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

2024
Views:

464

Pages:

3
Author:

Bailey Ayala

Question 1 Use the unit circle to evaluate the six trigonometric functions of: a) 90° b) 180° Answer Given: Use the unit circle to evaluate the six trigonometric functions: a) 90° b) 180° Any point in on the unit circle is given by: ⇒ (cos θ, sin θ) a) 90°: Point on the unit circle is given by: ⇒ (cos 90°, sin 90°) ⇒ (0,1) So, ⇒ sin 90° = 1 ⇒ cos 90° = 0 ⇒ tan 90° = sin 90°/cos 90° = 1/0 = not defined ⇒ csc 90° = 1/sin 90° = 1/1 = 1 ⇒ sec 90° = 1/cos 90° = 1/0 = not defined ⇒ cot 90° = 1/tan 90° = 0 b) 180°: Point on the unit circle is given by: ⇒ (cos 180°, sin 180°) ⇒ (-1,0) So, sin 180° = 0 cos 180° = -1 tan 180° = sin 180°/cos 180° = 0/-1 = 0 csc 180° = 1/sin 180° = 1/0 = not defined sec 180° = 1/cos 180° = 1/-1 = -1 cot 180° = 1/tan 180° = not defined Question 2 Use the Cofunction Theorem to fill in the blank so that the expression becomes a true statement. sec y = csc ___ Answer Trigonometry's co-function identities include: If a function g and f work together, then f(A) = g(B) when A and B are complementary angles. sinA= cos (90° - A) cosA = sin (90° - A) sinA= cosB, If A + B = 90° secA = csc (90° - A) cscA = sec (90° - A) secA = cscB, if A + B = 90° tanA = cot (90° - A) cotA = tan (90° - A) tanA = cotB, if A + B = 90° The co-function identities of trigonometry have been covered previously. sec y = csc (90-y) Question 3 Answer using ASTC chart: 5csc²θ - 4cscθ - 1 = 0 Answer Solution: 5csc²θ - 4cscθ - 1 = 0 5csc²θ - 5cscθ + cscθ - 1 = 0 5cscθ[cscθ - 1] + 1[cscθ - 1] = 0 (5cscθ + 1)(cscθ - 1) = 0 5cscθ + 1 = 0 or cscθ - 1 = 0 cscθ = -1/5 or cscθ = 1 sinθ = -5 or sinθ = 1 θ = not possible θ = 90° Question 4 Find all solution of the equation in the interval [0, 2π]. (cos x - 1)(sin x + 1) = 0 Answer (cos x - 1)(sin x + 1) = 0 The given equation's solution, i.e., must now be found. The entire expression will be equal to 0 if any one factor on the left side of the equation is equal to 0. cos x - 1 = 0 sin x + 1 = 0 Set the first factor equal to 0 cos x = 1 x = 2π - 0 Similarly sin (x) = -1 Taking the inverse sine of both sides of the equation to exact x from inside the sine x = arcsin (-1) The exact value of arcsin (-1) is -π/2 x = -π/2 but it should be between [0, 2π] x = 2π + (-π/2) x = 3π/2 Question 5 Find all solution of the equation in the interval [0, 2π]. sin 2x(csc 2x - 2) = 0 Answer sin 2x(csc 2x - 2) = 0 So. for the solution of the equation sin 2x = 0 2x = arcsin (0) 2x = 180° or 2x = π x = π/2 or π Or 2x = 2π x = π x= π/2 or π Similiarly, csc 2x - 2 = 0 csc 2x = 2 2x = arccos (2) 2x = π/6 or 2x = π - π/6 So, x ∈ π/12 or 5π/12