The Chemistry Maths Book, Second Edition

Solutions to exercises 633

Chapter 3

Section 3.2

1.c 1 = 113 ,

2. (i) 1 (ii) 1


3. (i)π 236 (ii) 29 π 260 (iii) 2 π 23

(iv) (v) 3 π (vi) 4 π

4. (i)18° (ii)45° (iii)30°

(iv)60° (v)67.5° (vi)157.5°

5. (i) 32 2 rad 1 ≈ 1 85.94° (ii) 2 π 251 ≈ 1 1.257

(iii) 2 π 1 ≈ 1 6.283 (iv) 8 π 1 ≈ 1 25.13

6. (i)sin 13 π 241 = 1 , cos 13 π 241 = 1 ,

tan 13 π 241 = 1 − 1

(ii)sin 15 π 241 = 1 , cos 15 π 241 = 1 ,

tan 15 π 241 = 1 + 1

(iii)sin 17 π 241 = 1 ,cos 17 π 241 = 1 ,

tan 17 π 241 = 1 − 1

10. (i)π (ii) 2 π 23

Section 3.3

12. (i)π 26 (ii)π 22 (iii)π 23

(iv)π

13.sin

− 1

(1 2 4) 1 ≈ 1 14.48°,sin

− 1

(1 2 2) 1 = 1 30°,

sin

− 1

(3 2 4) 1 ≈ 1 48.59°,sin

− 1

(1) 1 = 1 90°

Section 3.4

14.C 1 = 15 π 2121 = 1 75°,,

1

15.A 1 = 1 cos

− 1

(0.75) 1 ≈ 1 41.41°,

B 1 = 1 cos

− 1

(0.5625) 1 ≈ 1 55.77°,

C 1 = 1 π 1 − 1 A 1 − 1 B 1 ≈ 1 82.82°

16.c 1 ≈ 1 2.8336,A 1 ≈ 1 48.47°,B 1 ≈ 1 86.53°

17. 
    

B 1 = 1 π 1 − 1 π 241 − 1 A 1 ≈ 1 108.43°

18. (i)sin 17 θ 1 = 1 sin 15 θ 1 cos 12 θ 1 + 1 cos 15 θ 1 sin 12 θ

(ii)sin 13 θ 1 = 1 sin 15 θ 1 cos 12 θ 1 − 1 cos 15 θ 1 sin 12 θ

(iii)cos 17 θ 1 = 1 cos 15 θ 1 cos 12 θ 1 − 1 sin 15 θ 1 sin 12 θ

(iv)cos 13 θ 1 = 1 cos 15 θ 1 cos 12 θ 1 + 1 sin 15 θ 1 sin 12 θ

c== š

−

5252657

1

, cos ( ). , A

c=°≈2 75 1 3660sin.

b= 32

+ 12

− 12

− 12 − 12

− 12

+ 12

13

9

π

cosec B=,B=,B=

13

5

13

12

12

5

sec cot

sin =,cos =,tan BBB=,

5

13

12

13

5

12

cosec A=,A=,A=

13

12

13

5

5

12

sec cot

sin =,cos =,tan AAA=,

12

13

5

13

12

5

19. (i)sin 13 θ 1 = 131 sin 1 θ 1 − 141 sin

3

1 θ

(ii)cos 13 θ 1 = 141 cos

3

1 θ 1 − 131 cos 1 θ

20. (i)cos

2

12 x 1 − 1 sin

2

12 x (ii) 11 − 121 sin

2

12 x

(iii)2cos

2

12 x 1 − 11

(iv) 11 − 181 sin

2

1 x 1 + 181 sin

4

1 x

(v) 11 − 181 cos

2

1 x 1 + 181 cos

4

1 x

21.0.9396

22. (i)

(ii)

23. (i)

(ii)

24. (i)sin 1 (π 1 ± 1 θ) 1 = 1

3

sin 1 θ

(ii)cos 1 (π 1 ± 1 θ) 1 = 1 −cos 1 θ

25. (i)t 1 = 1 n 2 2,n 1 = 10 , 1 , 2 ,=

(ii)t 1 = 1 (2n 1 + 1 1) 24 ,n 1 = 10 , 1 , 2 ,=

Section 3.5

27. (i) (ii)


28. (i) (ii)


29. (i)

(ii)

30. (i)

(ii)

Section 3.6

32. (i)e

5

(ii) 1 (iii) 12 e

(iv)e (v)e

9

33. (i)

(ii)0.71653

34. (i)

(ii)0.999000499833,4 terms

(iii)0.999999000001,3 terms

(iv)0.999999999000,2 terms

(v)0.999999999999,2 terms

(vi)1.00000000000, 1 term

36. (i) (ii) 1

Section 3.7

37. (i) 2 (ii) 4 (iii)− 5 (iv)x

2

(v)−(ax

2

1 + 1 bx 1 + 1 c) (vi)−kt

nn e

ij

kT

ij

=

−−()εε

1

2 6 24 120

3

6 9 12 15

−+−+ −x

xxx x

1

3 18 162 1944 29160

23 4 5

−+ − + −

xx x x x

13 2 3 213 7

1

,

()

+≈.

−

tan π °

13 2 3 146 3

1

,−

()

+≈

−

tan π .°

13 2 3 2 326 3

1

,−

()

+≈

−

tan π .°

13 2 3 33 7

1

,

()

≈.

−

tan °

−,−

()

−, 32 3 32

()

32 3 32

32 3 32,−

()

32 3 32,

()

1

2

cos 28 xx+cos

()

1

2

cos 28 xx−cos

()

1

2

sin sin 82 xx−

()

1

2

sin sin 82 xx+

()