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

2024
Views:

50

Pages:

2
Author:

Bailey Ayala

Question 1 Calculate ∑(n=1 to ∞) arctan(2√2/(k²+1)) Answer Note that arctan(2√2/(k²+1)) = arctan((k+1/√2) - (k-1/√2)) / (1 + (k+1/√2)*(k-1/√2)) = arctan(k+1/√2) - arctan(k-1/√2) Therefore, ∑(k=1 to n) arctan(2√2/(k²+1)) = ∑(k=1 to n) [arctan((k+1)/√2) - arctan((k-1)/√2)] = arctan((n+1)/√2) + arctan(n/√2) - arctan(1/√2) - arctan(0/√2) Ergo, ∑(k=1 to ∞) arctan(2√2/(k²+1)) = π - arctan(1/√2) ≈ 2.52611 Question 2 How is the trig identity tan²(x) = sec²(x) - 1 derived? Answer 1 tan²x = sin²x/cos²x = (sin²x + cos²x - cos²x)/cos²x = (1- cos²x)/cos²x = 1/cos²x - cos²x/cos²x = 1/cos²x - 1 = sec²x - 1 Answer 2 Divide the identity sin²x + cos²x = 1 by cos²x (sin²x + cos²x)/cos²x = 1/cos²x sin²x/cos²x + cos²x/cos²x = 1/cos²x tan²x + 1 = sec²x Question 3 How to solve cos(θ + π/3) equation? Answer Your step is ok, but it misses the "second" solution: θ + π/3 = ±arccos(1/√3) Move the π/3 over and you are done! Question 4 General Sine and Cosine formula for sum of a finite number of angles I was wondering if there is a general formula for sin (x1 + x2 + x3 + ... + xn) as well as for the cosine function. I know that sin (x1 + x2) = sin(x1)cos (x2) + cos(x1)sin (x2) and cos (x1+x2) = cos (x1) cos (x2) - sin (x1)sin (x2) But I want to find a general formula for the sum of a finite number of angles for the Sine and the cosine but I didn't noticed any pattern. I suspect that it may have a recursive pattern. Answer Well, for n=3: cos (x + y + z) = cos (x + y)cos (z) - sin (x + y)sin (z) = (cos(x)cos (y) – sin (x)sin (y))cos (z) - (sin(x)cos (y) + cos(x)sin (y))sin (z) It should be easy to do this for higher n as well, and for sin (x + y + z)