Questions and Answers #73 Trigonometry

Questions and Answers Sheet 73 Trigonometry Question #1 What will be the minimum value of this trigonometric function? 5 sin2 𝑥 − 12 sin 𝑥 cos 𝑥 + 10 cos 2 𝑥 My doubt is that as this can be written as (2 sin 𝑥 − 3 cos 𝑥)2 + 1 and minimum value of 2 sin 𝑥 − 3 cos 𝑥 is -13, so will the minimum value be 14, same as maximum value? Answer: HINT you are minimizing 𝑓(𝑥) = 𝑔(𝑥)2 + 1 and the minimum of f will occur in the same place as the minimum of 𝑔2 . Clearly 𝑔(𝑥)2 ≥ 0 so if you can set g(x)=0 you will get the minimum you seek. Question #2 When is sin3 (𝑥) > cos 3(𝑥) ? Answer: As 𝑡 ↦ 𝑡 3 is continuous and strictly increasing we have 𝑥 < 𝑦 ⟺ 𝑥 3 < 𝑦 3 and so sin(𝑥)3 > cos(𝑥)3 ⟺ sin(𝑥) > cos(𝑥) Question #3 1 1 2 √5 Which is bigger tan−1 ( ) or ? Answer: 1 sin(𝑥) 2 cos(𝑥) Suppose tan(𝑥) = . Then in particular you know the last two equalities you find that sin(𝑥) = 1 √5 1 = . You also know sin2 (𝑥) + cos 2 (𝑥) = 1. Using 2 . Since 𝑥 ≥ sin(𝑥), you're done. Question #4 Why can't 1 sin 𝑥 cos 𝑥 be expressed in the form 𝐴 sin 𝑥 + 𝐵 cos 𝑥 1 A B = +  1 = A cos x + B sin x sin( x) cos( x) sin x cos x π I let 𝑥 = , getting A=-1. Then I let 𝑥 = π, getting B=1. This means that 1 = cos 𝑥 − sin 𝑥 which is 2 obviously wrong most of the time. Answer: we can write this as −4𝑖 (𝑒 −𝑖𝑥 − 𝑒 𝑖𝑥 )(𝑒 −𝑖𝑥 + 𝑒 𝑖𝑥 ) So by letting 𝑢 = 𝑒 𝑖𝑥 we can write this as −4𝑖 𝑢2 4𝑖 4𝑖 4𝑖 = −4𝑖 = − + 2 2 2 (1/𝑢 − 𝑢)(1/𝑢 + 𝑢) (1 − 𝑢 )(1 + 𝑢 ) 2(𝑢 + 1) 4(𝑢 + 1) 4(𝑢 − 1) assuming I did my calculations correctly. Note that this isn't going to combine together in the way you want. Partial fractions are fundamentally property of rational functions of polynomials. Question #5 Why sec2 𝑥 √sec2 𝑥 ≠ sec 𝑥? 𝑑 When I evaluate 𝑑𝑥 (sinh−1 (tan 𝑥)), i get: sec2 𝑥 √1+tan2 𝑥 At this point I want to use the sec 2 𝑥 = 1 + tan2 𝑥 identity to simplify to: sec 2 𝑥 = sec 𝑥 sec 𝑥 Answer: Formally, you get sec2 𝑥 √1+tan2 𝑥 = sec2 𝑥 √sec2 𝑥 sec2 𝑥 = |sec 𝑥| = |sec 𝑥| Question #6 Why should we have sin2 (𝑥) = 1−cos(2𝑥) 2 knowing that sin2 (𝑥) = 1 − cos 2(𝑥) ? Logically, can you not subtract cos 2 (𝑥) to the other side from this Pythagorean identity sin2 (𝑥) + cos 2 (𝑥) = 1 ? When I look up trig identities, however, it says sin2 (𝑥) = 1−cos(2𝑥) 2 Answer: Notice that cos 2(𝑥) ≔ (cos(𝑥))2 is not the same thing as cos(2𝑥). It is indeed true that sin2 (𝑥) = 1 − cos 2 (𝑥) and that sin2 (𝑥) = 1−cos(2𝑥) 2 Question #7 Why solving trigonometry equation for 𝑥𝑦 ′ = 𝑦 cos ln 𝑦 𝑥 Answer: With 𝑢 = ln 𝑦 − ln 𝑥 𝑢′ = 𝑦′ 1 1 − = (cos 𝑢 − 1) 𝑦 𝑥 𝑥 This has stationary solutions 𝑢 = 𝑐 = 𝑐𝑜𝑛𝑠𝑡, 𝑦 = 𝑒 𝑐 𝑥 for the roots of cos 𝑢 − 1 = 0 which are at 𝑐 = 2𝑘π . For all other points you can separate the equation to get 𝑢 cot ( ) = ln 𝑥 + 𝑐 2 Question #8 Would someone recommend a method of proving this relation? | e −2 x sin( x) | x , 1 + x2 x0 Unfortunately differentiation leads to messy transcendental expressions. Answer: |sin 𝑥| ≤ 𝑥, for 𝑥 ≥ 0 is well-known. On the other hand: 𝑒 2𝑥 = 1 + 2𝑥 + (2𝑥)2 + ⋯ = 1 + 2𝑥 + 2𝑥 2 + ⋯ ≥ 1 + 𝑥 2 2! therefore 𝑒 −2𝑥 ≤ 1 1+𝑥 2 taking products, 𝑒 −2𝑥 |sin 𝑥| ≤ 𝑥 1+𝑥 2 Question #9 Writing sin 𝑥+3 cos 𝑥 2 cos 𝑥 in a form that uses only one trig function (sin , cos , tan , 𝑒𝑡𝑐) Answer: sin 𝑥 + 3 cos 𝑥 sin 𝑥 3 cos 𝑥 1 3 = + = tan 𝑥 + 2 cos 𝑥 2 cos 𝑥 2 cos 𝑥 2 2 Question #10 𝑥 Writing tan2 (2 sec −1 ( )) in algebraic form 3 Answer: 𝑥 𝑥 3 3 3 𝑥 Let θ = sec −1 ( ) thus sec θ = cos θ = since sec = 1/ cos 9 3 9 𝑥 𝑥2 You have cos θ = . So sin θ = √1 − So tan2 (2θ) = ( x2 − 9 4 9 2 x −9 2 (1 − ) 9 2 tan θ 1−tan2 θ 2 ) = and tan θ = √1− 2 𝑥 3 𝑥 = √𝑥 2 −9 3