Questions and Answers #10 Trigonometry

Questions and Answers Sheet 10 Trigonometry Question #1 Evaluating lim(sin 𝑥) 𝑥 𝑥→0 Answer: Most probably you mean 𝑥 → 0+ , because for 𝑥 < 0 the expression is not defined for real functions. Using lim 𝑡 𝑡 = 1 you get 𝑡→0+ x x → 0+ ((sin x)sin x ) sin x → 11 = 1 Question #2 how to solve sin2 (𝜃)−3 cos2 (𝜃)+1 sin2 (𝜃)−cos2 (𝜃) ≡ 2 and similar equations Answer: You can proceed and solve question step by step using Trignometric identities. Let's take left hand side: = sin2 (𝜃)−3 cos2 (𝜃)+1 sin2 (𝜃)−cos2 (𝜃) sin2 (𝜃)−3 cos2 (𝜃)+sin2 (𝜃)+cos2 (𝜃) (Since sin2 sin2 (𝜃)−cos2 (𝜃) 2(sin2 𝜃−cos2 𝜃) = 𝜃 + cos 2 𝜃 = 1) sin2 𝜃−cos2 𝜃 = 2 = Right hand side Question #3 Find a solution to the equation 1 sin2 (2𝑥) + tan(𝑥) − 1 tan(𝑥) =2 Assuming sin(𝑥) ≠ 0 and cos(𝑥) ≠ 0 , I simplified the above to 4 sin2 (𝑥) − 1 − 4 sin(𝑥) cos(𝑥) = 0 which is equivalent to cos(2𝑥) + sin(2𝑥) = 1 2 I do not know what to do next. I would be grateful for any hints. Answer: We have that by linear combination of sine and cosine 𝜋 1 cos(2𝑥) + sin(2𝑥) = √2 sin (2𝑥 + ) = 4 2 Question #4 Let 𝑓(𝑥) = tan−1 𝑥 , 𝑥 ∈ 𝑅 , then the sequence {𝑓 (𝑛) (0)} is unbounded. True or false? Answer: True is the answer, and it gets short if the Taylor series 𝑓(𝑥) = arctan 𝑥 = ∑∞ 𝑘=0 (−1)𝑘 2𝑘+1 𝑥 2𝑘+1 is available right from scratch. Then 𝑓 (𝑛) (0) can be read of from the Taylor coefficients: The non-zero values satisfy 𝑓 (2𝑘+1) (0) (2𝑘+1)! = (−1)𝑘 2𝑘+1 thus one has n, is even, 0 f ( n ) (0) = ( n −1)/2 (n − 1)! n, is odd. (−1) Question #5 I have troubles with such limit lim ( x , y ) →(0,0) x3 y 4 + 2sin 2 x I approach on any line or curve I get limit equal to 0. I try polar coordinates - also nothing, still 0. I try to bound it somehow - I get infinity. However, wolfram tells that there is no limit for the function in (0,0) Answer: Call your function 𝑓(𝑥, 𝑦) . Note that |𝑓(𝑥, 𝑦)| ≤ |𝑓(𝑥, 0)| and lim 𝑓 (𝑥, 0) = lim 𝑥→0 𝑥3 𝑥→0 2 sin2 (𝑥) =0 Question #6 I cannot find how or why this, 5 sin 3 𝑥 − 1 = 3 Has one of two solutions