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

2024
Views:

155

Pages:

3
Author:

Bailey Ayala

Question 1 Use power reducing formula to simplify (cos 3x)⁴ to find powers of cosines Answer (cos3x)⁴ Knowing this cos(3x) = 4cos³x - 3cosx So, (cos3x)⁴ = (4cos³x - 3cosx)⁴ = (4cos³x - 3cosx)² (4cos³x - 3cosx)² = (16cos⁶x + 9cos²x - 24cos⁴x)(16cos⁶x + 9cos²x - 24cos⁴x) = 256cos¹²x + 144cos⁴x - 384cos¹⁰x + 144cos⁴x + 81cos⁴x - 216cos⁶x - 384cos¹⁰x - 216cos⁶x + 576cos⁸x = 256cos¹²x - 768cos¹⁰x + 864cos⁸x - 432cos⁶x + 225cos⁴x Question 2 Use the fundamental identities to find the value of the trigonometric function Find csc θ, given that sin θ = -1/23 and θ is in quadrant 4. Answer Given, sin θ = -1/23 and θ is in quadrant 4. We have to find the value of csc using the fundamental identities of trigonometry. Concept used: csc θ = 1/sin θ csc θ is negative in quadrant 4. Calculation sin θ = -1/23 ∴ csc θ = 1/sin θ = 1/(-1/23) = -23 But csc θ is negative in 4 quadrant. ⇒ csc θ = -23 Question 3 Prove that 2arcsin(x) = arccos(1-2x²) for x ∈ (0,1) Answer We have two ways to demonstrate this. The first is by employing function derivatives. Secondly, by applying trigonometric Let F(x) = 2arcsin(x), and G(x) = arccos(1-2x²) F'(x) = G'(x), so F'(x) = 2/√(1-x²) G'(x) = -1/√(1-(1-2x²)²) * (-4x) = 4x/√(1-(1-4x²+4x⁴) = 4x/√(4x²-4x⁴) = 4x/(2x√(1-x²) = 2/√(1-x²) therefore F'(x) = G'(x) In order to confirm that a constant is '0,' this means that F(x) and G(x) must differ by a constant. Replace with any value of the x test that you like, such as 0. F(0) = 2arcsin(0) = 0 while G(0) = arccos(1) = 0 constant is zero. So, F(x) = G(x) => 2arcsin(x) = arccos(1-2x²) Question 4 Two direction angles of a vector are given. Find the third direction angle, given that is either acute. α = π/3, γ = 2π/3, β is acute Answer According to given details α = π/3, γ = 2π/3 Using trigonometry identities cos²α + cos²β + cos²γ = 1 cos²(π/3) + cos²β + cos²(2π/3) = 1 (1/2)² + cos²β + (-1/2)² = 1 1/4 + cos²β + 1/4 = 1 cos²β = 1 - 1/2 = 1/2 cos β = √1/2 = 1/√2 β = π/4 β = 45° Question 5 Two direction angles of a vector are given. Find the third direction angle, given ths either obtuse α = 60°, β = 50°, γ is obtuse Answer According to given details α = 60°, β = 50° Using trigonometry identities cos²α + cos²β + cos²γ = 1 cos²(60°) + cos²(50°) + cos²γ = 1 (1/2)² + (0.6429)² + cos²γ = 1 1/4 + 0.4132 + cos²γ = 1 cos²γ = 1 - 0.6632 cos²γ = 0.3368 cos γ = √0.3368 γ = 125.47°