Questions and Answers #54 Trigonometry

Questions and Answers Sheet 54 Trigonometry Question #1 If 𝑎 sin2 𝑥 + 𝑏 cos 2 𝑥 = 𝑐 𝑏 sin2 𝑦 + 𝑎 cos 2 𝑦 = 𝑑 𝑎 tan 𝑥 = 𝑏 tan 𝑦 then prove that 𝑎2 (𝑑 − 𝑎)(𝑐 − 𝑎) = 𝑏 2 (𝑏 − 𝑐)(𝑏 − 𝑑) Answer: 𝑎2 tan2 𝑦 sin2 𝑦 cos 2 𝑥 = = 𝑏 2 tan2 𝑥 cos 2 𝑦 sin2 𝑥 also cos 2 𝑥 = 𝑐−𝑎 𝑏−𝑎 sin2 𝑥 = 𝑐−𝑏 𝑎−𝑏 cos 2 𝑦 = 𝑑−𝑏 𝑎−𝑏 sin2 𝑦 = 𝑑−𝑎 𝑏−𝑎 Therefore the result can easily be derived. 𝑎2 (𝑑 − 𝑎)(𝑐 − 𝑎)(𝑎 − 𝑏)(𝑎 − 𝑏) = 𝑏 2 (𝑏 − 𝑎)(𝑏 − 𝑎)(𝑐 − 𝑏)(𝑑 − 𝑏) Question #2 3 If 𝐴 + 𝐵 + 𝐶 = 180∘ , cot 𝐴 = 2 and cot 𝐵 = , what is sin 𝐶 ? 4 Answer: tan 𝐶 = − tan(𝐴 + 𝐵) = 125 121 ⇒ sin2 𝐶 = ⇒ sin 𝐶 = 121 125 11 5√5 Question #3 tan 𝐴+tan 𝐵 tan 𝐴 tan 𝐵−1 =⋯=− 11 2 ⇒ cot 𝐶 = − 2 11 ⇒ cot 2 𝐶 = 4 121 ⇒ csc 2 𝐶 = 1 + cot 2 𝐶 = If 𝑥𝑛 = cos √𝑛 + 1 − cos √𝑛, what is lim 𝑥𝑛 ? 𝑥→∞ I think it may be 0, because √𝑛 + 1 and √𝑛 are two very close angles. But I don't know how to prove it. If I use the formula for cos − 𝑐𝑜𝑠. I get −0 ⋅ ∞ Answer: Well cos √𝑛 + 1 − cos √𝑛 = 2 sin 1 1 (√𝑛 + 1 + √𝑛) sin (√𝑛 − √𝑛 + 1) 2 2 and as 𝑛 → ∞ the term in red goes to zero. Question #4 If x+y+z=xyz , find 3𝑥−𝑥 3 1−3𝑥 2 + 3𝑦−𝑦 3 1−3𝑦 2 + 3𝑧−𝑧 2 1−3𝑧 2 Answer: We can not make 𝐴 + 𝐵 + 𝐶 = 𝜋 as tan(𝐴 + 𝐵 + 𝐶) = ∑ tan 𝐴−tan 𝐴 tan 𝐵 tan 𝐶 1−∑ tan 𝐴 tan 𝐵 ∑ tan 𝐴 = tan 𝐴 tan 𝐵 tan 𝐶 ⇒ tan(𝐴 + 𝐵 + 𝐶) where n is any integer [Clearly, you have taken a special value of n] Now for any integer 𝑚, 𝑚(𝐴 + 𝐵 + 𝐶) = 𝑚𝑛𝜋 ⇒ tan(𝑚𝐴 + 𝑚𝐵 + 𝑚𝐶) = tan 𝑚 𝑛𝜋 = 0 ⇒ ∑ tan 𝑚 𝐴 = tan 𝑚 𝐴 tan 𝑚 𝐵 tan 𝑚 𝐶 Now set m=3 and apply tan 3 𝑥 formula Question #5 If 𝑦 = −𝑒 𝑥 cos 2 𝑥, show that 𝑑2 𝑦 𝑑𝑥 2 Answer: 4 3 4 sin 2 𝑥 + 4 cos 2 𝑥 = 5 ( sin 2 𝑥 + cos 2 𝑥) 5 5 Note we want 4 5 3 sin 2 𝑥 + cos 2 𝑥 = sin(2𝑥 + 𝛼) for some 𝛼 5 By compound angle formula, we know sin(2𝑥 + 𝛼) sin 2 𝑥 cos(𝛼) + cos 2 𝑥 sin(𝛼) So in order to fulfill the requirement, we only need to