Questions and Answers #9 Trigonometry

Questions and Answers Sheet 9 Trigonometry Question #1 What is the value of tan 𝜃 = 1 −1 ? Answer: Let 𝑧 = −1 + 𝑖 = 𝑥 + 𝑦𝑖, with 𝑥 = −1 and 𝑦 = 1. Clearly z is in the second quadrant, since 𝑥 < 0 and 𝑦 > 0. The polar form is 𝑧 = 𝑟(cos 𝜃 + 𝑖 sin 𝜃) As you noted, 𝑟 = √2 1 1 Therefore, cos 𝜃 = − and sin 𝜃 = , so we can take 𝜃 = Note that 𝜋 2 √2 3𝜋 √2 4 < 𝜃 < 𝜋, so 𝜃 is indeed in the second quadrant. Question #2 How to prove sin(𝑥 − 𝑎) + sin(𝑥 + 𝑎) = 2 sin(𝑥) cos(𝑎) ? Answer: HINT sin(𝑥 + 𝑦) = sin 𝑥 cos 𝑦 + cos 𝑥 sin 𝑦 Question #3 1398𝜋 ∫0 2𝑥 (sin2 𝑥)cos 𝑑𝑥 =? Answer: 1 𝑥 1−𝑥 +(1−𝑥)𝑥 It's a bit more pleasant to use e.g. ∫0 𝜋 2𝑥 ∫ ((sin2 𝑥)cos √𝑥(1−𝑥) 𝑑𝑥 than 2 + (cos 2 𝑥)sin 𝑥 )𝑑𝑥 0 Note: If we define 𝑝𝑛 (𝑥) as a polynomial of degree n , we get: 1 ∫ 𝑥 1−𝑥 + (1 − 𝑥) 𝑥 ∞ 𝑑𝑥 = 2√2 ∑ 𝑝2𝑘 (ln 2) 2𝑘 2 (2𝑘 + 1)! √𝑥(1 − 𝑥) 0 𝑘=0 𝑝0 (𝑥) = 1 𝑝2 (𝑥) = 𝑥 2 + 2𝑥 − 1 𝑝4 (𝑥) = 𝑥 4 + 4𝑥 3 − 6𝑥 2 + 60𝑥 + 65 𝑝6 = 𝑥 6 + 6𝑥 5 − 15𝑥 4 + 300𝑥 3 + 975𝑥 2 + 3750𝑥 + 6687 𝑝8 (𝑥) = 𝑥 8 + 8𝑥 7 − 28𝑥 6 + 4550𝑥 4 + 35000𝑥 3 + 187236𝑥 2 + 769272𝑥 + 1305089 Question #4 Limit of sin(√𝑥 + 2) − sin(√𝑥 + 4) as 𝑥 → ∞. Answer: Original question: sin(√𝑥 + 2 + √𝑥 + 4) = sin ( −2 √𝑥+2+√𝑥+4 ) → sin(0) = 0 Question #5 What is arctan(𝑧1 ) ± arctan(𝑧2 ) with 𝑧1 , 𝑧2 ∈ 𝐶 On wikipedia, there is the following identity: 𝑢±𝑣 arctan(𝑢) ± arctan(𝑣) = arctan ( ) 1 ∓ 𝑢𝑣 However when I try some 𝑢, 𝑣 ∈ 𝐶 to check, the formula does not hold. Is there an equivalent formula for complex numbers? Answer: The problem is that arctan is a multivalued function. If you want a specific function, you need to specify 𝜋 𝜋 which branch you are using. For example, the principal branch has real part in (− , ]. Other branches 2 2 will differ from that one by an integer multiple of 𝜋 . So the correct results are arctan(𝑢) ± arctan(𝑣) = arctan ( 𝑢±𝑣 1∓𝑢𝑣 ) + 𝑛𝜋 where n is an integer. If you are using the principal branch, it is the integer needed to put the real part of 𝜋 𝜋 the arctan on the right in the interval (− , ] 2 2 Question #6 What is the period of the 𝑓(𝑥) = sin 𝑥 + sin 3 𝑥 ? 