Questions and Answers #17 Trigonometry

Questions and Answers Sheet 17 Trigonometry Question #1 Simplifying (sin6 𝑥 + cos 6 𝑥) − (sin4 𝑥 + cos 4 𝑥) + sin2 𝑥 cos 2 𝑥 Answer: (sin6 𝑥 + cos 6 𝑥) − (sin4 𝑥 + cos 4 𝑥) + sin2 𝑥 cos 2 𝑥 = sin6 𝑥 + cos 6 𝑥 − sin4 𝑥 − cos 4 𝑥 + sin2 𝑥 cos 2 𝑥 = cos 6 𝑥 (tan6 𝑥 + 1 − tan4 𝑥 sec 2 𝑥 − sec 2 𝑥 + tan2 𝑥 sec 2 𝑥) = cos 6 𝑥 (tan6 𝑥 + 1 − tan4 𝑥 (1 + tan2 𝑥) − (1 + tan2 𝑥) + tan2 𝑥 (1 + tan2 𝑥)) = cos 6 𝑥 (tan6 𝑥 + 1 − tan4 𝑥 − tan6 𝑥 − 1 − tan2 𝑥 + tan2 𝑥 + tan4 𝑥) = cos 6 𝑥 (0) =0 Question #2 How to prove the following equation: tan2 ( 𝜋 3𝜋 5𝜋 7𝜋 ) + tan2 ( ) + tan2 ( ) + tan2 ( ) = 28 16 16 16 16 Answer: It can also be calculated by applying residue theorem to the function 𝑓(𝑧) = tan(8𝑧) tan2 (𝑧) and the box contour {𝑅𝑒(𝑧) = 0} ∪ {𝑅(𝑧) = 𝜋} ∪ {𝑡 + ±∞𝑖: 0 ≤ 𝑡 ≤ 𝜋}. Then, ∫R(𝑧)=0 𝑓(𝑧)𝑑𝑧 and ∫R(𝑧)=𝜋 𝑓(𝑧)𝑑𝑧 are cancelled. 𝜋 𝜋 lim ∫ 𝑓(𝑡 + 𝑖𝑦)𝑑𝑡 = ∫ (±𝑖)3 𝑑𝑡 = ∓𝜋𝑖 𝑦→±∞ 0 0 Sum of residues at 𝑧 = Residue at 𝑧 = 𝜋 2 2𝑘+1 16 𝜋, 𝑘 = 0,1, … ,7 is equal to − is equal to lim𝜋 𝑧→ 𝑑 𝑑𝑧 𝑆 4 where S is the given sum. 𝜋 2 (𝑧 − ) 𝑓(𝑧) = 8 2 2 𝑆 Combining them, by residue theorem we have 2𝜋𝑖 = 2𝜋𝑖 (− + 8) which gives S=28. 4 Question #3 Showing that 𝑓(4𝑥) + 4𝑓(2𝑥) = 8 cos 4 (𝑥) − 3, where 𝑓(𝑥) = cos 𝑥 Answer: You can use the property: cos(2𝑥) = cos 2(𝑥) − sin2 (𝑥) Since this will allow you to express cos 2 (2𝑥) − (1 − cos 2(2𝑥)) = 2 cos 2(2𝑥) − 1 By substituting again cos(2𝑥) by the aforementioned expression you will be able to express 4 cos(2𝑥) + cos(4𝑥) in terms of sin(𝑥) and cos(𝑥) raised to some powers. The sinus will cancell out and will obtained the desired result. Question #4 2𝜋 2𝜋 calculating 2 constants in a function 𝑀 = {𝑓 ∈ 𝐶[0,2𝜋], ∫0 𝑓(𝑥) sin 𝑥 𝑑𝑥 = 𝜋, ∫0 𝑓(𝑥) sin 2 𝑥 𝑑𝑥 = 2𝜋} 𝑎, 𝑏 ∈ Ⅎo𝑅, 𝑔 ∈ 𝑀, 𝑔(𝑥) = 𝑎 𝑠𝑖𝑛 𝑥 + 𝑏 𝑠𝑖𝑛 2𝑥, 𝑥 ∈ [0,2𝜋] I've read on the answers that a=1,b=2 and I don't know how to calculate them. Answer: 2𝜋 The overall pattern is ∫0 sin(𝑚𝑥) sin(𝑛𝑥) 𝑑𝑥 = 𝜋𝛿𝑚𝑛 𝑤ℎ𝑒𝑟𝑒𝛿𝑚𝑛 is 1 when m=n and 0 otherwise. The average value of sin2 is 1 2 and sines with different periods are orthogonal over a common period. 