set 4 3 cos(𝛼) = , sin(𝛼) = 5 5 3 = 5𝑒 𝑥 sin(2𝑥 + 𝛼) where 𝛼 = tan−1 ( ) 4 4 3 4 3 5 5 5 5 Such 𝛼 exists. Hence 5 ( sin 2 𝑥 + cos 2 𝑥) = 5 sin(2𝑥 + 𝛼) with cos(𝛼) = , sin(𝛼) = , that is tan(𝛼) = 3 4 Question #6 𝑥 2𝑧 2 1+𝑧 2 If 𝑧 = tan ( ), show that sin(𝑥) = Answer: 𝑥 2 tan ( ) 2 𝑥 = 1 + tan2 ( ) 2 We know that tan2 (𝑥) + 1 = sec 2 (𝑥), so: 𝑥 2 tan ( ) 2 𝑥 = sec 2 ( ) 2 𝑥 2 tan ( ) 2 = 1 𝑥 cos 2 ( ) 2 𝑥 𝑥 2 tan ( ) cos 2 ( ) = 2 2 𝑥 sin ( ) 2 cos 2 (𝑥 ) = 2 𝑥 2 cos ( ) 2 𝑥 𝑥 2 sin ( ) cos ( ) = 2 2 We know that sin(2𝑥) = 2 sin(𝑥) cos(𝑥), so: 𝑥 sin (2 ) = sin(𝑥) 2 Question #7 𝜋 If the quadratic equation 𝑥 2 + 2 cos(4𝜃) = 4𝑥 sin (𝜃 + ) has equal roots in x , find the general value of 𝜃 4 in radians. Answer: your discriminate is given by 1 − cos(4𝜃) + sin(2𝜃) and can be factorized in to 2 sin(𝜃) cos(𝜃) (2 sin(2𝜃) + 1) can you solve this? cos(4𝜃) = 8(cos(𝜃))4 − 8(cos(𝜃))2 + 1 and 1 + 8 cos(𝜃)4 − 8 cos(𝜃)2 + sin(2𝜃) − 1 this is 8 cos(𝜃)2 1 − cos(𝜃 2 ) + sin(2𝜃) 8 sin(𝜃)2 cos(𝜃)2 + sin(2𝜃) Question #8 Im trying to show that 1 1 + 𝑖𝑧 𝑧 = tan [ log (√ )] 𝑖 1 − 𝑖𝑧 Answer: Your idea of writing tan 𝑥 = 𝑒 𝑖𝑥 −𝑒 −𝑖𝑥 𝑖(𝑒 𝑖𝑥 +𝑒 −𝑖𝑥 ) 1 1+𝑖𝑧 𝑖 1−𝑖𝑧 is a good start. Substituting 𝑥 = log √ shows that 1 + 𝑖𝑧 1 − 𝑖𝑧 1 √1 − 𝑖𝑧 − √1 + 𝑖𝑧 tan 𝑥 = . 𝑖 1 + 𝑖𝑧 1 − 𝑖𝑧 √ +√ 1 − 𝑖𝑧 1 + 𝑖𝑧 Mutiplying the numerator and denominator by √(1 + 𝑖𝑧)(1 − 𝑖𝑧) yields tan 𝑥 = 1 (1+𝑖𝑧)−(1−𝑖𝑧) 𝑖 (1+𝑖𝑧)+(1−𝑖𝑧) = 1 2𝑖𝑧 𝑖 2 = 𝑧, which isn't as nasty a calculation as you might have feared. Question #9 Implict differentiation: 𝑥 cos(4𝑥 + 3𝑦) = 𝑦 sin 𝑥 Answer: sin(4𝑥 + 3𝑦) + 𝑥 cos(4𝑥 + 3𝑦) (4 + 3𝑑𝑦/𝑑𝑥) = (𝑑𝑦/𝑑𝑥) cos 𝑥 − 𝑦 sin 𝑥 sin(4𝑥 + 3𝑦) + 4𝑥 cos(4𝑥 + 3𝑦) + 3(𝑑𝑦/𝑑𝑥) cos(4𝑥 + 3𝑦) = (𝑑𝑦/𝑑𝑥) cos 𝑥 − 𝑦 sin 𝑥 sin(4𝑥 + 3𝑦) + 4𝑥 cos(4𝑥 + 3𝑦) + 𝑦 sin 𝑥 = (𝑑𝑦/𝑑𝑥) cos 𝑥 + 3(𝑑𝑦/𝑑𝑥) cos(4𝑥 + 3𝑦) sin(4𝑥 + 3𝑦) + 4𝑥 cos(4𝑥 + 3𝑦) + 𝑦 sin 𝑥 = (𝑑𝑦/𝑑𝑥)(cos 