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

2024
Views:

76

Pages:

3
Author:

Bailey Ayala

Question 1 Please, help to find a model for simple harmonic motion satisfying the specified conditions. Displacement (t = 0) 0, Amplitude 3 meters, Period 6 seconds. Answer 1 a=3 2π/ω = 6 ω = 2π/6 = π/3 d = 3sin(π/3)t Answer 2 A = 3, 2π/ω = 6. ω = π/3, THEN d = 3sin(π/3)t Question 2 Find the Value of each trigonometric ratio to the radical form if necessary. sin 45° Answer The value of sin 45° in decimal form is 0.7071067812 sin(x) = opposite side/hypotenuse sin 0° = √0/4 = 0 sin 30° = √1/4 = 1/2 sin 45° = √2/4 = 1/√2 sin 60° = √3/4 = √3/2 sin 90° = √4/4 = 1 From the above values of sin the value of sin 45° = 1/√2 Question 3 Determine the exact value of expression. tan (330) x cos (240) - 2cos (270) Answer Given, tan (330) × cos (240) - 2cos (270) Now, tan (330) × cos (240) - 2cos (270) = tan (360-30) × cos (180+60) - 2cos (270) = -tan (30) × (-cos (60)) - 2(0) = -1/√3 × (-1/2) - 0 = 1/2√3 Question 4 Determine the exact value of expression. tan (60) × 3sin (90) – sin (315) Answer Given, tan (60) × 3sin (90) – sin (315) Now, tan (60) × 3sin (90) - sin (315) = √3 x 3(1) - sin (360 – 45) = 3√3 + (-sin (45)) = 3√3 + 1/√2 = 3√3 × √2+1 / √2 = (3√6+1)/√2 = (3√6+1)/√2 × √2/√2 = (6√3+√2)/2 Question 5 Determine the exact value of expression. sec (210) x cot (300) + sin (225) Answer Given, sec (210) x cot (300) + sin (225) Now, sec (210) x cot (300) + sin (225) = sec (180+30) × cot (360-60) + sin (180+45) = -sec (30) × (-cot (60)) + (-sin (45)) = (-2/√3) x (-1/√3) + (-1/√2) = 2/3 - 1/√2 = (2√2 - 3)/(3√2) = (2√2-3)/3√2 *√2/√2 = (4-3√2)/6