1 1 2 2 You can prove the the first from sin2 𝑥 = − cos(2𝑥). The cos(2𝑥) integrates away. You can prove the first by using the function product identities, which gets you a sum of sine waves that integrate to zero as well. The same thing works for cosines. This works because the sines and cosines form an orthogonal basis for the 2𝜋 periodic functions and you are effectively expanding a function in that basis. Question #5 1−cos 𝑥 cos 2𝑥⋯ cos 𝑛𝑥 Evaluate lim 𝑥2 𝑥→0 It should be 1 12 𝑛(𝑛 + 1)(2𝑛 + 1) but I don't know how to prove that. I am also aware that lim 𝜃→0 but I don't know how to use that. Answer: Use 1 − ∏𝑛𝑘=1 cos 𝑘 𝑥 = 1 − ∏𝑛𝑘=1 (1 − 2 sin2 𝑘𝑥 2 Question #6 Solve 𝑑𝑥 𝑑𝑦 + 𝑥 √𝑥 2 +𝑦 2 =𝑦 Answer: Let x=yu, then ∴𝑦 𝑑𝑥 𝑑𝑦 =𝑦 𝑑𝑢 𝑑𝑦 +𝑢 𝑑𝑢 𝑦𝑢 +𝑢+ =𝑦 2 𝑑𝑦 √𝑦 𝑢2 + 𝑦 2 𝑦 𝑑𝑢 𝑢 +𝑢+ =𝑦 2 𝑑𝑦 √𝑢 + 1 𝑦 𝑑𝑢 𝑢 =𝑦−𝑢− 𝑑𝑦 √𝑢2 + 1 (𝑦 − 𝑢 − 𝑢 √𝑢2 𝑘𝑥 2 ) ∼ 2 ∑𝑛𝑘=1 ( ) = 𝑑𝑦 =𝑦 𝑑𝑢 +1 ) This belongs to an Abel equation of the second kind. 2 𝑥2 2 ∑𝑛𝑘=1 𝑘 2 1−cos 𝜃 𝜃2 = 1 2 Let 𝑣 = 𝑦 − 𝑢 − 𝑢 √𝑢2 +1 Then 𝑦 = 𝑣 + 𝑢 + 𝑢 √𝑢2 +1 𝑑𝑦 𝑑𝑣 1 = +1+ 3 𝑑𝑢 𝑑𝑢 (𝑢2 + 1)2 ∴ 𝑣( 𝑑𝑣 1 𝑢 +1+ 3) = 𝑣 + 𝑢 + 2 𝑑𝑢 √𝑢 + 1 (𝑢2 + 1)2 𝑣 𝑑𝑣 1 𝑢 + (1 + 3) 𝑣 = 𝑣 + 𝑢 + 2 𝑑𝑢 √𝑢 + 1 (𝑢2 + 1)2 𝑣 𝑑𝑣 𝑣 𝑢 =− 3+𝑢+ 2 𝑑𝑢 √𝑢 + 1 (𝑢2 + 1)2 Question #7 1/(tan(𝜋 sin 𝑥)) 𝜋 Idea for lim𝜋 (tan ( sin 𝑥)) 𝑥→ 4 2 Need help for this. I tried using trigonometric identities but nothing seems to fit. I'm tired and clueless. Answer: Warning: you'll want to double-check all my arithmetic. 𝑦 ln tan 4 𝑦→𝜋 tan 𝑦 With 𝑦 ≔ 𝜋 sin 𝑥 we can rewrite this as exp lim . Define 𝑡 ≔ tan 𝑦 4 so tan 𝑦 2 = 2𝑡 1−𝑡 2 , tan 𝑦 = 4𝑡(1−𝑡 2 ) 1−6𝑡 2 +𝑡 4 Then your limit is exp lim (1−6𝑡 2 +𝑡 4 ) ln 𝑡 𝑡→1 ln(1+𝜖) exp lim (1+𝜖)𝜖(2+𝜖) = exp lim 𝜖→0 4𝑡(1−𝑡 2 ) ln(1+𝜖) 𝜖→0 2𝜖 = 𝑒𝑥𝑝 − lim ln 𝑡 . Finally, write 𝑡 = 1 + 𝜖 so the limit is 𝑡→1 𝑡(1−𝑡 2 ) = √𝑒 Question #8 𝜋 If cos 6(𝑥) + sin4 (𝑥) = 1 find x if 𝑥 ∈ [0, ] 2 My attempt: cos 6 (𝑥) + sin4 (𝑥) = 1 cos 6 (𝑥) + (1 − cos 2 (𝑥))2 = 1 cos 6 (𝑥) + 1 − 2 cos 2 (𝑥) + cos 4 (𝑥) = 1 cos 6 (𝑥) + cos 4 (𝑥) − 2 cos 2 (𝑥) = 0 Answer: Hint If we rewrite 𝑢 = cos 2 𝑥 and factor, the equation becomes (𝑢 + 2)𝑢(𝑢 − 1) = 0 𝜋 Alternative hint We have cos 6 𝑥 ≤ cos 2 𝑥 and sin4 