Questions and Answers #14 Trigonometry

Questions and Answers Sheet 14 Trigonometry Question #1 Eliminate 𝜃 from from 𝑥−𝑘 sin 𝜃 cot 𝛼 𝑘 cos 𝜃 = 𝑦−𝑘 cos 𝜃 tan 𝛼 𝑘 sin 𝜃 = 𝑧 𝑐 Answer: Write your line as 𝑥−𝑘 sin 𝜃 cot 𝛼 𝑘 cos 𝜃 = 𝑧 𝑦−𝑘 cos 𝜃 tan 𝛼 and 𝑐 𝑘 sin 𝜃 = 𝑧 𝑐 Manipulate the two equations into 𝑐𝑥 = 𝑐 cot 𝛼 sin 𝜃 + 𝑧 cos 𝜃 𝑘 𝑐𝑦 = 𝑧 sin 𝜃 + 𝑐 tan 𝛼 cos 𝜃 𝑘 so solving for sin 𝜃 , 𝑐𝑜𝑠 𝜃: 𝑘(𝑐 2 − 𝑧 2 ) sin 𝜃 = 𝑐 tan 𝛼 ⋅ 𝑐𝑥 − 𝑧 ⋅ 𝑐𝑦 𝑘(𝑐 2 − 𝑧 2 ) cos 𝜃 = 𝑐 cot 𝛼 ⋅ 𝑐𝑦 − 𝑧 ⋅ 𝑐𝑥 Now remember sin2 𝜃 + cos 2 𝜃 = 1 Question #2 How to prove: cos 3 2 𝜃 + 3 cos 2 𝜃 = 4(cos 6 𝜃 − sin6 𝜃) Answer: Write cos 2 𝜃 = 𝑥. We have 𝑥 = 2 𝑐𝑜𝑠 2 𝜃 − 1 = 1 − 2sin2 𝜃 1+𝑥 = cos 2 𝜃 2 1−𝑥 = sin2 𝜃 2 Thus the RHS becomes 4 ( (1+𝑥)3 8 − (1−𝑥)3 8 1 ) = (1 + 3𝑥 + 3𝑥 2 + 𝑥 3 − (1 − 3𝑥 + 3𝑥 2 − 𝑥 3 )) = 𝑥 3 + 3𝑥 = 𝐿𝐻𝑆 2 Question #3 prove that 2 arctan(csc arctan 𝑥 − tan 𝑎𝑟𝑐𝑐𝑜𝑡𝑥) = arctan 𝑥 Answer: Hint: 𝜋 𝜋 2 2 Let 𝑎𝑟𝑐𝑡𝑎𝑛 𝑥 = 2𝑦, − < 2𝑦 < , 𝑦 ≠ 0 𝜋 tan(𝑎𝑟𝑐𝑐𝑜𝑡 𝑥) = tan ( − 2𝑦) = cot 2 𝑦 2 csc 2 𝑦 − cot 2 𝑦 = ⋯ = tan 𝑦 Let arctan 𝑥 = 𝑦 tan 𝑦 = 𝑥, − 𝜋 𝜋 <𝑦< 2 2 𝜋 1 tan(𝑎𝑟𝑐𝑐𝑜𝑡 𝑥) = tan ( − 𝑦) = cot 𝑦 = 2 𝑥 sec 𝑦 = +√1 + 𝑥 2 csc arctan 𝑥 = csc 𝑦 = sec 𝑦 =? tan 𝑦 Question #4 Proving 1+cos 𝑥 2+sin 𝑥 < 4 3 Answer: 3 4 5 5 If cos(𝜃) = and sin(𝜃) = (which is possible since 32 + 42 = 52 ), the inequality (with ≤ rather than <) is equivalent to cos(𝑥 − 𝜃) ≤ 1 Question #5 Prove angle addition holds for 𝑅𝑛 𝑢⋅𝑣 Define 𝜃(𝑢, 𝑣) = cos −1 (|𝑢||𝑣|) be the angle between 𝑢, 𝑣 ∈ 𝑅𝑛 , where 𝑢 ⋅ 𝑣 is the standard inner product and |𝑥| = √𝑥 ⋅ 𝑥 for all 𝑥 ∈ 𝑅𝑛 . Let 𝑤 = 𝑠𝑢 + 𝑡𝑣 where 𝑠, 𝑡 ≥ 0 are scalars. Show 𝜃(𝑢, 𝑣) = 𝜃(𝑢, 𝑤) + 𝜃(𝑣, 𝑤) for 𝑢, 𝑣 ≠ 0 Answer: Assume |u|=|v|=1 and w=su+tv, and denote 𝑢 ⋅ 𝑣 = 𝑥 . Then |𝑤| = √𝑠 2 + 𝑡 2 + 2𝑠𝑡𝑥, cos 𝜃 (𝑢, 𝑤) = 𝑠+𝑡𝑥 𝑡+𝑠𝑥 and cos 𝜃 (𝑣, 𝑤) = 2 2 . Also sin 𝜃 (𝑢, 𝑤) = √1 − cos 2 𝜃 (𝑢, 𝑤) = 2 2 √𝑠 +𝑡 +2𝑠𝑡𝑥 √ √𝑠 +𝑡 +2𝑠𝑡𝑥 𝑠 2 +𝑡 2 +2𝑠𝑡𝑥−𝑠 2 −𝑡 2 𝑥 2 −2𝑠𝑡𝑥 𝑠 2 +𝑡 2 +2𝑠𝑡𝑥 = 𝑡√1−𝑥 