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7 Analytic Trigonometry 7.1 Trigonometric Identities Let’s begin by listing the identities we already know. Reciprocal Identities: csc θ = 1 sin θ sec θ = 1 cos θ cot θ = tan θ = sin θ cos θ cot θ = cos θ sin θ 1 tan θ Pythagorean Identities: sin2 θ + cos2 θ = 1 tan2 θ + 1 = sec2 θ cot2 θ + 1 = csc2 θ Even/Odd Identities: sin(−θ) = − sin θ cos(−θ) = cos θ tan(−θ) = − tan θ Cofunction Identities: sin tan π 2 π 2 π sec 2 − θ = cos θ − θ = cot θ − θ = csc θ cos cot csc π π2 π2 2 − θ = sin θ − θ = tan θ − θ = sec θ Examples: 1. Simplify cos(t) + tan(t) sin(t) solution: sin(t) sin(t) cos(t) sin2 (t) cos(t) + cos(t) 2 cos (t) + sin2 (t) cos(t) 1 cos(t) sec(t) cos(t) + tan(t) sin(t) = cos(t) + = = = = 7 ANALYTIC TRIGONOMETRY 2. Simplify 99 sin θ cos θ + . cos θ 1 + sin θ solution: sin θ cos θ sin θ(1 + sin θ) + cos2 θ + = cos θ 1 + sin θ cos θ(1 + sin θ) sin θ + sin2 θ + cos2 θ = cos θ(1 + sin θ) sin θ + 1 = cos θ(1 + sin θ) 1 = = sec θ cos θ 3. Verify the identity cos θ(sec θ − cos θ) = sin2 θ solution: To verify identities we have to transform one side until it looks like the other - do not move terms from side to side. LHS = cos θ(sec θ − cos θ) 1 − cos θ = cos θ cos θ 1 − cos2 θ = cos θ cos θ 2 = 1 − cos θ = sin2 θ = RHS 4. Verify the identity 2 tan θ sec θ = 1 1 − 1 − sin θ 1 + sin θ solution: We work with the right hand side RHS = = = = = 1 1 − 1 − sin θ 1 + sin θ 1 + sin θ − (1 − sin θ) (1 − sin θ)(1 + sin θ) 2 sin θ 1 − sin2 θ 2 sin θ cos2 θ 2 tan θ sec θ = LHS 7 ANALYTIC TRIGONOMETRY 5. Verify the identity 100 cos u = sec u + tan u 1 − sin u solution: This one is a little trickier. We start with the right hand side: RHS = sec u + tan u 1 sin u = + cos u cos u 1 + sin u = cos u Now we appear to be stuck. We want this to look like the LHS which has 1 − sin u in the denominator, so we could try multiplying the RHS by this on the top and bottom to see what happens: RHS = = = = = 1 + sin u 1 − sin u · cos u 1 − sin u (1 + sin u)(1 − sin u) cos u(1 − sin u) 1 − sin2 u cos u(1 − sin u) cos2 u cos u(1 − sin u) cos u = LHS 1 − sin u 6. Verify the identity 1 + cos θ tan2 θ = cos θ sec θ − 1 solution: We should start on the right hand side and convert to sines and coses: 7 ANALYTIC TRIGONOMETRY 101 RHS = = = tan2 θ sec θ − 1 sin2 θ cos2 θ 1 − cos θ 2 sin θ cos2 θ 1−cos θ cos θ 2 1 sin θ cos θ · 2 cos θ 1 − cos θ (1 − cos2 θ) cos θ = cos2 θ(1 − cosθ ) (1 − cos θ)(1 + cos θ) = cos θ(1 − cosθ ) 1 + cos θ = = RHS cos θ = 7.2 Addition and Subtraction Formulas Formulas for sine: sin(s + t) = sin s cos t + cos s sin t sin(s − t) = sin s cos t − cos s sin t Formulas for cosine: cos(s + t) = cos s cos t − sin s sin t cos(s − t) = cos s cos t + sin s sin t Formulas for tangent: tan