Questions and Answers #33 Trigonometry

Questions and Answers Sheet 33 Trigonometry Question #1 How to prove ∑∞ 𝑛=1 sin 𝑛 𝑛 = 𝜋−1 2 Answer: Make use of the following fact: For |𝑧| ≥ 1 and 𝑧 ≠ 1, we have log(1 − 𝑧) = − ∑∞ 𝑘=1 Take 𝑧 = 𝑒 𝑖 and look at the imaginary part. We hence have ∞ log(1 − 𝑒 𝑖) = −∑ ∞ 𝑘=1 𝑒 𝑖𝑘 ⇒ Imag(log(1 − 𝑒 𝑖 )) 𝑘 𝑒 𝑖𝑘 = Imag (− ∑ ) 𝑘 𝑘=1 which gives us ∞ ∑ 𝑘=1 sin(𝑘) = −Imag (𝑙𝑜𝑔(1 − 𝑒 𝑖 )) 𝑘 log(1 − 𝑒 𝑖 ) = log(2 sin2 (1/2) − 2𝑖 sin(1/2) cos(1/2)) = log(2 sin(1/2) 𝑒 𝑖𝜋/2 𝑒 𝑖/2 ) Hence Imag(log(1 − 𝑒 𝑖 )) = 1−𝜋 2 which gives us ∑∞ 𝑘=1 sin(𝑘) 𝑘 = 𝜋−1 2 Question #2 What is derivative of 𝑥 2−tan 𝑥 Answer: ′ 𝑥 𝑥 ′ (2 − tan 𝑥) − 𝑥(2 − tan 𝑥)′ ) = (2 − tan 𝑥)2 2 − tan 𝑥 ′ 2 − tan 𝑥 − 𝑥(− tan 𝑥) = (2 − tan 𝑥)2 𝑥 2 − tan 𝑥 + 2𝑥 cos = (2 − tan 𝑥)2 2 cos 2 𝑥 − sin 𝑥 cos 𝑥 + 𝑥 = cos 2 𝑥 (2 − tan 𝑥)2 𝑦′ = ( Question #3 How to prove it? 2 sin 𝑥 ≤ 𝑥 − 𝜋 + √3 3 Answer: 𝜋 The following is a standard mechanical approach. Let 𝑓(𝑥) = 𝑥 − + √3 − 2 sin 𝑥 3 Then 𝑓 ′ (𝑥) = 1 − 2 cos 𝑥 1 𝜋 The derivative is negative until cos 𝑥 = , that is, until 𝑥 = , and then is positive. 2 3 𝑧𝑘 𝑘 𝜋 So f(x) reaches a minimum in our interval at 𝑥 = . Evaluate 𝑓(𝜋/3). We get 0. So in our interval, 𝑓(𝑥) ≥ 0, with equality only at 𝑥 = 3 𝜋 3 Question #4 2nd solution of cos 𝑥 cos 2 𝑥 cos 3 𝑥 = 1 4 Attempt explained: (2 cos 𝑥 cos 3 𝑥) cos 2 𝑥 = 1 2 (cos 4 𝑥 + cos 2 𝑥) cos 2 𝑥 = 1 . 𝐿𝑒𝑡, 𝑦 = 2𝑥 2 cos 2 𝑦 + cos 𝑦 (2 cos 2 𝑦 − 1) = 1 2 Answer: It's cos 2 𝑥(cos 2 𝑥 + cos 4 𝑥) = 1 2 Now, let cos 2 𝑥 = 𝑡 Thus, 𝑡(𝑡 + 2𝑡 2 − 1) = 1 2 or 4𝑡 3 + 2𝑡 2 − 2𝑡 − 1 = 0 or 2𝑡 2 (2𝑡 + 1) − (2𝑡 + 1) = 0 or (2𝑡 2 − 1)(2𝑡 + 1) = 0 which gives cos 4 𝑥 = 0 1 or cos 2 𝑥 = − and we can write the answer. 