2 √𝑠 2 +𝑡 2+2𝑠𝑡𝑥 and similarly sin 𝜃 (𝑣, 𝑤) = 𝑠√1−𝑥 2 √𝑠 2 +𝑡 2 +2𝑠𝑡𝑥 . Now cos(𝜃(𝑢, 𝑤) + 𝜃(𝑣, 𝑤)) = cos 𝜃 (𝑢, 𝑤) cos 𝜃 (𝑣, 𝑤) − sin 𝜃 (𝑢, 𝑤) sin 𝜃 (𝑣, 𝑤) = 𝑠𝑡 + 𝑠 2 𝑥 + 𝑡 2 𝑥 + 𝑠𝑡𝑥 2 − 𝑠𝑡(1 − 𝑥 2 ) (𝑠 2 + 𝑡 2 + 2𝑠𝑡𝑥)𝑥 = = 2 = 𝑥 = cos 𝜃 (𝑢, 𝑣) 𝑠 2 + 𝑡 2 + 2𝑠𝑡𝑥 𝑠 + 𝑡 2 + 2𝑠𝑡𝑥 Question #6 Can csc 𝛼+cos 𝛼 cos 𝛼−tan 𝛼−sec 𝛼 be simplified? Answer: If you multiply the numerator and denominator of your last expression by 4(1 − sin 𝛼) and use the identity sin 2 𝛼 = 2 sin 𝛼 𝑐𝑜𝑠 𝛼, you can rewrite your result as (1 − csc 𝛼)(2 csc 2 𝛼 + 1). This reduces the original 24 symbols to 17. Question #7 Proving 1+csc2 𝐴 tan2 𝐶 1+csc2 𝐵 tan2 𝐶 = 1+cot2 𝐴 sin2 𝐶 1+cot2 𝐵 sin2 𝐶 Answer: hose to manipulate the left hand side of the equation, by firstly replacing cot 2 𝐴 with csc 2 𝐴 − 1according to the identities. After doing the same with the denominator, 𝑅𝐻𝑆 = 1+(csc2 𝐴−1) sin2 𝐶 1+(csc2 𝐵−1) sin2 𝐶 1 + csc 2 𝐴 sin2 𝐶 − sin2 𝐶 1 + csc 2 𝐵 sin2 𝐶 − sin2 𝐶 1 + csc 2 𝐴 sin2 𝐶 − sin2 𝐶 (cos 2 𝐶 + csc 2 𝐴 sin2 𝐶) sec 2 𝐶 1 + csc 2 𝐴 tan2 𝐶 = = 1 + csc 2 𝐵 sin2 𝐶 − sin2 𝐶 (cos 2 𝐶 + csc 2 𝐵 sin2 𝐶) sec 2 𝐶 1 + csc 2 𝐵 tan2 𝐶 = Question #8 For proving 16 cos 4𝑥+7 = 1 sin4 𝑥+cos2 𝑥 + 1 sin2 𝑥+cos4 𝑥 I tried to use that: sin4 𝑥 + cos 4 𝑥 = sin4 𝑥 + 2 sin2 𝑥 cos 2 𝑥 + cos 4 𝑥 − 2 sin2 𝑥 cos 2 𝑥 = (sin2 𝑥 + cos 2 𝑥)2 − 2 sin2 𝑥 cos 2 𝑥 1 = 12 − (2 sin 𝑥 cos 𝑥)2 2 1 = 1 − sin2 (2𝑥) 2 1 1 − cos 4 𝑥 =1− ( ) 2 2 = 3 1 + cos 4 𝑥 4 4 Answer: sin4 𝑥 + cos 2 𝑥 = (1 − cos 2 𝑥)2 1 + cos 2 𝑥 3 + cos 2 2 𝑥 3 + + = = 4 2 4 1 + cos 4 𝑥 7 + cos 4 𝑥 2 = 4 8 𝜋 7 + cos 4 ( − 𝑥) 7 + cos 4 𝑥 𝜋 𝜋 2 2 sin 𝑥 + cos 𝑥 = sin ( − 𝑥) + cos ( − 𝑥) = = 2 2 8 8 2 4 4 Question #9 𝑘𝜋 4𝜋 16 16 Evaluate (∑7𝑘=1 tan2 ( )) − tan2 ( ) Answer: Using tan2 𝑥 + cot 2 𝑥 = sin4 𝑥+cos4 𝑥 sin2 𝑥⋅cos2 𝑥 = 1−2 sin2 𝑥 cos2 𝑥 sin2 𝑥 cos2 𝑥 So we have = 4 csc 2 (2𝑥) − 2 = 2 + 4 cot 2 (2𝑥) 𝜋 𝜋 3𝜋 So our expression is 2 + 4 cot 2 ( ) + 2 + 4 