⎭ 2. 3. ⎧ −3π π 7π , , − ⎪⎪ 2 6 x ∈⎨ 2 ⎪− 11π , π , 5π ⎪⎩ 6 6 6 ⎫ ⎪ ⎬ ⎪ ⎪⎭ 14. ⎧−5.553, − 3.871, 0.730,⎫ ⎪ ⎪ x ∈⎨ ⎬ 3π π ⎪2.412, − 2 , 2 ⎪ ⎩ ⎭ 15. x ∈{−45°} 16. x ∈{ ±4.373, ± 1.911} 17. ⎧19π , ⎪⎪ 18 A ∈⎨ ⎪ 35π , ⎪⎩ 18 18. ⎧0°, 75°, 90°, 165°, 180°,⎫ ⎪ ⎪ ⎪255°, 270°, 345°, 360°, ⎪ x ∈⎨ ⎬ ⎪435°, 450°, 525°, 540°, ⎪ ⎪615°, 630°, 705°, 720° ⎪ ⎩ ⎭ 19. ⎪⎧ π ⎪⎫ x ∈⎨± , 0⎬ ⎪⎩ 3 ⎪⎭ 20. ⎧ 7π 11π 19π , , , ⎪⎪ 18 18 18 x ∈⎨ ⎪ 23π , 31π , 35π ⎪⎩ 18 18 18 21. x ∈⎨ ⎪⎧14°, 54°, 134°, ⎪⎫ θ ∈⎨ ⎬ ⎪⎩174°, 254°, 294° ⎪⎭ ⎪⎧11π 4. x ∈⎨ 5. x ∈⎨± ⎪⎩ 12 , 19π ⎪⎫ ⎬ 12 ⎪⎭ ⎧⎪ 2π ⎫⎪ ⎬ ⎪⎩ 3 ⎪⎭ 6. ⎧⎪30°, 90°, 150°, ⎫⎪ x ∈⎨ ⎬ ⎪⎩210°, 270°, 330° ⎪⎭ 7. ⎧⎪ π π 7π 4π ⎫⎪ x ∈⎨ , , , ⎬ 3 ⎪⎭ ⎪⎩ 6 3 6 8. x ∈⎨ 9. θ ∈{25°, 295°} 10. θ ∈{−165°, − 15°} 11. ⎧⎪ π ⎫⎪ x ∈ ⎨− ⎬ ⎪⎩ 4 ⎪⎭ 12. 13. ⎧ ⎪ ⎪ ⎪ x=⎨ ⎪ ⎪ ⎪ ⎩ ,⎪ ⎧⎪ 5π 7π 17π 19π ⎫⎪ , , ⎬ 12 ⎪⎭ ⎪⎩ 12 12 12 , t ∈{90°, 210°, 330°} 156 π ± 2π n 6 5π ± 2π n 6 π ± 2π n 2 ⎧⎪ π 23π , 18 43π , 18 5π ⎫⎪ ⎬ ⎪⎩12 12 ⎪⎭ , 31π , 18 47π 18 ⎫ ⎪⎪ ⎬ ⎪ ⎪⎭ ⎫ ⎪⎪ ⎬ ⎪ ⎪⎭ ⎪⎧ 2π ⎪⎫ ⎬ ⎪⎩ 3 ⎪⎭ 22. x ∈⎨ 23. x ∈{0.554, 2.125, 3.696, 5.266} 31. 24. ⎪⎧ π 3π ⎪⎫ x ∈⎨ , ⎬ ⎩⎪ 2 2 ⎭⎪ 25. x ∈⎨ 26. ⎪⎧ 2π ⎪⎫ a ∈⎨ ⎬ ⎪⎩ 3 ⎪⎭ 27. ⎧ ±15π ±13π ±11π ±9π ⎫ , , , ,⎪ ⎪⎪ ⎪ 8 8 8 8 x ∈⎨ ⎬ ⎪ ±7π , ±5π , ±3π , ±π ⎪ ⎪⎩ 8 ⎪⎭ 8 8 8 ⎪⎧ 5π ⎪⎩ 6 , 11π ⎪⎫ ⎬ 6 ⎪⎭ 28. x ∈ ⎡⎣ −2π , 2π ) 29. ⎧⎪−63.435°, − 26.565°,⎫⎪ θ ∈⎨ ⎬ ⎪⎩116.525°, 153.435° ⎪⎭ 30. θ ∈{±67.5°, 112.5°, 247.5°} ⎧ 3π 7π 11π 15π 19π ⎫ , , , ,⎪ ⎪⎪ , ⎪ 16 16 16 16 16 x ∈⎨ ⎬ ⎪ 23π , 27π , 31π ⎪ ⎪⎩ 16 ⎪⎭ 16 16 π ± 2π n 2 32. A= − 33. θ ∈{±60°, 0°, 180°} 34. θ ∈{ ±60°} 35. ⎧11π , ⎪ ⎪ 36 x ∈⎨ ⎪ 43π , ⎪ ⎩ 36 36. ⎧ 11π 7π π ⎫ , − , ,⎪ ⎪⎪− ⎪ 12 12 12 x ∈⎨ ⎬ ⎪ 5π , ± π , ± 3π ⎪ ⎪⎩ 12 4 4 ⎪⎭ 37. No solutions 35π 59π 19π ⎫ , , , 36 36 36 ⎪⎪ ⎬ 67π ⎪ ⎪ 36 ⎭ 3-6 Multiple Choice Homework 1. 2. 3. 4. 5. 156 C C C D A Trigonometric Identities Practice Test Answer Key Page 1 1. 33 65 2. − 7 25 3. −7 4. 65 63 5. − 120 119 6. − 7. 