2 Question #5 A closed curve in the (x,y) -plane is represented by the functions 1 𝑥(𝜃) = (cos 𝜃 + √2(sin 𝜃)) 2 1 𝑦(𝜃) = (− cos 𝜃 + √2(sin 𝜃)) 2 Answer: Clearly 1) → 2𝑥 = cos(𝜃) + √2 sin(𝜃) 2) → 2𝑦 = − cos(𝜃) + √2 sin(𝜃) Thus by 1)+2) we get, 2𝑥 + 2𝑦 = 2√2 sin(𝜃) 𝑥+𝑦 So = sin(𝜃) √2 Thus by 1)-2) we get, 2𝑥 − 2𝑦 = 2 cos(𝜃) So 𝑥 − 𝑦 = cos(𝜃) Now clearly 𝑥+𝑦 2 sin2 (𝜃) + cos 2 (𝜃) = ( ) + (𝑥 − 𝑦)2 = 1 √2 ⇒ 3𝑥 2 − 2𝑥𝑦 + 3𝑦 2 = 2 Question #6 A problem in trigonometry If tan(𝑥 + 𝑦) = 𝑎 + 𝑏 and tan(𝑥 − 𝑦) = 𝑎 − 𝑏. Then prove that: 𝑎 tan(𝑥) − 𝑏 tan(𝑦) = 𝑎2 − 𝑏 2 I have used the formulae for tan(𝑥 + 𝑦) and tan(𝑥 − 𝑦) and then cross multiplied and then found a and b individually and then tried to put the values of a and b in " 𝑎 tan(𝑥) − 𝑏 tan(𝑦) "(l.h.s) but it did not lead to the r.h.s. Answer: In addition to David's: Let 𝑢 = tan 𝑥 and 𝑣 = tan 𝑦. You have 𝑢+𝑣 𝑎+𝑏 = 1 − 𝑢𝑣 and 𝑢−𝑣 𝑎−𝑏 = 1 + 𝑢𝑣 What if you add two equations: 𝑢+𝑣 𝑢−𝑣 + = 1 − 𝑢𝑣 + 1 + 𝑢𝑣 = 2 𝑎+𝑏 𝑎−𝑏 ⇒ (𝑎 − 𝑏)(𝑢 + 𝑣) + (𝑎 + 𝑏)(𝑢 − 𝑣) = 2(𝑎2 − 𝑏 2 ) ⋯ Question #7 a problem with condition sin(2𝜃) = tan(𝜃) − cos(2𝜃) Answer: Hint: Using The Weierstrass Substitution, 𝑡 = tan 𝜃 2𝑡 + 1 − 𝑡 2 = 𝑡(1 + 𝑡 2 ) ⟺ 0 = 𝑡 3 + 𝑡 2 − 𝑡 − 1 = 𝑡 2 (𝑡 + 1) − (𝑡 + 1) = (𝑡 + 1)2 (𝑡 − 1) ⇒ 𝑡 = ±1 Question #8 Without induction. How to prove that sin(2𝑛𝑥) = 2𝑛 sin 𝑥 cos 𝑥 ∏𝑛−1 𝑘=1 (1 − sin2 (𝑥) sin2 𝑘𝜋 2𝑛 ) for Natural n I Tried by several way and the last try is to use euler formula which lead me to sin(2𝑛𝑥) = 𝑘 2𝑘 sin 𝑥 cos 2𝑚+1 𝑥 ∑𝑛−1 𝑘=0((−1) ((2𝑛), (2𝑘 + 1))(cos 𝑥 sin 𝑥) ) and no idea how to continue or if this method lead to the answer Answer: first by use: sin2 𝑥 = 1−cos 2𝑥 2 𝑘𝜋 cos 2 𝑥 − cos sin2 (𝑥) 𝑛) (1 − )=( 𝑘𝜋 𝑘𝜋 2 2 sin 2 sin 2𝑛 2𝑛 then by use: cos 