𝑥 + 3 cos(4𝑥 + 3𝑦)) Question #10 𝑎 In a △ 𝐴𝐵𝐶, = 2 + √3 and ∠𝐶 = 60∘ . Then find the ordered pairs (∠𝐴, ∠𝐵) = 𝑏 Answer: sine rule sin 𝐴 sin 𝐵 sin 𝐶 = = 𝑎 𝑏 𝑐 𝑎 sin 𝐴 = = 2 + √3 𝑏 sin 𝐵 𝐴 + 𝐵 = 180 − 𝐶 = 120 𝐵 = 120 − 𝐴 sin 𝐴 = 2 + √3 sin 𝐵 sin 𝐴 = 2 + √3 sin(120 − 𝐴) now solve for A sin 𝐴 = 2 + √3 sin 1 20 cos 𝐴 − cos 1 20 sin 𝐴 but it seem to solve for B is easier ! sin(120 − 𝐵) = 2 + √3 sin 𝐵 1 √3 cos 𝐵 − sin 𝐵 2 2 = 2 + √3 sin 𝐵 1 √3 cot 𝐵 − = 2 + √3 2 2 Question #11 In a triangle ABC, AB=10 cm and AC=5 cm. The area is 15𝑐𝑚2 and the angle BAC is equal to 𝜃. Give two possible values of 𝜃 I was able to find one of the values of 𝑐𝑜𝑠 𝜃 which is 0.8 but how can I get a second value for 𝑐𝑜𝑠 𝜃 ? Answer: The area of the triangle will be equal to (𝐴𝐵)(𝐴𝐶) sin 𝜃 2 meaning that sin 𝜃 = 0.6, as you probably found. This means that cos 2 𝜃 = 1 − sin2 𝜃 = 0.64 and cos 𝜃 = ±0.8 The positive value corresponds to 𝜃 an acute angle, and then negative value corresponds to obtuse 𝜃 . Question #12 In any triangle, if cos 𝐴+2 cos 𝐶 cos 𝐴+2 cos 𝐵 = sin 𝐵 sin 𝐶 , prove that the triangle is either isosceles or right angled. My Attempt: Given: cos 𝐴 + 2 cos 𝐶 sin 𝐵 = cos 𝐴 + 2 cos 𝐵 sin 𝐶 𝑏 2 + 𝑐 2 − 𝑎2 𝑎2 + 𝑏 2 − 𝑐 2 +2 𝑏 2𝑏𝑐 2𝑎𝑏 = 𝑏 2 + 𝑐 2 − 𝑎2 𝑎2 + 𝑐 2 − 𝑏 2 𝑐 +2 2𝑏𝑐 2𝑎𝑐 On simplification, 𝑎𝑏 2 + 𝑎𝑐 2 − 𝑎3 + 2𝑎2 𝑐 + 2𝑏 2 𝑐 − 2𝑐 3 𝑏 = 𝑎𝑏 2 + 𝑎𝑐 2 − 𝑎3 + 2𝑎2 𝑏 + 2𝑏𝑐 2 − 2𝑏 3 𝑐 Answer: Alternate approach: Cross multiply cos 𝐴 sin 𝐶 + sin 2 𝐶 = sin 𝐵 cos 𝐴 + sin 2 𝐵 cos 𝐴 (sin 𝐶 − sin 𝐵) − (sin 2 𝐵 − sin 2 𝐶) = 0 cos 𝐴 ∗ 2 ∗ sin 𝐶−𝐵 𝐵+𝐶 cos − 2 ∗ sin(𝐵 − 𝐶) cos(𝐵 + 𝐶) = 0 2 2 cos 𝐴 ∗ 2 ∗ sin 𝐶−𝐵 𝐵+𝐶 cos + 2 ∗ sin(𝐵 − 𝐶) cos 𝐴 = 0 2 2 2 cos 𝐴 [sin 𝐶−𝐵 𝐵+𝐶 𝐶−𝐵 𝐵−𝐶 cos + 2 sin cos ]=0 2 2 2 2 𝐵−𝐶 𝐵+𝐶 𝐵−𝐶 [cos − 2 cos ]=0 2 2 2 2 cos 𝐴 ∗ sin IF: cos 𝐵+𝐶 2 − 2 cos 𝐵−𝐶 2 = 0 ⇒ tan 𝐵 2 tan 𝐶 2 =− 1 3 Question #13 In triangle