𝑥 ≤ sin2 𝑥 nd in both cases equality holds (for 𝑥 ∈ [0, ] ) only for x=0, 𝜋 2 Question #9 2 𝑥 Show that when written in terms of t, where 𝑡 = tan ( ) the expression 2(1 + cos(𝑥))(5 sin(𝑥) + 2 12 cos(𝑥) + 13) is a perfect square. Answer: 2(1 + cos(𝑥))(5 sin(𝑥) + 12 cos(𝑥) + 13) = 2 (1 + = 1 − 𝑡2 10𝑡 12 − 12𝑡 2 ) ( + + 13) 1 + 𝑡2 1 + 𝑡2 1 + 𝑡2 4 (𝑡 + 5)2 (1 + 𝑡 2 )2 Question #10 How many solutions has the equation tan(3𝑥) = 1 ? I've come to the following solutions: 𝑥1 = 15 , so the angle is 45∘ 𝑥2 = 75 , so the angle is 225∘ But i've been told that there're up to six posible solutions fo x, and I don't think so. I've been watching the unit circle and there're only two possible tangents that has a value of 1. Answer: Just 3𝑥 = 45∘ + 180∘ 𝑘, where k is an integer number. Now, solve the following system: 0∘ ≤ 15∘ + 60∘ 𝑘 ≤ 360∘ Question #11 Evaluating ∑∞ 𝑛=1 sin(𝑛𝑥) 𝑛 𝑛𝑥 without integrating ∑∞ 𝑛=1 𝑒 𝑛𝑥 My solution proceeds by integrating ∑∞ = 𝑛=1 𝑒 𝑒 𝑖𝑥 1−𝑒 𝑖𝑥 Answer: Note that ∑∞ 𝑛=1 ∞ = Im (∑ 𝑛=1 sin(𝑛𝑥) 𝑛 = Im (∑∞ 𝑛=1 𝑒 𝑖𝑛𝑥 𝑛 ) (𝑒 𝑖𝑥 )𝑛 ) 𝑛 𝑛 If 𝑥 ∈ 𝑅 {0}, then 𝑒 𝑖𝑥 ∈ 𝑆 1 {1} and therefore ∑∞ 𝑛=1 (𝑒 𝑖𝑥 ) 𝑛 = log ( 1 1−𝑒 𝑖𝑥 ) (where 𝑙𝑜𝑔 is the principal branch of the logarithm), since log(1 + 𝑥) = 𝑥 − − (𝑥 + 𝑥2 2 + 𝑥3 3 + 𝑥4 4 + ⋯) 1 𝑥2 𝑥3 𝑥4 ⇒ log ( )=𝑥+ + + + 1−𝑥 2 3 4 Therefore ∑∞ 𝑛=1 sin(𝑛𝑥) 𝑛 = − arctan ( − sin 𝑥 1−cos(𝑥) ) 𝑥 = arctan (cot ( )) 2 Of course, if x=0, then this equality also holds. 𝑥2 2 + 𝑥3 3 − 𝑥4 𝑥 + ⋯ ⇒ log(1 − 𝑥) = Question #12 𝑡 How do I show that cos(𝑡) + 1 = 2 cos 2 ( ) 2 𝑡 I have been trying to show that cos(𝑡) + 1 is equal to 2 cos 2 ( ). Here's what I have done so far: 2 cos 𝑡 + 1 Using the identity: 1 = 2 cos 2(𝑡) − cos(2𝑡) cos(𝑡) + (2 cos 2(𝑡) − cos(2𝑡)) cos(𝑡) − cos(2𝑡) + 2 cos 2 (𝑡) cos(2𝑡) can be written as cos 2(𝑡) − sin2 (𝑡) cos(𝑡) − (cos 2(𝑡) − sin2 (𝑡)) + 2 cos 2 (𝑡) cos(𝑡) − cos 2 (𝑡) + 2 cos 2 (𝑡) + sin2 (𝑡) cos(𝑡) + cos 2 (𝑡) + sin2 (𝑡) cos(𝑡) + 1 I keep returning to what I started with Answer: Notice that 𝑐𝑜 𝑠(2𝛼) = 𝑐𝑜𝑠 2 𝛼 − sin2 𝛼 Therefore, with 𝛼 = 𝑡/2 one has cos(𝑡) = cos 2 (𝑡/2) − sin2 (𝑡/2) since sin2 (𝑡/2) = 1 − cos 2 (𝑡/2) one has the result. Question #13 How do you integrate ∫ cos(4𝑥) cos(𝑥) 𝑑𝑥 I tried using trigonometric formulas for turning it into 2 ∫ one, but still no idea of how to proceed with ∫ cos2 (2𝑥) cos(𝑥) cos2 (2𝑥) cos(𝑥) 𝑑𝑥 − ∫ 1 cos(𝑥) 𝑑𝑥 and can solve the second 𝑑𝑥. Answer: I would suggest that you instead use the cosine’s quadruple-angle formula (which can be derived by applying the double-angle formula twice): cos(4𝑥) = 8 cos 4(𝑥) − 8 cos 2(𝑥) + 1 which would turn your integral into ∫ cos(4𝑥) cos(𝑥) 𝑑𝑥 = 8 ∫ cos 3 (𝑥) 𝑑𝑥 − 8 ∫ cos(𝑥) 𝑑𝑥 + ∫ sec(𝑥) 𝑑𝑥 and you probably know how to evaluate each of these integrals. Question #14 1 Atypical way to find angle between unit vectors: 𝜃 = 2 sin−1 ( |𝐴̂ − 𝐵̂ |) 2 At my work, I have come across code with the following way of calculating the angle between two vectors. 1 𝜃 = 2 sin−1 ( |𝐴̂ − 𝐵̂ |) 2 (Note the physics convention: 𝑣̂ indicates the normalization of v; ie, 𝑣̂ ≔ 𝑣/|𝑣| ). Answer: This works when A and B are unit vectors. You have |𝐴 − 𝐵|2 = (𝐴 − 𝐵) ⋅ (𝐴 − 𝐵) = |𝐴|2 + |𝐵|2 − 2 𝐴 ⋅ 𝐵 = 2 − 2 cos 𝜃 = 4 sin2 Solving, 𝜃 = 2 arcsin 𝜃 2 |𝐴−𝐵| 2 Question #15 Integral of 𝑥2 √𝑥 2 +5 Answer: Hint: (Use hyperbolic trigonometric substitution to) 𝑥2 √𝑥 2 + 5 = √𝑥 2 + 5 − 5 √𝑥 2 + 5 Question #16 𝜋 If cos 4 𝛼 + 4 sin4 𝛽 − 4√2 cos 𝛼 sin 𝛽 + 2 = 0 then find 𝛼, 𝛽 in (0, ) 2 Try: I am trying to convert it into sum of square of quantity like (cos 2 𝛼)2 + (2 sin2 𝛽)2 − 2 ⋅ cos 2 𝛼 ⋅ 2 sin2 𝛽 − 4√2 cos 𝛼 sin 𝛽 + 2 + 4 cos 2 𝛼 ⋅ sin2 𝛽 (cos 2 𝛼 − 2 sin2 𝛽)2 − 4√2 cos 𝛼 sin 𝛽 + 4 cos 2 𝛼 ⋅ sin2 𝛽 Now i did not know how to solve it Answer: Hint: For real cos 2 𝛼 , sin2 𝛽 cos 4 𝛼 + 4 sin4 𝛽 + 1 + 1 4 ≥ √cos 4 𝛼 ⋅ 4 sin4 𝛽 4 The equality will occur if cos 4 𝛼 = 4 sin4 𝛽 = 1 Question #17 How to solve cos(2𝑛 𝑥) = cos(𝑥) I am given the function 𝐹: 𝑅 → 𝑅, 𝑥 ↦ 4𝑥(1 − 𝑥) and I have to find all n-cycles of the function. I have already reduced this to the problem of solving cos(2𝑛 𝑥) = cos(𝑥) for 𝑥 ∈ [0, 𝜋]. Using Mathematica and 2𝑘𝜋 2𝑘𝜋 𝑛 ∈ {1,2,3,4}, I concluded that the solutions must be { 𝑛 , 𝑛 } ∩ [0, 𝜋]. Since the equation obviously has 2 −1 2 +1 2𝑛 