being this, 42.99 + 𝑛 × 120∘ I am lost on how to get this solution. I have found the other one, so I will not mention it. Answer: Note that sin 3 𝑥 = sin(180∘ − 3𝑥) = 45 = sin 5 3.1∘ , which gives two sets of solutions, 3𝑥 = 53.1∘ + 𝑛360∘ ∘ ∘ 3𝑥 = (180 − 53.1) + 𝑛360 with n being integers. Thus, the complete sultions are 𝑥 = 17.7∘ + 𝑛120∘ , 𝑥 = 42.3∘ + 𝑛120∘ Question #7 let be 𝑥 ∈ 𝑅 I want to show that if lim sin(𝑛𝑥) = 0 𝑛→∞ then sin(𝑥) = 0 Answer: sin(𝑛 + 1) 𝑥 = sin(𝑛𝑥) cos 𝑥 + cos(𝑛𝑥) sin 𝑥 . So sin(𝑛𝑥) → 0 implies that cos(𝑛𝑥) sin 𝑥 → 0 . If sin 𝑥 ≠ 0 this gives cos(𝑛𝑥) → 0. But then 1 = sin2 (𝑛𝑥) + cos 2 (𝑛𝑥) → 0 , a contradiction. Question #8 Evaluate 𝑑𝑥 ∫ sin 𝑥−cos 𝑥 I know it can be done by Weierstrass substitution. But I am looking for new/simple approach. For example I tried: 1 sin 𝑥+cos 𝑥 ∫ sin 𝑥−cos 𝑥 ⋅ sin 𝑥+cos 𝑥 𝑑𝑥 = ∫ sin 𝑥+cos 𝑥 − cos 2𝑥 𝑑𝑥 but I can't continue from here. Answer: Write sin 𝑥 − cos 𝑥 = 1 1 = √2 ( sin 𝑥 − cos 𝑥) = √2 √2 𝜋 = √2 sin (𝑥 − ) 4 Question #9 Solve equation 1 + sin2 𝜃 = 3 sin 𝜃 cos 𝜃 ( given tan 7 1∘ 34′ = 3) Answer: 1 − 2 sin 𝜃 cos 𝜃 + sin2 𝜃 − sin 𝜃 cos 𝜃 = 0 (sin 𝜃 − cos 𝜃)2 + sin 𝜃 (sin 𝜃 − cos 𝜃) = 0 (sin 𝜃 − cos 𝜃)(2 sin 𝜃 − cos 𝜃) = 0 1 tan 𝜃 = 1 or tan 𝜃 = 2 𝜋 𝜃 = 𝑛𝜋 + 4 Question #10 A parametric curve is described by 𝜋 𝑥 = sin 𝑡 , 𝑦 = sin (𝑡 + ) 6 Is there a way to show that 𝑦= √3 2 𝑥+ √1−𝑥 2 2 without using the property sin(𝑎 + 𝑏) = sin 𝑎 cos 𝑏 + sin 𝑏 cos 𝑎 ? Answer: 𝜋 3 6 √2 Since sin (𝑡 + ) = 1 3 2 √2 sin 𝑡 + cos 𝑡, we have 𝑦 = 1 𝑥 + √1 − 𝑥 2 2 Without using the formula, I feel like 𝜋 arcsin 𝑦 − = 𝑡 = arcsin 𝑥 6 Question #11 How cos (2 x )) = 2 2 1 (1 + cos(4 x)) 2 4 Can you please give me the proof for cos (2 x )) = 2 I know that 1 2 2 1 (1 + cos(4 x)) 2 4 (cos(2𝑥) + 1) = cos 2 𝑥 Answer: From the identity you already know, we can use that cos(4𝑥) = cos 2(2𝑥) − sin2 (2𝑥) = 2 cos 2 (2𝑥) − 1 then 1 + cos(4𝑥) = 2 cos 2 (2𝑥) Question #12 𝜋 I need to solve ∫02 cos 6 𝑥 𝑑𝑥 I tried to use cos(3𝑥) = 4 cos 3 𝑥 − 3 cos 𝑥 but not succeeded Answer: Hint: Use