1 1 tan x + cot x 2 2 1 1 sin x cos x 2 + 2 = 1 1 cos x sin x 2 2 1 1 sin 2 x + cos 2 x 2 2 = 1 1 sin x cos x 2 2 1 = 1 1 sin x cos x 2 2 1 1 = csc x sec x 2 2 8. ⎧⎪ 3π π ⎫⎪ , − ⎬ 4 ⎭⎪ ⎩⎪ 4 9. x ∈⎨ 10. x ∈⎨ ⎧ 7π 11π 19π 23π 31π 35π ⎫ , , , , ⎬ 18 18 18 18 ⎪⎭ ⎪ ⎩ 18 18 , 156 2 5 1+ tan 2 x 1− tan 2 x sin 2 x 1+ cos 2 x = sin 2 x 1− cos 2 x sin 2 x 1+ 2 cos 2 x cos x = i sin 2 x cos 2 x 1− cos 2 x cos 2 x + sin 2 x = cos 2 x − sin 2 x 1 = 2 cos x − sin 2 x 1 = cos 2x = sec 2x Page 2 1. − 56 65 2. − 24 25 3. 4. − 65 16 5. − 119 120 6. 7. 1 2 tan x 2 1 1+ tan 2 x 2 1 2 tan x 2 = 1 sec 2 x 2 1 2sin x 2 = 1 1 cos x sec 2 x 2 2 1 2sin x 2 = 1 1 1 cos x sec x sec x 2 2 2 1 2sin x 2 = 1 sec x 2 1 1 = 2sin x cos x 2 2 ⎛ 1 ⎞ = 2sin ⎜ 2 i x ⎟ ⎝ 2 ⎠ 8. − 7 73 1 5 1 − tan 2 β 1 + tan 2 β sin 2 β 1− cos 2 β = sin 2 β 1+ cos 2 β cos 2 β − sin 2 β cos 2 β = cos 2 β + sin 2 β cos 2 β cos 2 β − sin 2 β cos 2 β cos 2 β = i cos 2 β + sin 2 β cos 2 β cos 2 β = cos 2 β − sin 2 β = cos 2 β = sin x 156 9. ⎧ 5π 9π ⎫ x ∈ ⎨− , − ⎬ 4 ⎭ ⎩ 4 10. ⎧ π 5π 13π ⎫ ⎪⎪− 18 , − 18 , − 18 , ⎪⎪ x ∈⎨ ⎬ 17π 25π 29π ⎪− ⎪ , − , − ⎪⎩ 18 18 18 ⎪⎭ Page 3 1. 63 65 2. − 4. 65 33 5. 24 7 7. 25 7 = = = )( ) ( ( )( ) 1 6. 1 1 tan x + cot x 2 2 sin x 1 + cos x = + 1 + cos x sin x sin x sin x + 1 + cos x 1 + cos x = sin x 1 + cos x ( 5 3. 5 2 ) sin 2 x + 1 + 2cos x + cos 2 x ( sin x 1 + cos x 2 + 2cos x ( sin x 1 + cos x ( 2 1 + cos x ( ) sin x 1 + cos x ) ) ) 2 sin x = 2csc x = 8. (1+ tanδ )( tan 2δ ) ⎛ 2 tanδ ⎞ = (1+ tanδ ) ⎜ ⎝ 1− tan 2 δ ⎟⎠ = = 2 tanδ (1+ tanδ ) (1− tanδ )(1+ tanδ ) 2 tanδ 1− tanδ 156 ⎧ 5π π⎫ , − ⎬ 4 ⎭⎪ ⎩⎪ 4 9. x ∈⎨− 10. ⎧ 25π 29π 37π ⎫ , − , − , ⎪ ⎪⎪− ⎪ 18 18 18 x ∈⎨ ⎬ ⎪− 41π , − 49π , − 53π ⎪ ⎪⎩ 18 18 18 ⎪⎭ Page 4 1. − 16 65 2. − 4. − 65 56 5. 7 24 7. 25 24 3. 1 5 6. − 7 5 2 ( ) ( ) sin ( A + B ) − sin ( A − B ) sin A + B + sin A − B sin Acos B + cos Asin B + sin Acos B − cos Asin B sin Acos B + cos Asin B − sin Acos B + cos Asin B 2sin Acos B = 2cos Asin B sin A cos B = i cos A sin B = tan Acot B = 8. ( ) ( ) cos a + b cos b + sin a + b sin b = cos b ⎡⎣cos acos b − sin asin b ⎤⎦ + sin b ⎡⎣sin acos b + cos asin b ⎤⎦ = cos acos 2 b − sin asin bcos b + sin asin bcos b + cos asin 2 b ( = cos a cos 2 b + sin 2 b ) = cos a ⎧⎪ 3π 9. x ∈⎨ 10. x ∈⎨ ⎩⎪ 4 ⎧ 7π , 7π ⎫⎪ ⎬ 4 ⎭⎪ 11π π 5π 13π 17π ⎫ , − , − , − , − ⎬ 18 18 18 18 ⎪⎭ ⎪⎩ 18 18 , 156