solutions and the solution set has 2𝑛 elements, it only remains to verify that they are indeed solutions to the equation. I tried using cos(2𝑥) = 2 cos 2 (𝑥) − 1 and proving this by induction on n, but I was not able to. Answer: Using the identity cos 𝐴 − cos 𝐵 = 2 sin ( 2 sin ( 2𝑛 −1 2 𝑥= 2𝑛 −1 2 𝑥) sin ( 𝑥 = 𝑘𝜋 OR 2𝑘𝜋 2𝑛 −1 OR 𝑥 = 2𝑛+1 2 2𝑛+1 2 2𝑘𝜋 2𝑛+1 𝐵−𝐴 2 ) sin ( 𝑥), which holds iff sin ( 2𝑛 −1 2 𝐵+𝐴 2 ), we have 0 = cos(𝑥) − cos(2𝑛 𝑥) = 𝑥) = 0 OR sin ( 2𝑛+1 2 𝑥) = 0 𝑥=𝑘 for some 𝑘 ∈ 𝑍 Question #18 1 sin(5𝑥/2) 2 2 sin(𝑥/2) Precalc Trig Identity, verify: 1 + cos(𝑥) + cos(2𝑥) = + Answer: For sin 𝑥 2 ≠ 0 we obtain: 1 + cos 𝑥 + cos 2 𝑥 = = 2 sin 2 sin 𝑥 𝑥 𝑥 + 2 sin cos 𝑥 + 2 sin cos 2 𝑥 2 2 2 = 𝑥 2 sin 2 𝑥 3𝑥 𝑥 5𝑥 3𝑥 5𝑥 + sin − sin + sin − sin sin 2 2 2 2 2 =1+ 2 𝑥 𝑥 2 2 sin 2 sin 2 2 I used the following formula. 1 sin 𝛼 cos 𝛽 = (sin(𝛼 + 𝛽) + sin(𝛼 − 𝛽)) 2 For example, 2 sin 𝑥 2 1 𝑥 𝑥 3𝑥 2 2 2 2 cos 𝑥 = 2 ⋅ (sin ( + 𝑥) + sin ( − 𝑥)) = sin − sin 𝑥 2 Question #19 Find all 4 complex roots to 3𝑧 4 − 𝑧 3 + 2𝑧 2 − 𝑧 + 3 = 0 using a trig substitution. A previous part of the question made us prove that 2 cos 𝑛 𝜃 = 𝑧 𝑛 + 𝑧 −𝑛 and it wants us to prove that the polynomial can be expressed as 6 cos 2 𝜃 − 2 cos 𝜃 + 2 = 0 . I'm assuming that using 6 cos 2 𝜃 − 2 cos 𝜃 + 2 = 0 we will be able to find all the roots of the equation but I'm unable to see how to get the polynomial into the trig equation and then how to solve the trig equation. Answer: Hint: The given equation is equivalent to 3 (𝑧 2 + 1 𝑧2 1 ) − (𝑧 + ) + 2 = 0 𝑧 1 By letting 𝑧 + = 𝑤 we get 3(𝑤 2 − 2) − 𝑤 + 2 = 0 𝑧 1 Solve this quadratic equation for w and subsequently solve 1 + = 𝑤 .Since w is real and |w| < 2 , we 𝑧 can substitute 𝑧 = 𝑒 𝑖𝜃 , 𝑤 = 2𝑐𝑜𝑠 𝜃 and solve for 𝜃 Question #20 Finding the graph defined by 𝑥 = 𝑠𝑖𝑛 𝜃 and 𝑦 = 3 − 2 cos(2𝜃) We are supposed to use the identity that sin2 𝜃 + cos 2 𝜃 = 1 Answer: 𝑥(𝜃) = sin 𝜃 𝑦(𝜃) = 3 − 2 cos(2𝜃) Note that cos(2𝜃) = 1 − 2 sin2 (𝜃) The expression for 𝑦(𝜃) simplifies to 𝑦(𝜃) = 3 − 2 cos(2𝜃) = 1 + 4 sin2 (𝜃) = 1 + 4𝑥 2 Thus your parabola is simply 𝑦 = 1 + 4𝑥 2 where −1 ≤ 𝑥 ≤ 1