𝑥 − cos 𝑦 formula I got (1 − sin2 (𝑥) ) = −( 2 𝑘𝜋 sin 2𝑛 𝑘𝜋 𝑘𝜋 sin(𝑥+ 2𝑛) sin(𝑥−2𝑛) sin2 𝑘𝜋 2𝑛 ) so R.H.S = 2𝑛 sin 𝑥 cos 𝑥 ∏𝑛−1 𝑘=1 (− sin(𝑥+ 𝑘𝜋 𝑘𝜋 ) sin(𝑥− ) 2𝑛 2𝑛 𝑘𝜋 sin2 2𝑛 ) 𝑛−1 sin(2𝑛𝑥) 𝑘𝜋 = ∏ sin (2𝑥 + ) 𝑛−1 2 sin 𝑥 cos 𝑥 2 𝑛 𝑘=1 so R.H.S = 𝑛 sin(2𝑛𝑥) 22𝑛−2 and Finally R.H.S = ∏𝑛−1 𝑘=1 ( 1 sin2 𝑛 sin(2𝑛𝑥) 22𝑛−2 ( 𝑘𝜋 2𝑛 ) 1 𝑛22−2𝑛 ) = sin(2𝑛𝑥) =L.H.S Question #9 𝑥 𝑥4 2 2 Why cos (4 cos −1 ( )) = − 2𝑥 2 + 1 Answer: Note that you may apply the double angle formula for cosine: cos(2𝑥) = 2 cos 2 (𝑥) − 1 Thus 𝑥 𝑥 cos (4 cos −1 ( )) = 2 cos 2 (2 cos −1 ( )) − 1 2 2 2 𝑥 2 −1 = 2 (2 cos (cos ( )) − 1) 2 2 𝑥 2 = 2 (2 ( ) − 1) − 1 2 2 2 𝑥 = 2 ( − 1) − 1 2 𝑥4 = 2 ( − 𝑥 2 + 1) − 1 4 4 𝑥 = − 2𝑥 2 + 1 2 Question #10 Why is 2𝑛/2 (5/3)𝑛 (cos(𝑛𝜋/4) + sin(𝑛𝜋/4)) an alternate form of the complex number (5/3)𝑛 (1 + 𝑖)𝑛 ? Answer: Let z = 1 + i = 2( 1 1 1 ( +i ) 2 2 2 𝜋 𝜋 = 21/2 (cos ( ) + 𝑖 sin ( )) 4 4 Now use de Moivre: 𝑧 𝑛 = 2𝑛/2 (cos ( 𝑛𝜋 𝑛𝜋 ) + 𝑖 sin ( )) 4 4 And finally: 5 𝑛 5 𝑛 ( ) (1 + 𝑖)𝑛 = ( ) 𝑧 𝑛 3 3 5 n n = 2n /2 ( ) n (cos( + i sin( )) 3 4 4 Question #11 𝜋 Why is ∫−𝜋(2 − |𝑥|) cos(𝑥) = 4? Answer: 𝜋 𝜋 Because ∫−𝜋(2 − |𝑥|) cos(𝑥) 𝑑𝑥 = 2 ∫0 (2 − 𝑥) cos(𝑥) 𝑑𝑥 so 𝜋 𝜋 𝜋 ∫ (2 − |𝑥|) cos(𝑥) 𝑑𝑥 = 2 ∫ cos 𝑥 𝑑𝑥 − ∫ 𝑥 cos 𝑥 𝑑𝑥 0 = 𝜋 2 sin 𝑥|𝜋0 − 𝑥 sin 𝑥|𝜋0 0 0 + ∫ sin 𝑥 𝑑𝑥 0 = − cos 𝑥|𝜋0 = 2 so 𝜋 ∫ (2 − |𝑥|) cos(𝑥) 𝑑𝑥 = 4 −𝜋 Question #12 Why is 0.5 cos2 (30) = tan(30) 0.5 cos(30) cos2 (13) , but ≠ tan(13) cos(13) ? Answer: It's simply because sin(30∘ ) = 1/2 Question #13 Why is cos 2 (2𝜋/5) + cos 2(4𝜋/5) = 3/4 ? Answer: 1 2 1 2 Let 𝜁 = 𝑒 𝑖𝜃 . Then you have cos 2(2𝜃) + cos 2 (4𝜃) = (𝑒 2𝜃 + 𝑒 −2𝜃 ) + (𝑒 4𝜃 + 𝑒 −4𝜃 ) or, recalling the 4 definition of 𝜁 , 1 2 ( +  −2 ) 2 + (( 4 +  −4 ) 2 ) 4 