University:
Washington State University
Course:
MATH 172 | Calculus II
Academic year:

2024
Views:

435

Pages:

2
Author:

Bailey Ayala

Question 1 12 sin (5x/2) Precalc Trig Identity, verify: 1 + cos(x) + cos (2x) = + 2sin (x/2) Answer For sin(x/2) ≠ 0 we obtain: 1 + cos(x) + cos 2x = (2sin(x/2) + 2sin(x/2)cos(x) + 2sin(x/2)cos(2x))/(2sin(x/2)) = (2sin(x/2) + 2sin(x/2)cos(x) + 2sin(x/2)cos(2x))/(2sin(x/2)) = (2sin(x/2) + sin(3x/2) + sin(-x/2) + sin(5x/2) + sin(-3x/2))/(2sin(x/2)) = (2sin(x/2) + sin(3x/2) - sin(x/2) + sin(5x/2) - sin(3x/2))/(2sin(x/2)) = (sin(x/2) + sin(5x/2))/(2sin(x/2)) = (1/2) * 2(sin(5x/2) + sin(x/2))/2sin(x/2) = sin(3x/2)*cos(x)+sin(3x/2)*cos(2x) = 12 sin (5x/2)/1 + 2sin (x/2) I used the following formula. sin αcos β = 1/2(sin (α + β) + sin (α – β)) For example, 2sin(x/2)cos(x) = 2 * (1/2)(sin(x/2 + x) + sin(x/2 - x)) = sin(3x/2) + sin(-x/2) = sin(3x/2) - sin(x/2) Question 2 Proving an upper bound for Σ(k=0 to N) (k²/N³)*cos(kx/N) I'm struggling to prove, that there exists a constant C, so that for every N∈ N and every x ∈ [0, Nπ], the following holds: |Σ(k=0 to N) (k²/N³)*cos(kx/N)| ≤ C • (x⁻¹ + x⁻² + x⁻³) Answer Let us consider f(y) = y²cos(xy). Then ∫f(y),dy = (2y²sin(xy) + (xy² - 2)cos(xy) + 2ysin(xy))/x³ and the difference between this integral and the associated Riemann sum Σ(k=0 to N) (k²/N³)*cos(kx/N) is bounded by x/N² due to the Lipschitz-continuity of cos and summation/integration by parts. It follows that your bound holds as soon as N|x| Question 3 Approximating cos (47°) Given that cos (45°) = √2/2 what would cos (47°) be. Using differential approximation, I get cos (47°) is about cos(45π/180) - (2sin(47π/180) * (π/180)) = -0.755600622 which is of course not right as cos (47°) = 0.68199836 cos (47°) ≈ 0.68199836 Answer 1 When you write cos'(θ) = -sin (θ) (dropping the minus sign) you need to measure θ in radians. That comes out in your formula in the times 2, which should be times 2 * π/180 So cos (47°) ≈ cos (45°) - ((2π/180)sin (45°)) ≈ (√2/2)(1-0.035) ≈ 0.682 Answer 2 Check your units. The general form for differential approximation is f(x₀) + (x-x₀) * (df/dx) You convert your 47, which I'm assuming is in degrees to radians by multiplying by π/180 This is fine. But then you use 2 degrees (I'm assuming) as your x - x₀ term. You need this to be in radians. Try your computation again, using the same method you've already used to convert from degrees to radians for 2 degrees. The answer you then get is off by about 0.004. But I'll leave it to you to check which way it is off (±)