ABC, the angle(BAC) is a root of the equation √3 cos 𝑥 + sin 𝑥 = Answer: 1 2 We have 1 1 1 √3 cos(𝑥) + sin(𝑥) = ⇒ sin(𝑥 + 𝜋/3) = 2 2 4 4 We hence obtain 𝑥 + 𝜋/3 = arcsin(1/4) or 𝑥 + 𝜋/3 = 𝜋 − arcsin(1/4). Note that arcsin(1/4) < arcsin(1/2) = 𝜋 6 𝜋 < . Since 𝑥 ∈ (0, 𝜋), we have 3 𝑥 + 𝜋/3 = 𝜋 − arcsin(1/4) ⇒ 𝑥 = 𝜋 − (arcsin(1/4) + 𝜋/3) > 𝜋/2 Hence, the triangle is obtuse. Question #14 In which point the tangent of 𝑦 = sin 𝑥 − cos 𝑥 is parallel with y=x? Answer: You need to solve (sin 𝑥 − cos 𝑥)′ = 1 which is cos 𝑥 + sin 𝑥 = 1 or 𝜋 1 sin ( + 𝑥) = … 4 √2 Question #15 Injectivity and range of arctan (√ 1+𝑥 1−𝑥 ) According to the definition, for a function to be injective 𝑓(𝑎) = 𝑓(𝑏) → 𝑎 = 𝑏 for all 𝑎, 𝑏 ∈ 𝐷𝑓 Using this I get: arctan (√ ⟺ (√ ⟺ 1+𝑎 1+𝑏 ) = arctan (√ ) 1−𝑎 1−𝑏 1+𝑎 1+𝑏 ) = (√ ) 1−𝑎 1−𝑏 1+𝑎 1+𝑏 = 1−𝑎 1−𝑏 ⟺ 2𝑎 = 2𝑏 𝑎=𝑏 Hence the function is injective. Now this could be total bs but I also am not able to find the proper range 𝜋 for this one. WolframAlpha says 0 ≤ 𝑦 < . But how does one come up with this range? 2 Answer: 𝜋 The domain is −1 ≤ 𝑥 < 1, lim− 𝑓(𝑥) = , 𝑓(−1) = 0 and f is a continuous function. 2 𝑥→1 𝜋 Thus, the range is [0, ) 2 Question #16 Integral 𝜋/2 ∫ 0 6 sin 𝑥 arctan2 ( ) 𝑑𝑥 3 + cos 2 𝑥 Answer: I will refer to the following result 𝜋 2 𝐼(𝑟, 𝑠) = ∫ arctan(𝑟 sin 𝜃) arcsin(𝑠 sin 𝜃) 𝑑𝜃 0 = 𝜋𝜒2 ( √1 + 𝑟 2 − 1 √1 + 𝑠 2 − 1 × ) 𝑟 𝑠 where 𝜒2 is the Legendre chi function. Using the addition formula for the arctangent, it follows that 3 sin 𝑥 6 sin 𝑥 1 arctan ( ) = arctan ( 2 ) = arctan(sin 𝑥) + arctan ( sin 𝑥) 1 3 + cos 2 𝑥 2 1 − sin2 𝑥 2 So it follows that 𝜋 2 6 sin 𝑥 1 1 1 ∫ arctan2 ( ) 𝑑𝑥 = 𝐼(1,1) + 2𝐼 (1, ) + 𝐼 ( , ) 3 + cos 2 𝑥 2 2 2 0 which reduces to a combination of Legendre chi functions 𝜋 [𝜒2 (3 − 2√2) + 𝜒(9 − 4√5) + 2𝜒 ((√2 − 1)(√5 − 2))] Question #17 Integrate tan2 ( 𝜋 