Now, if you expand those out you get 1 4 (𝜁 + 2 + 𝜁 −4 + 𝜁 8 + 2 + 𝜁 −8 ) 4 1 or, recalling that 𝜁 5 = 1 this is (4 + 𝜁 4 + 𝜁 + 𝜁 2 + 𝜁 3 ) 4 Rearranging gives 1 (3 + (1 + 𝜁 + 𝜁 2 + 𝜁 3 + 𝜁 4 )) 4 and by de Moivre we have 1 + 𝜁 + 𝜁 2 + 𝜁 3 + 𝜁 4 , so this is just 3/4 Question #14 Why is (1 − cot 3 7∘ )(1 − cot 8∘ ) = 2 Answer: cos(𝑥) sin(𝑥) sin(𝑥) − cos(𝑥) = sin(𝑥) 1 − cot(𝑥) = 1 − 4 𝜋 sin ( − 𝑥) 4 = −√2 sin(𝑥) Therefore, 𝜋 𝜋 sin ( − ( − 𝑥)) 𝜋 4 4 1 − cot ( − 𝑥) = −√2 𝜋 4 sin ( − 𝑥) 4 sin(𝑥) = −√2 𝜋 sin ( − 𝑥) 4 Hence, 𝜋 sin ( − 𝑥) 𝜋 sin(𝑥) 4 (1 − cot(𝑥)) ⋅ (1 − cot ( − 𝑥)) = [−√2 ] ⋅ [−√2 ] 𝜋 4 sin(𝑥) sin ( − 𝑥) 4 Question #15 Why did not I find all the solutions to this equation? sin 3 𝑥 + cos 2 𝑥 − sin 𝑥 = 1 x needs to be in [0, 𝜋] What I tried sin 3 𝑥 + cos 2 𝑥 − sin 𝑥 = 1 sin 3 𝑥 − sin 𝑥 = 1 − cos 2 𝑥 2 sin 𝑥 cos 2 𝑥 = 1 − (cos 2 𝑥 − sin2 𝑥) 2 sin 𝑥 cos 2 𝑥 = sin2 𝑥 + cos 2 𝑥 − cos 2 𝑥 + sin2 𝑥 2 sin 𝑥 cos 2 𝑥 = 2 sin2 𝑥 cos 2 𝑥 = sin 𝑥 𝜋 𝜋 𝜋 2 2 2 cos 2 𝑥 = cos − 𝑥 ⇒ 2𝑥 = − 𝑥 + 2𝑘𝜋 ∨ 2𝑥 = 𝑥 − + 2𝑘 𝑥= 𝜋 2𝑘𝜋 𝜋 (𝑘 ∈ 𝑍)(1) ∨ 𝑥 = − + 2𝑘𝜋(𝑘 ∈ 𝑍)(2) + 6 3 2 𝜋 5𝜋 6 6 (1): 𝑘 = 0 ⇒ 𝑥 = , 𝑘 = 1 ⇒ 𝑥 = But 0 and 𝜋 are solutions too. Why did not I find 0 and 𝜋? Answer: Your idea is correct except when you simplify sin 𝑥 in 2 sin 𝑥 cos 2 𝑥 = 2 sin2 𝑥 A good way is: 2 sin 𝑥 cos 2 𝑥 − 2 sin2 𝑥 = 0 (sin 𝑥)(2 cos 2 𝑥 − 2 sin 𝑥) = 0 What give you sin 𝑥 = 0 or 2 cos 2 𝑥 − 2 sin 𝑥 = 0 Then, solve both equations and get the complete solution. Question #16 Why can I cancel this ∑𝑖 2 cos 3 (𝛼𝑖 ) sin 𝛼𝑖 term for pairs 180degrees apart? 