Final Exam Review PDF

## Final Exam Review ### 1) - degree radians - 30° → π/6 - 45 ← π/4 ### 2) - Convert to degrees; θ = 3.75 - 3.75 x 180/π = 214° ### 3) - Find Cos θ if the terminal side passes through the point (15, -20) - r = √15² + (-20)² = 25 - Cos θ = 15/25 = 3/5 ### 4) - A right triangle with sides 5, 1...

## Final Exam Review ### 1) - degree radians - 30° → π/6 - 45 ← π/4 ### 2) - Convert to degrees; θ = 3.75 - 3.75 x 180/π = 214° ### 3) - Find Cos θ if the terminal side passes through the point (15, -20) - r = √15² + (-20)² = 25 - Cos θ = 15/25 = 3/5 ### 4) - A right triangle with sides 5, 12 and 13. - Sec θ = 13/12 - Cot θ = 12/5 ### 5) - See #4 ### 6) - Solve the right triangle, a = 4, b = 5. - c = √4² + 5² = √41 - tan A = 5/4 - A = tan⁻¹(5/4) = 39° - B = 90 - 39 = 51° ### 7) - Find Cos θ given sin θ = 4/√15, θ in QII - Cos θ = -√15/4 ### 8) - Find the period - y = -3 Cos(3x + π/3) - 2π/3 or 120° ### 9) - Graph y = 1/2 Cos (2x + π) + 1 - Period = 2π/2 = π - Phase Shift = -π/2 - Amplitude = 1/2 ### 10) - Write in terms of sin and cos - Csc x + Cot x = 1/sin x + cos x/sin x = 1 + cos x/sin x ### 11) - Simplify Cos x - 1/sin(-x) - Sin² x + Cos² x = 1 - Cos² x - 1 = -Sin² x ### 12) - & See #11 (Basic Identity) ### 13) - Cos A = 1/√2, Sin B = -3/√13. A in QIV, B in QIII - Find Sin (A+B) - Sin (A + B) = Sin A Cos B + Cos A Sin B = (1/√2)(-2/√13) + (1/√2)(-3/√13) = -2√30 - 1/2√13 ### 14) - Find the exact value of Cos x Cos 25 - Sin x Sin 25 = Cos (x + 25) - = Cos 30° = √3/2 ### 15) - Given Sin 195° = ± √1 - Cos² 390°/2 - Positive or negative - 195° in QIII - Sin 195° is negative ### 16) - Identity or not and Identity - Tan x = Sec x - Sin x / Cos x = 1/ Cos x - Identity ### 17) - Find θ given Sin θ = 0.261 - θ = Sin⁻¹(0.261) = 15. one or 180 - 15 - 1/2 = 165°. - θ = 15 + 360k° or θ = 165 + 360k° ### 18) - Find θ if Cot θ = 3 - Tan θ = 1/3 - θ = Tan⁻¹(1/3) = 18.43° ### 19) - Solve 2 Cos x = 1 - Cos x = 1/2 - X = 60° or X = 300° - X = π/3 or X = 5π/3 - {x = π/3 + 2πk or x = 5π/3 + 2πk} ### 20) - Solve the triangle ABC - c = 10, A = 25°, B = 53°. - ASA - Law of Sines - C = 180° - 25° -53° = 102° - a/sinA = c/sinC and b/sinB = c/sinC ### 21) - Determine the number of triangles - a) a = 7, b = 6, c = 15 - 0 triangles - b) a = 8, b = 12, A = 30°. - Sin B = b Sin A / a = 12 Sin 30°/8 = 6/8 = 3/4 - B = Sin⁻¹(3/4) = 49° or 131°. - 2 triangles ### 22) - Find the area a = 3, b = 4, c = 5 - A = √s(s-a)(s-b)(s-c) = √6(6-3)(6-4)(6-5) - =√6(3)(2)(1) = 6 ### 23) - Find |Vx| and |Vy| given |v| = 40 θ = 60° - |Vx| = |v| Cos θ = 40 Cos 60° = 20 - |Vy| = |v| Sin θ = 40 Sin 60° = 40√3/2 = 20√3 = 34.64 ### 24) - Convert to Rectangular Coordinates - (-6, -π/2) - x = r Cos θ = -6 (Cos (-π/2)) = -6 - y = r Sin θ = -6 Sin (-π/2) = .164 ### 25) - Write equivalent rectangular equation - r = 7 Cos θ. - r² = 7 x - r² = x² + y² - x² + y² = 7x