University:
University of North Georgia
Course:
MATH 1113 | Precalculus
Academic year:

2021
Views:

422

Pages:

6
Author:

CometConductor

Chapter 5 Test Review HPC Simplify the expression: csc²(θ)(1 - cos²(θ)) = csc²(θ)sin²(θ) = 1 (sec²x + tan²x)(sec²x - tan²x) / (sec²x + tan²x) = sec²x - tan²x = 1 (cosx + sinx) / (cosx + sinx) = 1 (sec²x - 1) / (secx - 1) = (secx + 1)(secx - 1) / (secx - 1) = secx + 1 (tanx + tany) / (cotx + coty) = (sinx/cosx + siny/cosy) / (cosx/sinx + cosy/siny) = (sinxcosy + sinycosx) / (cosxcosy + cosxsiny) = (sinxcosy + cosxsiny) / (cosxcosy + sinxcosy) = tanx * tany (sin(-x)cotx) / sin(π/2 - x) = (-sinx * cosx/sinx) / cosx = -cosx / cosx = -1 Verify the identity: sec²(x)cotx - cotx = tanx = cotx(sec²(x) - 1) = cotx * tan²(x) = tanx cot²(x) - cos²(x) = cot²(x)cos²(x) = cos²(x)/sin²(x) - cos²(x) = (cos²(x) - cos²(x)sin²(x)) / sin²(x) = cos²(x)(1 - sin²(x)) / sin²(x) = cos²(x)cos²(x) / sin²(x) = cot²(x)cos²(x) cos³xsin²x = (sin²x - sin⁴x)cosx = cosx(1 - sin²x)sin²x = cosx(cos²x)sin²x = cos³xsin²x sec²(x)(tan²x + 1) = sec⁴x = sec²(x)sec²(x) = sec⁴x 11. (1 + sec(x)) / (sin(x) + tan(x)) = -csc(x) (1 + 1/cosx) / (sinx + sinx/cosx) = (cosx + 1) / (sinxcosx + sinx) = (cosx + 1) / sinx(cosx + 1) = 1 / sinx = -csc(x) 12. (cotx - secx) / secx = csc(x) (cosx/sinx - 1/cosx) / (1/cosx) = (cos²x - sinx) / sinx = cosx/sinx - sinx/sinx = cotx - 1 13. (sec²x - tan²x + 1) / sinx = csc(x) (1/cos²x - sin²x/cos²x + 1) / sinx = (1 - sin²x + cos²x) / (cos²x * sinx) = (2cos²x) / (cos²x * sinx) = 2 / sinx = 2csc(x) 14. (cos²x + tan²x - 1) / sin²x = tan²(x) (cos²x - 1 + tan²x) / sin²x = (-sin²x + tan²x) / sin²x = -1 + tan²x / sin²x = -1 + sec²x = tan²x 15. csc(x - 3π/2) = sec(x) 1 / sin(x - 3π/2) = 1 / -cos(x) = -sec(x) 16. cos(x + π/2) = -sin(x) = cos(x)cos(π/2) - sin(x)sin(π/2) = 0 - sin(x) = -sin(x) 17. cos(x) - 1 = (cos(2x) - 1) / 2(cos(x) + 1) = (2cos²(x) - 2) / 2(cos(x) + 1) = 2(cos²(x) - 1) / 2(cos(x) + 1) = (cos(x) + 1)(cos(x) - 1) / (cos(x) + 1) = cos(x) - 1 18. (sin(x) + cos(x))² = 1 + sin(2x) = sin²(x) + 2sinxcosx + cos²(x) = 1 + 2sinxcosx = 1 + sin(2x) Given sin A = 3/5 and cos B = -7/25, with both A and B in Quadrant 2. 32. sin(A + B) sin(A + B) = sinAcosB + cosAsinB = (3/5)(-7/25) + (-4/5)(24/25) = -21/125 - 96/125 = -117/125 33. tan(A - B) tan(A - B) = (tanA - tanB) / (1 + tanAtanB) = (-3/4 + 24/7) / (1 + (-3/4)(24/7)) = 75/100 = 3/4 34. sin 2A sin 2A = 2sinAcosA = 2(3/5)(-4/5) = -24/25 35. cos(B/2) cos(B/2) = ±√((1 + cosB)/2) = ±√((1 + (-7/25))/2) = ±√(9/25) = ±3/5 Find the exact values of sin 2u, cos 2u, tan 2u. 36. sin u = 2/5, π/2 < u < π cos u = -√(1 - sin²u) = -√21/5 sin 2u = 2sinucosu = 2(2/5)(-√21/5) = -4√21/25 cos 2u = cos²u - sin²u = (-√21/5)² - (2/5)² = 17/25 tan 2u = sin 2u / cos 2u = (-4√21/25) / (17/25) = -4√21/17 37. tan u = -9/4, π/2 < u < π sin u = 9/√97 cos u = -4/√97 sin 2u = 2sinucosu = 2(9/√97)(-4/√97) = -72/97 cos 2u = cos²u - sin²u = (-4/√97)² - (9/√97)² = -63/97 tan 2u = sin 2u / cos 2u = (-72/97) / (-63/97) = 8/7 Find the exact values of sin u/2, cos u/2, tan u/2. 38. tan u = 1, π < u < 3π/2 sin u = 1/√2 cos u = -1/√2 sin(u/2) = ±√((1 - cosu)/2) = ±√((1 + 1/√2)/2) = ±√(2 + √2)/2 cos(u/2) = ±√((1 + cosu)/2) = ±√((1 - 1/√2)/2) = ±√(2 - √2)/2 tan(u/2) = sin(u/2) / cos(u/2) = ±√(2 + √2) / ±√(2 - √2) = ±(√2 + 1) 39. sec u = -6, π/2 < u < π cos u = -1/6 sin u = √(1 - cos²u) = √35/6 sin(u/2) = ±√((1 - cosu)/2) = ±√(7/12) cos(u/2) = ±√((1 + cosu)/2) = ±√(5/12) tan(u/2) = sin(u/2) / cos(u/2) = ±√7/√5 = ±√35/5