12 ⋅ 𝑦) Answer: Let 𝜋 12 ⋅𝑦 =𝑡 ⇒ 𝜋 12 𝑑𝑦 = 𝑑𝑡 ⇒ 𝑑𝑦 = We have ∫ tan2 ( 𝜋 12 ⋅ 𝑦) 𝑑𝑦 = ∫ tan2 (𝑡) 𝑑𝑡 12 𝜋 12 𝜋 𝑑𝑡 = 12 ∫ tan2 (𝑡) 𝑑𝑡 𝜋 = 12 12 ∫ sec 2 (𝑡) 𝑑𝑡 − ∫ 𝑑𝑡 𝜋 𝜋 = 12 12 tan(𝑡) − ⋅𝑡+𝐶 𝜋 𝜋 Setting 𝑡 = 𝜋 12 ⋅𝑦 = 12 𝜋 12 𝜋 tan ( ⋅ 𝑦) − ⋅ ⋅𝑦+𝐶 𝜋 12 𝜋 12 = 12 𝜋 tan ( ⋅ 𝑦) − 𝑦 + 𝐶 𝜋 12 Question #18 Integrate the function 2𝑥 3 + 5𝑥 2 + 8𝑥 + 4 (𝑥 2 + 2𝑥 + 2)2 Answer: Let us start with 2𝑥 3 + 5𝑥 2 + 8𝑥 + 4 𝐴 + 𝐵𝑥 𝐶 + 𝐷𝑥 = 2 + 2 2 2 (𝑥 + 2𝑥 + 2) 𝑥 + 2𝑥 + 2 (𝑥 + 2𝑥 + 2)2 Remove the denominator, expand and group the terms. You should arrive to 2𝑥 3 + 5𝑥 2 + 8𝑥 + 4 = 𝐵𝑥 3 + (𝐴 + 2𝐵)𝑥 2 + (2𝐴 + 2𝐵 + 𝐷)𝑥 + (2𝐴 + 𝐶) which gives the four equations 𝐵 = 2, 𝐴 + 2𝐵 = 5 so 𝐴 = 1, 2𝐴 + 2𝐵 + 𝐷 = 8 𝑠𝑜 𝐷 = 2, 2𝐴 + 𝐶 = 4 so 𝐶=2 So we have 2𝑥 3 + 5𝑥 2 + 8𝑥 + 4 1 + 2𝑥 2 + 2𝑥 = 2 + 2 2 2 (𝑥 + 2𝑥 + 2) 𝑥 + 2𝑥 + 2 (𝑥 + 2𝑥 + 2)2 = 2𝑥 + 2 − 1 2𝑥 + 2 + 𝑥 2 + 2𝑥 + 2 (𝑥 2 + 2𝑥 + 2)2 So, setting 𝑢 = 𝑥 2 + 2𝑥 + 2, you can notice that 2x+2 is just u' which make things very simple for two parts and what is basically left is ∫ 𝑑𝑥 𝑑𝑥 =∫ (𝑥 + 1)2 + 1 𝑥 2 + 2𝑥 + 2 I am sure that you can take from here. Question #19 Inverse Trigonometry System of Equations 2 tan−1 (√𝑥 − 𝑥 2 ) = tan−1 (𝑥) + tan−1 (1 − 𝑥) I have a feeling solution includes drawing triangles but cannot make the leap to get the solution Answer: Take tangent of both sides (apply double angle and sum of angle formula and remember that tan(tan−1 𝑦) = 𝑦 2√𝑥 − 𝑥 2 𝑥+1−𝑥 = 2 1 − (𝑥 − 𝑥 ) 1 − 𝑥(1 − 𝑥) Question #20 Is a following function periodic 𝑓(𝑥) = tan 𝑥 + tan(√2𝑥) Answer: It's easy to show that a differentiable function f is periodic only if f' is periodic. So 𝑓 ′ (𝑥) = 1 + tan2 (𝑥) + √2 + √2 tan2 (√2𝑥) If f' has a period T>0 then 1 + √2 = 1 + tan2 (𝑇) + √2 + √2 tan2 (√2𝑇) This implies that tan 𝑇 and tan(√2𝑇) are both 0, which is impossible for the same reason as for 𝑐𝑜𝑠