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

2024
Views:

325

Pages:

2
Author:

Bailey Ayala

Question 1 Verify the identity: a) cos 2x = cos²x - sin²x b) cos 2x = 1 - 2sin²x Answer 1 Step 1 Use cos (A + B) = cos Acos B – sin Asin B cos 2x = cos (x + x) = cos x cos x – sin x sin x = cos²x - sin²x Step 2 cos²x - sin²x = 1 - sin²x - sin²x = 1 - 2sin²x Answer 2 Step 1 cos 2x = cos²x - sin²x Left Side: cos (A + B) = cos Acos B - sin Asin B cos 2x = cos (x + x) = cos x cos x - sin x sin x = cos²x - sin²x cos 2x = cos²x - sin²x cos 2x = cos (x + x) = cos x cos x - sin x sin x = cos²x - sin²x (using pythagorian identity) cos²x = 1 - sin²x = 1 - sin²x - sin²x = 1 - 2sin²x Question 2 How to evaluate limit without LHospitals rule, but using lim (sin x)/x = 1 x->0 I am having hard time evaluating this limit: lim (cos 3x - 1)/(cos 2x - 1) x->0 I know, that this fact can help lim (sin x)/x = 1 x->0 But how? We can transform the first expression to lim (cos 3x - 1)/(-2sin²x) x->0 but what should I do next? I cannot use LHopitals rule for solving this. Answer 1 lim (cos 3x - 1)/(cos 2x - 1) = lim ((1 - cos 3x)/(1 - cos 2x)) x->0 x->0 = lim ((9x²/2)/(4x²/2)) = lim (9/4) x->0 x->0 = 9/4 Answer 2 Hint: It is standard (from the limit of (sin x)/x) that lim (1 - cos x)/x² = 1/2, so 1 - cos x ~ x²/2 x->0 and you may proceed by substitution. Question 3 Verify the trig identity (cot²θ - 1)/csc²θ = cos 2θ - 1 Answer Yes indeed we have (cot²θ - 1)/csc²θ = (cos²θ/sin²θ - 1)/(1/sin²θ) = (cos²θ - sin²θ)/(1/sin²θ * sin²θ) = cos²θ - sin²θ = cos(2θ) The identity provided is incorrect. (cot²θ - 1)/csc²θ = cos(2θ) Question 4 Where is the angle for 7π/8 in a unit circle? I am having difficulty finding the angle for 7π/8 in a unit circle. I have found the coordinates for π/8 and have to use the symmetry to locate the coordinates for 7π/8. Is the angle just below the (1,0) and symmetry to π/8? Answer 1 7π/8 = π - π/8. So you must rotate your point clockwise through an angle π. This will take you to a point directly opposite the point you started from. Answer 2 7π/8 is in the third quadrant. So, using π/8 as your reference angle, make both coordinates negative. (It is also coterminal with 9π/8 since they differ by 2π.)