that 1 cos 6 𝑥 = (15 cos(2𝑥) + 6 cos(4𝑥) + cos(6𝑥) + 10) 32 Question #13 𝜋 1 4 2 Proving arctan(√𝑥 2 + 1) = − arctan(𝑥) Answer: Let 𝜃 = arctan(√𝑥 2 + 1 − 𝑥) , then tan 𝜃 = √𝑥 2 + 1 − 𝑥 and we can easily solve this for x, as 𝑥= 1−tan2 𝜃 2 tan 𝜃 𝜋 = cot 2 𝜃 = tan ( − 2𝜃) 2 which implies 𝜋 1 𝜃 = − arctan(𝑥) 4 2 Question #14 Solving 2 sin(𝑥 + 30∘ ) = cos(𝑥 + 150∘ ) for x between 0∘ and 360∘ Answer: HINT The equation is equivalent to 2 sin 𝑥 cos 3 0∘ + 2 sin 3 0∘ cos 𝑥 = cos 𝑥 cos 1 50∘ − sin 𝑥 sin 1 50∘ √3 sin 𝑥 + cos 𝑥 = − √3 2 1 cos 𝑥 − sin 𝑥 2 Question #15 Solving cos 165 2 Answer: 4 Note that √2√2 = √ so you have √2 √2√2(2√2 − 1 − √3) 𝑓= 4 1 2√2 − 1 − √3 = √ 2 √2 1 1+√3 2 √2 = √2 − so remains to prove that 1+√3 √2 = √2 + √3 To do that,multiply by √2 and square both sides to get 12 + 3 + 2√3 = 2(2 + √3) which are obviously equal to 4 + 2√3 Question #16 Determine the polar form of sin 𝜃 + 𝑖(1 + cos 𝜃) Answer: Hint for z 𝜃 cos(𝜃) = 2 cos 2 ( ) − 1 2 𝜃 𝜃 sin(𝜃) = 2 sin ( ) cos ( ) 2 2 Hint for w: Use the \cos and \sin addition formulae and 𝑤 = cos 𝜃 + sin 𝜃 + 𝑖(sin 𝜃 − cos 𝜃) 𝜋 𝜋 𝜋 𝜋 4 4 4 4 = √2 (cos(𝜃) cos ( ) + sin(𝜃) sin ( ) + 𝑖 (sin(𝜃) cos ( ) − cos(𝜃) sin ( ))) Question #17 How to simplify the ratio Answer: HINT Recall that 𝑒 𝑖𝑡 + 𝑒 −𝑖𝑡 cos 𝑡 = 2 sin 𝑡 = 𝑒 𝑖𝑡 −𝑒 −𝑖𝑡 2𝑖 Question #18 1−cos 2𝑥+𝑖 sin 2𝑥 1+cos 2𝑥−𝑖 sin 2𝑥 How to solve 𝐿𝑛(csc(𝑧)) = 𝐿𝑛(2) + 𝑖𝑧? Answer: Hint csc(𝑧) 2 = 𝑒 𝑖𝑧 If 𝑒 𝑖𝑧 = 𝑎 , Using Intuition behind euler's formula 2𝑎 = 2𝑖 𝑎− as 𝑎 ≠ 0 1 𝑎 𝑎2 − 1 = 𝑖 𝜋 𝑎2 = √2𝑒 𝑖 4 Question #19 Prove that tan 𝜋 9 + 4 sin 𝜋 9 = √3 Answer: 𝜋 Multiplying by cos ( ) , this is equivalent to 9 sin 𝜋 9 + 2 sin 2𝜋 9 = √3 cos 𝜋 9 (1) But 𝜋 𝜋 𝜋 𝜋 𝜋 𝜋 𝜋 𝜋 2𝜋 − sin = 2 (sin cos − cos sin ) = 2 sin ( − ) = 2 sin 9 9 3 9 3 9 3 9 9 which proves (1) √3 cos Question #20 Find the exact value of sec 𝜃. given the value of tan 𝜃. Answer: You can use the identity 1 + tan2 𝜃 = sec 2 𝜃. 𝜋 Since you know that −𝜋 < 𝜃 < − from the sign of \tan , you can determine that 𝑠𝑒𝑐 is negative. 2 Now, you have that sec(𝜃) = −√50 = −5√2