cot 2 ( ) + 2 + 4 cot 2 ( ) 8 4 8 𝜋 𝜋 6 + 4 + 4 [cot 2 ( ) + tan2 ( )] 8 8 𝜋 = 10 + 4 [2 + 4 cot 2 ( )] 4 = 10 + 4[2 + 4] = 34 Question #10 How do I show (csc 𝜃 − cot 𝜃)2 ≡ Answer: 1−cos 𝜃 1+cos 𝜃 ? There is (csc 𝜃 − cot 𝜃)2 = ( 1 sin 𝜃 − cos 𝜃 2 sin 𝜃 ) =( 1−cos 𝜃 2 sin 𝜃 ) Since sin2 𝜃 = 1 − cos 2 𝜃 the expression becomes: (csc 𝜃 − cot 𝜃)2 ≡ 1−cos 𝜃 1+cos 𝜃 and the identity is verified Question #11 sin(𝑥) + 𝑝 cos(𝑥) = 2𝑝, 𝑝 ∈ 𝑅 has solutions for p=? Answer: sin 𝑥 + 𝑝 cos 𝑥 = √𝑝2 + 1 sin(𝑥 + 𝛼) for some 𝛼 sin(𝑥 + 𝛼) = 2𝑝 √𝑝2 +1 . We need | 𝑠𝑖𝑛(𝑥 + 𝛼)| ≤ 1 What you need is |2𝑝| ≤ √𝑝2 + 1 Question #12 2𝜋 Show ∫0 cos 2 𝑡 𝑑𝑡 = 𝜋 using Pythagorean Theorem The 'traditional' way I always see this integral calculated is with the identity cos 2 𝑡 = 2𝜋 My alternative method uses cos 2 𝑡 + sin2 𝑡 = 1. It's obvious that ∫0 1+cos(2𝑡) 2 2𝜋 cos 2 𝑡 + sin2 𝑡 𝑑𝑡 = ∫0 1 𝑑𝑡 = 2 The interval of integration is an integer multiple of the periods of each function ( cos 2 𝑡 has a period of 𝜋 ), and so it seems reasonable to me that given the Pythagorean identity used above, cos 2 𝑡 and sin2 𝑡 contribute, for lack of a better word, equally to this final answer of 2𝜋 above, and so the integral in the title should be half of 2𝜋, or just 𝜋 Is this method valid? Answer: 𝑎+𝑇 Your argument is definitely valid. To add more explanation, we can say that ∫𝑎 2𝜋 any T-periodic function f(x) (try to prove this rigorously), and then ∫0 2𝜋− ∫−𝜋 𝜋 2 cos 2 𝑡 𝑑𝑡 = 2 2𝜋 ∫0 cos 2 𝑡 𝑑𝑡 Question #13 Different ways gives different results - solving tan 2 𝑎 = √3 𝜋 Case 1) tan 2 𝑎 = √3 = tan ( ) 3 2𝑎 = 𝑛𝜋 + 𝑎= 𝜋 3 𝑛𝜋 𝜋 + ,𝑛 ∈ 𝑍 2 6 Case 2) 2 tan 𝑎 = √3 1 − tan2 𝑎 Solving above equation gives tan 𝑎 = 1 √3 or − √3 𝑇 𝑓(𝑥)𝑑𝑥 = ∫0 𝑓(𝑥)𝑑𝑥 for 2𝜋 sin2 𝑡 𝑑𝑡 = ∫0 𝜋 cos 2 (𝑡 − ) 𝑑𝑡 = 2 𝜋 Thus either 𝑎 = 𝑛𝜋 + , 𝑛 ∈ 𝑍 6 𝜋 Or 𝑎 = 𝑚𝜋 − , 𝑚 ∈ 𝑍 3 Or either I did something wrong or both these results are same. Answer: 𝑛𝜋 