𝑓(𝑥) = sin 𝑥 + sin 3 𝑥 = 2 3𝑥 + 𝑥 𝑥 − 3𝑥 cos = 2 sin 2 𝑥 cos 𝑥 = 4 sin 𝑥 cos 2 𝑥 2 2 𝑓(𝑥 + 𝑇) = 4 sin(𝑥 + 𝑇) cos 2(𝑥 + 𝑇) = 4 sin 𝑥 cos 2 𝑥 Answer: In general, if T is the period of a function f(x) then the period of the function f(ax) is 𝑇 𝑎 . Suppose two periodic functions 𝑓1 (𝑥) and 𝑓2 (𝑥) have periods 𝑇1 and 𝑇2 . Then the period of the function 𝑔(𝑥) = 𝑓1 (𝑥) ± 𝑓2 (𝑥) is LCM (least common multiple) of 𝑇1 and 𝑇2 (although, this certainly isn't true for all periodic functions, as explained inside this answer.) 2𝜋 2𝜋 In your question the periods of sinx and sin3x are calculated as = 2 and respectively. 1 So the period of the function 𝑓(𝑥) = sin 𝑥 + sin 3 𝑥 is the LCM (2𝜋, Question #7 Prove trigonometric an identity 2 cos 𝛼−1 √3−2 sin 𝛼 𝛼 𝜋 2 6 = tan ( + ) Answer: Using Prosthaphaeresis Formulas 1 𝜋 𝜋 = sin = cos 2 6 3 3 2𝜋 3 ) = 2𝜋. 𝜋 𝜋 √3 = cos = sin 2 6 3 𝜋 𝜋 𝛼 𝜋 𝛼 cos 𝛼 − cos = 2 sin ( − ) sin ( + ) 3 sin 𝜋 6 𝜋 2 𝛼 6 2 𝜋 𝛼 6 2 6 𝜋 2 𝛼 6 2 − sin 𝛼 = 2 sin ( − ) cos ( + ) 3 Remember we need sin ( − ) ≠ 0 to reach at the identity required Alternatively, we can use Weierstrass Substitution in the left hand side and use tan(𝐴 + 𝐵) formula Question #8 Prove tan(sin−1 𝑡) = 𝑡 √1−𝑡 2 Answer: 𝜋 𝜋 Let sin 𝜃 = 𝑡, 𝑤𝑖𝑡ℎ𝜃 ∈ [− , ] the range of 𝑠𝑖𝑛 ^{−1}𝑡 . Then, 2 2 cos 𝜃 = √1 − Thus sin2 𝜃 = √1 − 𝑡 2 sin 𝜃 tan(sin−1 𝑡) = tan 𝜃 = = cos 𝜃 𝑡 √1−𝑡 2 Question #9 𝑥 Can you solve sin 𝑥 = sin ( ) without a sum/double/half angle identity? 2 Answer: We can simply use that by definition 𝑥 𝑥 𝑥 2 2 2 sin(𝑥) − sin ( ) ⇔ 𝑥 = + 2𝑘𝜋 ∨ 𝑥 = 𝜋 − + 2𝑘𝜋. Question #10 Verify the trig identity cot2 𝜃−1 csc2 𝜃 = csc 𝜃 − 1 Answer: Yes indeed we have cot2 𝜃−1 csc2 𝜃 cos2 𝜃 −1 sin2 𝜃 1 sin2 𝜃 = = cos 2 𝜃 − sin2 𝜃 = cos(2𝜃) ≠ csc 𝜃 − 1 Question #11 How to solve sum of cos(𝑘𝑥) for the case cos(𝑥) = 1 I have the solution for ∑𝑛𝑘=1 cos(𝑘𝑥) : ∑𝑛𝑘=1 cos(𝑘𝑥) = ℜ(∑𝑛𝑘=1 𝑒 𝑖𝑘𝑥 ) = ℜ (𝑒 𝑖𝑥 = ℜ( 𝑒 𝑖𝑛𝑥 − 1 ) 𝑒 𝑖𝑥 − 1 𝑒 𝑖𝑥 𝑒 𝑖𝑛𝑥 2 𝑖𝑥 𝑒2 𝑒 𝑖𝑛𝑥 2 𝑖𝑥 −𝑒 𝑒2 −𝑒 −𝑖𝑛𝑥 2 −𝑖𝑥 2 ) = ℜ (𝑒 = 𝑖(𝑛+1)𝑥 2 𝑛𝑥 2 𝑥 sin 2 sin cos ( 𝑛𝑥 2) 𝑥 sin 2 sin (𝑛+1)𝑥 2 ) but if 𝑒 𝑖𝑥 = 1 i am dividing trough 0 in line 2. So i know that in this case 𝑒 𝑖𝑥 ≠ 1 . So my question is: What is the ∑𝑛𝑘=1 cos(𝑘𝑥) if 𝑒 𝑖𝑥 = 1 Answer: 𝑒 𝑖𝑥 = 1 means 𝑥 = 2𝑚𝜋, with 𝑚 ∈ 𝑁 . And in that case the sum you’re looking for is equal to n. Question #12 Solving tan 2 𝑥 = 1 + 2 sin 4 𝑥 Answer: By tangent half angle identities we have that by 𝑡 = tan(2𝑥) 4𝑡 tan 2 𝑥 = 1 + 2 sin 4 𝑥 ⇔ 𝑡 = 1 + ⇔ 𝑡 3 − 𝑡 2 − 3𝑡 − 1 = 0 1 + 𝑡2 ⇔ (𝑡 + 1)(𝑡 2 − 2𝑡 − 1) = 0 Question #13 Solve 2 cos 2 𝑥 + sin 𝑥 = 1 for all possible x Answer: 2 cos 2 (𝑥) + sin(𝑥) = 1 ⇔ 2(1 − sin 𝑥)(1 + sin 𝑥) + sin 𝑥 − 1 = 0 ⇔ (sin 𝑥 − 1)(−2 − 2 sin 𝑥) + (sin 𝑥 − 1) = 0 ⇔ (sin 𝑥 − 1)(−1 − 2 sin 𝑥) = 0 which indeed leads to the same solutions, or also from here by 𝑡 = sin 𝑥 2 − 2 sin2 𝑥 + sin 𝑥 = 1 ⇔ 2𝑡 2 − 𝑡 − 1 = 0 𝑡1,2 = 1±√9 4 = 1, − 1 2 Question #14 7𝜋 How do you find the arcsin (sin ( ))? 6 Answer: The answer is: 𝜋 𝜋 6 6 arcsin (𝑠𝑖𝑛 (7 )) = − The range of a function arcsin(𝑥) is, by definition. 𝜋 𝜋 − ≤ arcsin(𝑥) ≤ 2 2 It means that we have to find an angle 𝛼 that lies between − 𝜋 2 and 𝜋 2 and whose sin(𝛼) equals to a 𝜋 𝑠𝑖𝑛 (7 ). 6 From trigonometry we know that sin(𝜙 + 𝜋) = − sin(𝜙) for any angle 𝜙 . This is easy to see if use the definition of a sine as an ordinate of the end of a radius in the unit circle that forms an angle \phi with the X-axis (counterclockwise from the X-axis to a radius). We also know that since is an odd function, that is sin(−𝜙) = − sin(𝜙) We will use both properties as follows: 𝜋 𝜋 𝜋 𝜋 6 6 6 6 sin (7 ) = sin ( + 𝜋) = − sin ( ) = sin (− ) As we see, the angle 𝛼 = − 𝜋 6 first our conditions. It is in the range from − 𝜋 sin (7 ) . Therefore, it's a correct answer to a problem. 6 Question #15 5 How do you calculate tan (arccos ( )) ? 13 Answer: 5 5 If you set 𝛼 = arccos ( ) ⇒ cos 𝛼 = than we have to calculate: 13 13 sin 𝛼 tan 𝛼 = cos 𝛼 cos 𝛼 = =√ 5 25 ⇒ sin 𝛼 = √1 − cos 2 𝛼 = √1 − 13 169 169−25 169 =√ 144 169 12 = 13 So: tan 𝛼 = 12 13 5 13 = 12 13 ⋅ 13 5 = 12 5 Question #16 1 How do you find sin (sin−1 ( )) ? 