s + tan t 1 − tan s tan t tan s − tan t tan(s − t) = 1 + tan s tan t π Example: Find cos(75◦ ) and cos 12 tan(s + t) = solution: cos(75◦ ) = cos(45◦ + 30◦ ) = cos(45◦ ) cos(30◦ ) − sin(45◦ ) sin(30◦ ) √ √ √ 2 3 21 = − √2 2 √ 2 2 6− 2 = 4 7 ANALYTIC TRIGONOMETRY cos π 12 102 = = = = π π cos − 3 4 π π π π cos cos + sin sin 3 4 √ 3 √ √4 3 2 1 2 + 2√ 2 √ 2 2 2+ 6 4 Expressions of the form A sin x + B cos x We can use the sum and difference identities to rewrite expressions of the form A sin x + B cos x as something simpler. The trick is to simultaneously transform the A into something that looks like sin φ and transform the B into something that looks like cos φ. Then we can use a sum identity. We do this by imagining a point in the xy-plane with coordinates (A, B). If φ is the angle between the line connecting (A, B) with the origin and the x-axis, then B A sin φ = √ cos φ = √ A2 + B 2 A2 + B 2 Thus we can rewrite, using the sum identity: √ A B 2 2 A +B √ sin x + √ cos x A sin x + B cos x = A2 + B 2 A2 + B 2 √ = A2 + B 2 (cos φ sin x + sin φ cos x) √ = A2 + B 2 sin(x + φ) Example: Express 3 sin x + 4 cos x in the form k sin(x + φ) √ √ solution: Here, k = A2 + B 2 = 32 + 42 = 5 also, sin φ = 45 , cos φ = radians. Thus, 3 sin x + 4 cos x = 5 sin(x + 0.927) 7.3 Double-Angle, Half-Angle, and Product-sum Formulas We state the double-angle identites: sin 2x = 2 sin x cos x 3 5 so φ = 0.927 7 ANALYTIC TRIGONOMETRY 103 cos 2x = cos2 x − sin2 x = 1 − 2 sin2 x = 2 cos2 x − 1 tan 2x = 2 tan x 1 − tan2 x Example: If cos x = − 23 and x is in quadrant II, find cos 2x and sin 2x. solution: 2 2 8 1 cos 2x = 2 cos x − 1 = 2 − −1= −1=− 3 9 9 2 sin 2x = 2 sin x cos x √ Since x is in quadrant II, we can sub in sin x = 1 − cos2 x: r r √ s 2 4 2 2 4 4 5 4 5 =− sin 2x = 2 − 1− − 1− =− =− 3 3 3 9 3 9 9 Formulas for Lowering Powers sin2 x = 1 − cos 2x 2 tan2 x = cos2 x = 1 + cos 2x 2 1 − cos 2x 1 + cos 2x Half-angle formulas sin u 2 r =± r u 1 − cos u 1 + cos u cos =± 2 2 2 u 1 − cos u tan = 2 sin u The ± is chosen in the first two depending on what quadrant u 2 is in. Example: Find the exact value of sin 22.5◦ . solution: Since 22.5◦ is half of 45◦ , we use the half-angle formula for sin with u = 45◦ . 7 ANALYTIC TRIGONOMETRY 104 Since 22.5◦ is in the first quadrant, we choose the + sign. ◦ 45 ◦ sin 22.5 = sin 2 r 1 − cos 45◦ = 2 s √ 1 − 22 = 2 s √ 2− 2 = 4 Example: Find tan u 2 if sin u = 2 5 and u is in quadrant II. solution: From the formula, tan u 2 = 1 − cos u sin u We know sin u = 25 , and from the usual pythagorean identity we know p p cos u = ± 1 − sin2 u = − 1 − sin2 u We choose the negative because cos is negative in the second quadrant. Hence u 1 − cos u tan = 2 sinpu 1 + 1 − sin2 u = qsin u 1 − ( 52 )2 1+ = 1+ = 21 25 2/5 5 = 2 = 2/5 q 5+ 5+ √ ! 