Math 103: College Algebra and TrigonometryExam 4 review answersAnswer 1. Several of these answers may be written in different forms. Two equivalentforms of the answer are given below where appropriate. Other forms of these answers maybe possible.(a)√32(b)√3 (c) 1 (d) 1(f) −12(g) −s21 +√32= −4√2 +√6(h)√22=1√2(i)p2 −√32=√6 −√24(j)π4= 45◦(k)5π6= 150◦(l)π3= 60◦(m) 7(n)2π3= 120◦(o) 2 (p) −√22= −1√2(q)√22=1√2(r)p2 +√22Answer 2. Quadrant IVAnswer 3. sin θ =6597; cos θ = −7297; tan θ = −6572; cot θ = −7265; sec θ = −9772; csc θ =9765Answer 4.Function Domain Range Even/oddsin θ (−∞, ∞) [−1, 1] Oddcos θ (−∞, ∞) [−1, 1] Eventan θ All real numbers except π/2 + kπ for integers k (−∞, ∞) Oddcot θ All real numbers except π/2 + kπ for integers k (−∞, ∞) Oddsec θ All real numbers except π/2 + kπ for integers k (−∞, −1] ∪ [1, ∞) Evencsc θ All real numbers except kπ for integers k (−∞, −1] ∪ [1, ∞) OddAnswer 5. While not strictly an error, there was a typo in the problem on the questionsheet. Writing the inverse trigonometric functions as sin−1θ, cos−1θ, etc. is misleading, sinceθ usually represents an angle, but the arguments to the inverse trigonometric functions arenot angles. (The outputs of the inverse trigonometric functions are angles, not the inputs.)It would have been better to have written sin−1x, cos−1x, etc.Function Domain Rangesin−1x [−1, 1] [−π/2, π/2]cos−1x [−1, 1] [0, π]tan−1x (−∞, ∞) (π/2, π/2)cot−1x (−∞, ∞) (π/2, π/2)sec−1x (−∞, −1] ∪ [1, ∞) [0, π/2) ∪ (π/2, π]csc−1x (−∞, −1] ∪ [1, ∞) [−π/2, 0) ∪ (0, π/2]Answer 6. The amplitude is 8, and the period is6π5.Page 1Answer 7. An equation for the graph on the left is y = −3 sinπx2. An equation for thegraph on the right is y = cos(3x) + 2.Answer 8.Function y-interceptsin x 0cos x 1tan x 0cot x no y-intercept (cot x is undefined at x = 0)sec x 1csc x no y-intercept (csc x is undefined at x = 0)Answer 9.cos θ1 − sin θ= sec θ + tan θAnswer 10.cos2θ − 1cos2θ − cos θ= 1 + sec θAnswer 11. Other methods are possible.(cos θ)(tan θ + cot θ) = (cos θ)sin θcos θ+cos θsin θ(a)= (cos θ)sin2θsin θ cos θ+cos2θsin θ cos θ= (cos θ)sin2θ + cos2θsin θ cos θ= (cos θ)1sin θ cos θ=cos θsin θ cos θ=1sin θ= csc θ.cos(α + β)cos α cos β=cos α cos β − sin α sin βcos α cos β(b)=cos α cos βcos α cos β−sin α sin βcos α cos β= 1 −sin αcos αsin βcos β= 1 − tan α tan β.9 sec2θ − 5 tan2θ = 4 sec2θ + 5 sec2θ − 5 tan2θ(c)= 4 sec2θ + 5(sec2θ − tan2θ)= 4 sec2θ + 5(1)= 5 + 4 sec2θ.Page 2sec2u − (sin2u)(sec2u + 2) = sec2u − sin2u sec2u − 2 sin2u(d)= (sec2u)(1 − sin2u) − 2 sin2u=1cos2u(cos2u) − 2 sin2u=cos2ucos2u− 2 sin2u= 1 − 2 sin2u= cos(2u).tan θ − cot θtan θ + cot θ=sin θcos θ−cos θsin θsin θcos θ+cos θsin θ(e)=sin2θsin θ cos θ−cos2θsin θ cos θsin2θsin θ cos θ+cos2θsin θ cos θ=sin2θ−cos2θsin θ cos θsin2θ+cos2θsin θ cos θ=sin2θ−cos2θsin θ cos θ1sin θ cos θ=sin2θ − cos2θsin θ cos θsin θ cos θ1= sin2θ − cos2θ.cot2θ − 12 cot θ=cos2θsin2θ− 12cos θsin θ(f)=cos2θsin2θ−sin2θsin2θ2 cos θsin θ=cos2θ−sin2θsin2θ2 cos θsin θ=cos2θ − sin2θsin2θsin θ2 cos θ=(cos2θ − sin2θ)(sin θ)2 sin2θ cos θ=cos2θ − sin2θ2 sin θ cos θ=cos(2θ)sin(2θ)= cot(2θ).cos θ + cos(3θ)2 cos(2θ)=2 cosθ+3θ2cosθ−3θ22 cos(2θ)(g)=2 cos4θ2cos−2θ22 cos(2θ)=2 cos(2θ) cos(−θ)2 cos(2θ)= cos(−θ)= cos θ.Answer 12.(a) Solution set:2π3,5π3(b) Solution set:0,2π5,4π5,6π5,8π5(c) Solution set:π2,2π3,4π3,3π2(d) Solution set:π4,π2,5π4,3π2(e) Solution set:π4,5π4Page 3Answer 13.20◦70◦99 cos 20◦≈ 8.45729 sin 20◦≈ 3.0782Answer 14. Wichita is approximately 377 km from Grand Island. A crow flying directlyfrom Grand Island to Wichita must fly on a bearing of about S10.26◦E, that is, 10.26◦eastof south.Answer 15. To the nearest tenth of a square inch, the area of the shaded region is 17.9 in.2Answer 16. To the nearest foot, the diameter of the wheel was 252 feet.Answer 17.(a) B = 110◦, b ≈ 3.6800, c ≈ 1.3394(b) No solution(c) c ≈ 1.6905, A = 65◦, B = 65◦(d) a ≈ 3.5128, A ≈ 43.78◦, C ≈ 36.22◦(e) Two solutions:C ≈ 74.62◦, A ≈ 65.38◦, a ≈ 2.8286 or C ≈ 105.38◦, A ≈ 34.62◦, a ≈ 1.7676(f) A ≈ 36.34◦, B ≈ 26.38◦, C ≈ 117.28◦(g) A ≈ 30.51◦, B ≈ 59.49◦, C = 90◦Answer 18. The area of the triangle is3√2554≈ 11.9765.Answer 19. The area of the triangle is approximately 0.2939 m2, or approximately2939 cm2, or approximately 455.52 in.2, or approximately 3.1633 ft2.Answer 20. The area of the triangle is3√914square cubits, or approximately 7.1545 squarecubits.Answer 21. The area of the hendecagon is approximately 29 735. [Note that the area ofthe circle is approximately 31 416, so our answer for the area of the hendecagon makes sense,since it covers nearly all of the circle.]Bonus. −1Page
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