𝜋 𝜋 + = 𝑚𝜋 + ⇔ 𝑛 = 2𝑚 2 6 6 𝑛𝜋 𝜋 𝜋 + = 𝑚𝜋 − ⇔ 3𝑛 + 1 = 6𝑚 − 2 ⇔ 𝑛 = 2𝑚 − 1 2 6 3 Question #14 Finding the infinite sum involving \coth function using contour integration I am looking to show: ∑∞ 𝑛=1 coth(𝑛𝜋) 𝑛3 = 7𝜋3 180 Answer: Observe that the integrand has simple poles at the points 𝑧 = 𝑛𝜋 and 𝑧 = 𝑖𝑛 𝜋 , with 𝑛 ∈ 𝑍, 𝑛 ≠ 0. The residues at the poles are computed as lim (𝑧 − 𝑛𝜋) cot 𝑧 coth 𝑧 𝑧→𝑛𝜋 lim (𝑧 − 𝑖𝑛𝜋) cot 𝑧 cot ℎ𝑧 𝑧3 𝑧→𝑖𝑛𝜋 cot(𝜁+𝑖𝑛𝜋) coth(𝜁) (𝜁+𝑖𝑛𝜋)3 = = lim 𝜁 𝜁→0 cot(𝜁) coth(𝜁+𝑛𝜋) (𝜁+𝑛𝜋)3 = lim 𝜁 coth(𝑛𝜋) (𝑛𝜋)3 𝑧3 𝜁→0 , where we used lim 𝑥 cot 𝑥 = lim 𝑥 coth 𝑥 = 1 𝑥→0 𝑥→0 cot(𝑥 + 𝑛𝜋) = cot 𝑥 cot h(𝑥 + 𝑖𝑛𝜋) = cot h(𝑥) cot(𝑖𝑥) = −𝑖 cot h(𝑥) With this and a suitable choice of the integration contour you will obtain: ∞ ∞ 𝑛=1 𝑛=1 cot h(𝑛𝜋) 4∑ = ∑[𝑅𝑒𝑠(𝑓, −𝑛𝜋) + 𝑅𝑒𝑠(𝑓, 𝑛𝜋) + 𝑅𝑒𝑠(𝑓, −𝑖𝑛 𝜋) + 𝑅𝑒𝑠(𝑓, 𝑖𝑛 𝜋)] (𝑛𝜋)3 = −𝑅𝑒𝑠(𝑓, 0) = 7 45 Question #15 If cos(𝑎 − 𝑏) = 3 sin(𝑎 + 𝑏) then find 1 1−3 sin 2𝑎 Answer: Hint Generalization: cos(𝑎 − 𝑏) = 𝑚 sin(𝑎 + 𝑏) cos 𝑎 (cos 𝑏 − 𝑚 sin 𝑏) = sin 𝑎 (𝑚 cos 𝑏 − sin 𝑏) tan 𝑎 =? sin 2 𝑎 = 2 tan 𝑎 =? 1 + tan2 𝑎 + 1 1−3 sin 2𝑏 = coth(𝑛𝜋) (𝑛𝜋)3 and We can prove 1 1−𝑚 sin 2𝑎 + [2} 1 1−𝑚 sin 2𝑏 1−𝑚2 Here m=3 Question #16 1 √2 √3 √3 sin (arccos ( )) = proof this equation Answer: 1 1 √3 √3 Let 𝜃 = arccos ( ), so cos 𝜃 = You want to find 𝑠𝑖𝑛 𝜃 Use sin2 𝜃 + cos 2 𝜃 = 1 Question #17 If 𝑥 ∈ [−1,0) , then what is the value of cos −1 (2𝑥 2 − 1) − 2 sin−1 𝑥 ? Answer: 𝜋 Let 𝑥 = sin(𝑢) , where 𝑢 ∈ [− , 0]. Then 1 − 2𝑥 2 = cos(2𝑢) 2 Then 2𝑥 2 − 1 = cos(𝜋 + 2𝑢) and 𝜋 + 2𝑢 ∈ [0, 𝜋]. Thus, cos −1 (2𝑥 2 − 1) − 2 sin−1 (𝑥) = 𝜋 + 2𝑢 − 2𝑢 =𝜋 Question #18 Why tan−2 (𝑥) = 1 tan2 (𝑥) ? Answer: As Arturo Magidin said in the comments, it is a matter of notation. Usually, sin𝑛 (𝑥) , cos 𝑛 (𝑥) , tan𝑛 (𝑥) denote (sin(𝑥))𝑛 , (cos(𝑥))𝑛 , (tan(𝑥))𝑛 respectively, when n=1,2,…. Unfortunately, due to various historical reasons, when n=-1, the notation sin−1 (𝑥) , cos −1 (𝑥) , tan−1 (𝑥) do not at all mean 1 divided by the respective functions, but rather represent the inverse function. Your confusion probably comes from your teacher/textbook using tan−2 (𝑥) to mean (tan(𝑥))−2 instead of (tan−1 (𝑥))2 or tan−1 (tan−1 (𝑥)) . This confusing notation unfortunately rather common. The good news is that there's nothing conceptual you're missing, it's just an instance of particularly confusing notation. Question #19 Finding the fundamental period of −6 cos(5𝜋𝑥) I am trying to find the fundamental period of 𝑓(𝑥) = −6 cos(5𝜋𝑥) . I know a periodic function satisfies 𝑓(𝑥) = 𝑓(𝑥 + 𝑝). I know that cos(𝑥)'s periodicity is 2𝜋 as cos(𝑥 + 2𝜋) = cos(𝑥) so I just assumed that −6 cos(5𝜋 ⋅ 1 + 2𝜋) = −6 cos(5𝜋), which it is but apparently this is not correct for the fundamental period? Answer: Another way of thinking of the periodicity is this: Let 𝑓(𝑥) = cos(𝑥). It is easy to show that 𝑓(𝑥) = 𝑓(𝑥 + 2𝜋𝑛) for 𝑛 ∈ 𝑁. In addition, 𝑓 ′ (𝑥) = 𝑓 ′ (𝑥 + 2𝜋𝑛). Thinking about this problem whilst considering the derivative (rate of change) will show you that the period will involve an additional factor of 5 i.e. cos(5𝜋𝑥) will oscillate more rapidly than cos(𝑥) , leading to a smaller period. As a note, the constant out front (-6) will only change the scale of the function, not the period. You can also consider cos(𝑥) has a period of 2𝜋, meaning the length in x required for the cos(𝑥) waveform to complete one "iteration". With an argument of 5𝜋𝑥, what value (length) of x will make the overall argument 2𝜋? 5𝜋𝑥 = 2𝜋 ⇒ 𝑥 = 2 5 Question #20 Find the value of log 𝑎+𝑏𝑖 𝑎−𝑏𝑖 Answer: Write 𝑎 + 𝑏𝑖 = 𝑟𝑒 𝑖𝜃 ; then Thus log 𝑎+𝑏𝑖 𝑎−𝑏𝑖 = 𝑟𝑒 𝑖𝜃 𝑟𝑒 −𝑖𝜃 𝑎+𝑏𝑖 𝑎−𝑏𝑖 = 𝑟𝑒 𝑖𝜃 𝑟𝑒 −𝑖𝜃 = 𝑒 2𝑖𝜃 = log 𝑒 2𝑖𝜃 = 2𝑖𝜃 + 2𝑘𝜋𝑖 It's generally wrong to say that 𝜃 = arctan 𝜃= 2𝜋 3 , but arctan 𝑏 𝑎 = arctan(−√3) = − 𝜋 3 𝑏 𝑎 1 √3 2 2 . A simple example is 𝑎 + 𝑏𝑖 = − 𝑖 where the argument is