21 5 √ 21 2 7 ANALYTIC TRIGONOMETRY 105 Product-to-Sum Formulas 1 [sin(u + v) + sin(u − v)] 2 1 cos u sin v = [sin(u + v) − sin(u − v)] 2 1 cos u cos v = [cos(u + v) + cos(u − v)] 2 1 sin u sin v = [cos(u − v) − cos(u + v)] 2 sin u cos v = Example: Express sin 3x sin 5x as a sum of trig functions. solution: We use the fourth product-to-sum formula: 1 1 1 sin 3x sin 5x = [sin(3x − 5x) − sin(3x + 5x)] = (cos(−2x) − cos 8x) = (cos(2x) − cos 8x) 2 2 2 Sum-to-Product Formulas sin x + sin y = sin x − sin y = cos x + cos y = cos x − cos y = x+y x−y 2 sin cos 2 2 x+y x−y 2 cos sin 2 2 x+y x−y 2 cos cos 2 2 x+y x−y −2 sin sin 2 2 Example: Write sin 7x + sin 3x as a product. solution: We use the first formula: 7x − 3x 7x + 3x sin 7x + sin 3x = 2 sin cos = 2 sin 5x cos 2x 2 2 7 ANALYTIC TRIGONOMETRY 7.4 106 Inverse Trigonometric Functions From looking at the graphs of the trig functions, we see that they fail the horizontal line test spectacularly. However, if you restrict their domain, you can find an inverse for these functions on this domain only. Inverse Sine The function sin is one-to-one when restricted to sin : − π2 , π2 → [−1, 1]. Thus there exists a function sin−1 or arcsin h π πi sin−1 : [−1, 1] → − , 2 2 arcsin x(= sin−1 x) sin x The arcsin function satisfies arcsin(x) = y ⇔ sin(y) = x π π ≤x≤ 2 2 sin(arcsin x) = x if − 1 ≤ x ≤ 1 arcsin(sin x) = x if − Examples: Find (a)sin−1 ( 12 ), (b)arcsin(− 21 ), (c)sin−1 ( 32 ) and (d)cos(sin−1 ( 35 )). solution: (a) We know that sin( π6 ) = 12 , so sin−1 ( 12 ) = π6 . (b) Likewise, sin(− π6 ) = − 21 , so arcsin(− 12 ) = − π6 . 7 ANALYTIC TRIGONOMETRY 107 (c) We know sin x is never 23 (it is never greater than 1), so sin−1 ( 32 ) is undefined. p (d) For u ∈ − π2 , π2 , we have cos(u) = 1 − sin2 u, so s 3 3 −1 2 −1 cos sin = 1 − sin sin 5 5 s 2 3 = 1− 5 r 9 1− = 25 r 16 = 25 4 = 5 Inverse Cosine The function cos is one-to-one when restricted to cos : [0, π] → [−1, 1]. Thus there exists a function cos−1 or arccos cos−1 : [−1, 1] → [0, π] cos x arccos x(= cos−1 x) The arccos function satisfies arccos(x) = y ⇔ cos(y) = x arccos(cos x) = x if 0 ≤ x ≤ π 7 ANALYTIC TRIGONOMETRY 108 cos(arccos x) = x if − 1 ≤ x ≤ 1 √ 3 −1 Examples: Find (a) cos and (b) cos−1 (0). 2 solution: (a) cos( π6 ) = √ 3 , 2 thus cos−1 √ 3 2 (b) cos( π2 ) = 0, thus cos−1 (0) = = π6 . π 2 Examples: √ 1. Show that sin(cos−1 (x)) = 1 − x2 . p solution: sin u = 1 − cos2 (u) for u ∈ [0, π]. Thus p √ sin(cos−1 (x)) = 1 − cos2 (cos−1 (x)) = 1 − x2 2. Show that tan(cos−1 (x)) = solution: tan(u) = sin u , cos u √ 1−x2 . x so sin(cos−1 (x)) = tan(cos (x)) = cos(cos−1 (x)) −1 √ 1 − x2 x √ 3. Show that sin(2 cos−1 (x)) = 2x 1 − x2 solution: sin 2u = 2 sin u cos u, so √ √ sin(2 cos−1 (x) = 2 sin(cos−1 (x)) cos(cos−1 (x)) = 2 1 − x2 x = 2x 1 − x2 Inverse Tangent The function tan is one-to-one when restricted to tan : − π2 , π2 → R. Thus there exists a function tan−1 or arctan h π πi −1 tan : R → − , 2 2 7 ANALYTIC TRIGONOMETRY 109 arctan x(= tan−1 x) tan x The arctan function satisfies arctan(x) = y ⇔ tan(y) = x π π ≤x≤ 2 2 tan(arctan x) = x if x ∈ R arctan(tan x) = x if − Examples: Find (a) tan−1 (1), (b) tan−1 √ 3 and (c)arctan(−20) solution: (a) tan( π4 ) = 1, thus tan−1 (1) = π4 . √ √ (b) tan( π3 ) = 3, thus tan−1 ( 3) = π 3 (c) Using a calculator, we find arctan(−20) ≈ 1.52084 Other Inverse Trig Functions The other trig functions cot, csc and sec also have inverses when restricted to suitable domains, namely cot−1 , csc−1 and sec−1 . You don’t need to worry about graphing these. Just keep in mind: 1 cot−1 6= tan−1 1 csc−1 6= sin−1 1 sec−1 6= cos−1 7 ANALYTIC TRIGONOMETRY 7.5 110 Trigonometric Equations One frequently has to solve equations involving trig functions. Sometimes the values of x you look at are restricted, while others you are asked to find all the values of x that make a given equation true. In the latter case, there are usually an infinite number of solutions whose form depends on the period of the trig functions involved. Examples: 1. Solve (a) tan2 (x) − 3 = 0 (b) sin(x) = cos(x) (c) 1 + sin x = 2 cos2 (x) (d) sin 2x − cos x = 0 (e) cos(x) + 1 = sin x with t ∈ [0, 2π] solution: √ (a) tan2 (x) − 3 = 0 ⇒ tan(x) = ± 3. √ tan(x) = 3 when x = π3 . But tan has period π, so √ π tan(x) = 3 ⇒ x = + kπ k ∈ Z 3 Likewise, √ π tan(x) = − 3 ⇒ x = − + kπ k ∈ Z 3 Thus the full set of solutions is o n π π S = − + kπ, x = + kπ | k ∈ Z 3 3 (b) If sin(x) = cos(x), cos(x) 6= 0 because sin and cos are not 0 in the same places. Thus we can divide both sides by cos and get sin(x) = 1 ⇒ tan(x) = 1 cos(x) This happens when x = π4 . Once again, tan has a period of π so π x = + kπ k ∈ Z 4 (c) We can get the equation 1 + sin x = 2 cos2 (x) into an easier form to deal with by subbing in 1 − sin2 x for cos2 x: 1 + sin x = 2 cos2 (x) 1 + sin x = 2 − 2 sin2 x 2 sin2 x + sin x − 1 = 0 7 ANALYTIC TRIGONOMETRY 111 This is a quadratic in sin x that factors as (2 sin x − 1)(sin x + 1) = 0 Which means either 2 sin x − 1 = 0 or sin x + 1 = 0. The first gives us that sin x = 21 , and the second gives sin x = −1. • sin x = −1 ⇒ x = • sin x = 1 2 ⇒x= π 6 3π 2 + 2kπ, k ∈ Z + 2kπ, k ∈ Z or x = 5π 6 + 2kπ, k ∈ Z (d) We use a double angle identity on sin 2x − cos x = 0 to get 2 sin x cos x − cos x = 0 cos x(2 sin x − 1) = 0 Hence either • cos x = 0 ⇒ x = • sin x = 1 2 ⇒x= π 2 π 6 + 2kπ, k ∈ Z or x = + 2kπ, k ∈ Z or x = 3π 2 5π 6 + 2kπ, k ∈ Z. + 2kπ, k ∈ Z. (e) Here we have to be a little trickier and square both sides. Unfortunately, squaring both