4 3 2 2 I have the equation, 𝐾11 ∑𝑖 cos 𝛼𝑖 + 𝐾12 ∑𝑖 2cos 𝛼𝑖 sin 𝛼𝑖 + 𝐾22 ∑𝑖 sin2 𝛼𝑖 cos 𝛼𝑖 = ∑𝑖 𝜌𝑖 cos 𝛼𝑖 In the paper it states, "For equal intervals in 180 degree segments the sums of odd powers are zero; therefore," 4 2 2 𝐾11 ∑𝑖 cos 𝛼𝑖 + 𝐾22 ∑𝑖 sin2 𝛼𝑖 cos 𝛼𝑖 = ∑ 𝜌𝑖 cos 𝛼𝑖 𝑖 The way I interpret what they mean is that, if I sum the values of cos 3 𝛼𝑖 for 0 degree and 180 degree I get a value of 0. Similarly, if I sum the values of cos 3 𝛼𝑖 for 1 degrees and 181 degree I get a value of 0. And so on and so forth all the way up to 179 degree and 359 degree, the sum of the values of cos 3 𝛼𝑖 will be 0. Therefore, we can cancel out the 𝐾1 2 ∑𝑖 2 cos 3 𝛼𝑖 sin 𝛼𝑖 term of Eqn(1). However, if I were to make the same sums with the term ∑𝑖 2 cos 3 𝛼𝑖 sin 𝛼𝑖 I see that I do not get values of zero. Answer: ∘ ∘ ∘ 180 +𝑥 Note that ∑180 cos(𝛼) = ∑180 𝛼=𝑥 𝛼−𝑥=0 cos(𝛼 − 𝑥) = ∑𝑦=0 cos 𝑦 and this last sum telescopes to 0 by the identity cos(180∘ − 𝑦) = − cos 𝑦 The same happens with ∑ cos 3 𝛼 𝑠𝑖𝑛 𝛼 since the sign stays the same because of the identity sin(180∘ − 𝑦) = sin 𝑦 Question #17 Why are these ∑ 𝑐𝑜𝑠 and 𝑐𝑠𝑐equivalent? Mathematica 'simplifies' this formula ∑𝑅𝑘=1 cos 1 𝜋𝑥 2 𝑅 to this (csc sin (2𝑅+1)𝜋𝑥 𝑅 2𝑘𝜋𝑥 𝑅 − 1) Answer: Assuming you're a bit familiar with complex numbers: 𝑒 𝑖𝑥 = cos(𝑥) + 𝑖 sin(𝑥) 2𝜋 Let 𝑎 = then you want to find the sum 𝑅 𝑅 ∑ cos(𝑎𝑘𝑥) 𝑘=1 Notice that it is way easier to first compute (which we later can relate to 𝑅 𝑅 ∑ 𝑒 𝑎𝑘𝑖𝑥 = ∑(𝑒 𝑎𝑖𝑥 )𝑘 𝑘=1 𝑘=1 which is a geometric series. This evaluates to 𝑅 ∑(𝑒 𝑎𝑖𝑥 )𝑘 = 𝑘=0 𝑒 𝑎𝑖𝑥(𝑅+1) − 1 𝑒 𝑎𝑖𝑥 − 1 But since your series starts at k=1 we have to subtract the first term, so that 𝑅 ∑(𝑒 𝑎𝑖𝑥 )𝑘 = 𝑘=1 𝑒 𝑎𝑖𝑥(𝑅+1) − 1 −1 𝑒 𝑎𝑖𝑥 − 1 Now substitute eq. (1) back into (5) to get ∑𝑅𝑘=1 cos(𝑎𝑘𝑥) + 𝑖 ∑𝑅𝑘=1 sin(𝑎𝑘𝑥) = ⋯ Question #18 1 Why cos 3 𝑥 − 2 cos(𝑥) sin2 (𝑥) = (cos(𝑥) + 3 cos(3𝑥)) 4 Answer: Here's a