4 Answer: Recall that sin[arcsin(𝑥)] = 𝑥 for x in the domain of arcsin(𝑥) (The domain of arcsin(𝑥) is: −1 ≤ 𝑥 ≤ 1 ) 1 1 4 4 Therefore sin [arcsin ( )] = Question #17 How do you evaluate sec[cot −1 (−6)] ? Answer: Explanation sec[cot −1 (−6)] 1 = cos −1 (−6) Using the following rule , we can simplify the denominator: 1 = −6 ( ) √1 + (−6)2 √1 + 36 = −6 √37 =− 6 Question #18 How do you find the exact value of arctan(2) + arctan(3) ? 𝜋 2 to 𝜋 2 and its sine equals to Answer: In this way: arctan(2) + arctan(3) = 𝛼. Let's take now the tangent of both members: tan(arctan 2 + arctan 3) = tan 𝛼 ⇒ tan arctan 2 + tan arctan 3 = tan 𝛼 ⇒ 1 − tan arctan 2 tan arctan 3 2+3 5 = tan 𝛼 ⇒ = tan 𝛼 ⇒ 1−2⋅3 1−6 3 tan 𝛼 = −1 ⇒ 𝛼 = 𝜋 + 𝑘𝜋. 4 Question #19 3 What is sin (2 arcsin ( )) ? 5 Answer: 3 𝜋 5 2 arcsin ( ) is some 𝜃 between − 𝜋 𝜋 2 2 Furthermore, with − ≤ 𝜃 ≤ and 𝜋 2 with sin 𝜃 = 3 5 and sin 𝜃 a positive number, we conclude that 𝜃 is between 0 and 𝜋 2 . We want to find sin 2 𝜃 and we already know sin 𝜃. So if we find cos 𝜃, then we can ue the double angle formula for sine. 3 You've probably done this kind of problem many times by now. 𝜃 is in the first quadrant and sin 𝜃 = , 5 find cos 𝜃. Use your favorite method -- draw a triangle, or a unit circle, or an angle in standard position, or skip the picture and use cos 𝜃 = ±√1 − sin2 𝜃 (recall that our \theta is in Quadrant 1, so its cosine is positive.) All of the above is really explanation of our thought process. All we really need to write is something like: 3 3 Let 𝜃 = arcsin ( ) , then sin 𝜃 = and 5 5 4 cos 𝜃 = 5 And sin(2𝜃) = 2 sin 𝜃 cos 𝜃. So, putting it all together we get: 3 3 4 24 5 5 5 25 sin (2 arcsin ( )) = 2 ( ) ( ) = Question #20 How do you simplify tan2 4 𝑏 − sec 2 4 𝑏 Answer: Call 𝑥 = 4𝑏 𝑓(𝑥) = (tan 𝑥 − sec 𝑥)(tan 𝑥 + sec 𝑥) sin 𝑥 1 sin 𝑥 1 =( − )( + ) cos 𝑥 cos 𝑥 cos 𝑥 cos 𝑥 (sin 𝑥 − 1)(sin 𝑥 + 1) 1 − sin2 𝑥 = = − cos 2 𝑥 cos 2 𝑥 cos 2 𝑥 =− = −1 cos 2 𝑥