sides may produce so-called extraneous solutions, so after we are done we must check each solution in the original equation. cos x + 1 = sin x (cos x + 1)2 = 1 − cos2 x cos2 x + 2 cos x + 1 = 1 − cos2 x 2 cos2 x + 2 cos x = 0 2 cos x(cos x − 1) = 0 Hence either cos x = 0 or cos x = −1. Between 0 and 2π, the first only happens at π2 and 3π and the second happens at −π. Hence the three solutions are 2 x= π 3π , , π. 2 2 We now check each of these numbers in the original equation cos This is true, so π 2 π π + 1 = sin 2 2 1 = 1. is a solution. cos 3π 3π + 1 = sin 2 2 1 = −1. 7 ANALYTIC TRIGONOMETRY This is false, so 3π 2 112 is not a solution. cos π + 1 = sin π 0 = 0. This is true, so π is a solution. 2. Consider the equation 2 sin(3x) − 1 = 0. (a) Find all the solutions to the equation. (b) Find all the solutions to the equation in the interval [0, 2π). solution: (a) The equation rearranges to sin(3x) = 12 . As we’ve seen before, this means that π 5π + 2kπ, k ∈ Z or 3x = + 2kπ, k ∈ Z. 6 6 π 2kπ 5π 2kπ x= + , k ∈ Z or x = + , k ∈ Z. 18 3 18 3 3x = (b) We solve the inequalities π 2kπ + < 2π 18 3 2k 1 + <2 0≤ 18 3 2k 1 35 1 <2− = − ≤ 18 3 18 18 1 35 − ≤ 2k < 6 6 1 35 − ≤k< ≈ 2.91 12 12 The ks in this range are k = 0, k = 1 and k = 2. Hence 0≤ x= π 13π 25π , , 18 18 18 For the other solutions we have 0≤ 5π 2kπ + < 2π 18 3 5 2k + 18 3 5 2k − ≤ <2− 18 3 5 − ≤ 2k < 6 0≤ <2 5 31 = 18 18 31 6 7 ANALYTIC TRIGONOMETRY 113 5 31 ≤k< ≈ 2.58 12 12 The ks in this range are k = 0, k = 1 and k = 2. Hence − x= 3. Consider the equation √ 5π 17π 29π , , 18 18 18 3 tan( x2 ) − 1 = 0. (a) Find all the solutions of the equation. (b) Find all the solutions in the interval [0, 4π). solution: (a) This rearranges to tan( x2 ) = √1 3 This gives us π x = + kπ, k ∈ Z 2 6 π x = + 2kπ, k ∈ Z 3 (b) As before, we find the ks we need: π 0 ≤ + 2kπ < 4π 3 1 0 ≤ + 2k < 4 3 1 1 11 − ≤ 2k < 4 − = 3 3 3 1 11 − ≤k< ≈ 1.83 6 6 The ks in this range are k = 0 and k = 1. Hence x = π 3 or x = 7π . 3 There is another way of doing problems of this type, see class for details. 4. Solve the equation tan2 x − tan x − 2 = 0. solution: This is a quadratic in tan x, so tan2 x − tan x − 2 = 0 (tan x − 2)(tan x + 1) = 0 So tan x = 2 or tan x = −1. There isn’t a convenient angle with tan x = 2, but we can use tan−1 to write x = tan−1 (2) + kπ, k ∈ Z The second one of course gives us x=− π + kπ, k ∈ Z 4 7 ANALYTIC TRIGONOMETRY 114 5. Solve the equation 3 sin θ − 2 = 0. solution: This rearranges to sin θ = 32 . Once again there is no easy angle that gives us sin θ = 23 . We know that sin is positive in the first and second quadrants, and thus the two solutions in [0, 2π) are 2 2 −1 −1 sin and π − sin 3 3 Hence the full solution is −1 θ = sin 2 2 −1 + 2kπ or θ = π − sin + 2kπ 3 3