detailed step by step: 1 = (cos(𝑥) + 3 cos(3𝑥)) 4 1 = (cos(𝑥) + 3 cos(𝑥 + 2𝑥)) 4 1 = (cos(𝑥) + 3[cos(𝑥) cos(2𝑥) − sin(𝑥) sin(2𝑥)]) 4 1 = (cos(𝑥) + 3[cos(𝑥) (1 − 2 sin2 (𝑥)) − 2 sin2 (𝑥) cos(𝑥)]) 4 1 = (cos(𝑥) + 3[cos(𝑥) − 4 cos(𝑥) sin2 (𝑥)]) 4 1 = (cos(𝑥) + 3 cos(𝑥) − 12 cos(𝑥) sin2 (𝑥)) 4 1 = (4 cos(𝑥) − 12 cos(𝑥) sin2 (𝑥)) 4 = cos(𝑥) − 3 cos(𝑥) sin2 (𝑥) = cos(𝑥) − cos(𝑥) sin2 (𝑥) − 2 cos(𝑥) sin2 (𝑥) cos(𝑥) (1 − sin2 (𝑥)) − 2 cos(𝑥) sin2 (𝑥) = cos 3 (𝑥) − 2 cos(𝑥) sin2 (𝑥) Question #19 When working with spherical coordinates I have come across cos −1 (cos(𝑎) × cos(𝑏)) and was wondering if there was some kind of identity for this that I could use instead of computing the inside of the acrccos function first. Answer: The problem with cos(𝑥) is that near 𝑥 = 0, cos 𝑥 = 1 − 𝑥 2 /2 + 𝑂(𝑥 4 ). Thus cos −1 (𝑥) loses accuracy near x=1. An equivalent formula without such accuracy loss is the following: cos −1 (cos(𝑎) cos(𝑏)) = 2 sin−1 (√sin(𝑎) sin(𝑏) /2 + sin2 ((𝑎 + 𝑏)/2)) Notice that near 𝑥 = 0, sin(𝑥) = 𝑥 + 𝑂(𝑥 3 ).Thus for x small enough, sin(𝑥) ≈ 𝑥 and we can use the simple approximation cos −1 (cos(𝑎) cos(𝑏)) ≈ √𝑎2 + 𝑏 2 . Notice that in a spherical right triangle with legs a,b and hypoteneuse c, the formula cos 𝑐 = cos 𝑎 cos 𝑏 holds and if the triangle has small enough sides, the Pythagorean approximation 𝑐 2 ≈ 𝑎2 + 𝑏 2 holds. Question #20 When does arcsin2 𝑥 + arccos 2 𝑥 = 1 for x real or comlex? Answer: Note that on reals arcsin 𝑥 + arccos 𝑥 = 𝜋 2 thus 2 𝜋 𝜋2 arcsin2 𝑥 + arccos 2 𝑥 = 𝑎𝑟𝑐𝑠𝑖𝑛2 𝑥 + ( − arcsin2 𝑥) = 2 arcsin2 𝑥 − 𝜋 arcsin 𝑥 + 2 4 Let 𝑦 = arcsin 𝑥 and consider 𝑓(𝑦) = 2𝑦 2 − 𝜋𝑦 + 𝜋2 4 𝑦 ∈ [−1,1] 𝑓 ′ (𝑦) = 4𝑦 − 𝜋 = 0 ⇒ 𝑦 = 𝜋 4 𝜋 𝜋2 𝜋 𝜋2 𝜋2 𝑓( ) = 2 −𝜋 + = >1 4 16 4 4 8