Multivariable Calculus 9th Edition by James Stewart solution manual


cos
  + 
2

. Using this sum­to­product identity, we have each grouped sum equal to 0, since
sin
  + 
2

= sin = 0 is always a factor of the right side. Since sin
100
100
= sin = 0 and sin
200
100
= sin2 = 0, the
sum of the given expression is 0.
Another approach: Since the sine function is odd, sin(−) = −sin. Because the period of the sine function is 2, we have
sin(− + 2) = −sin. Multiplying each side by −1 and rearranging, we have sin = −sin(2 − ). This means that
sin

100
= −sin

2 −

100

= −sin
199
100
, sin
2
100
= −sin

2 −
2
100

= sin
198
100
, and so on, until we have
sin
99
100
= −sin

2 −
99
100

= −sin
101
100
. As before we rearrange terms to write the given expression as

sin

100
+ sin
199
100

+

sin
2
100
+ sin
198
100

+ ··· +

sin
99
100
+ sin
101
100

+ sin
100
100
+ sin
200
100
Each sum in parentheses is 0 since the two terms are opposites, and the last two terms again reduce to sin and sin2,
respectively, each also 0. Thus, the value of the original expression is 0.
14. (a) (−) = ln

− +

(−) 2 + 1

= ln

− +
√  2
+ 1 ·
− −
√  2
+ 1
− −
√  2
+ 1

= ln

 2 −
  2
+ 1 
− −
√  2
+ 1

= ln

−1
− −
√  2
+ 1

= ln

1
 +
√  2
+ 1

= ln1 − ln   +
√  2
+ 1

= −ln   +
√  2
− 1

= −()
(b)  = ln   +
√  2
+ 1
 . Interchanging  and , we get  = ln   +   2
+ 1

⇒   =  +
  2
+ 1 ⇒
  −  =
  2
+ 1 ⇒  2 − 2  +  2 =  2 + 1 ⇒  2 − 1 = 2  ⇒  =
 2 − 1
2 
=  −1 ().
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CHAPTER 1 PRINCIPLESOF PROBLEMSOLVING
15. ln   2 − 2 − 2  ≤ 0 ⇒  2 − 2 − 2 ≤  0 = 1 ⇒  2 − 2 − 3 ≤ 0 ⇒ ( − 3)( + 1) ≤ 0 ⇒  ∈ [−13].
Since the argument must be positive,  2 − 2 − 2  0 ⇒
  −  1 − √ 3   −  1 + √ 3 
 0 ⇒
 ∈
 −∞1 − √ 3 
∪
 1 + √ 3∞  . The intersection of these intervals is  −11 − √ 3 
∪
 1 + √ 33  .
16. Assume that log 2 5 is rational. Then log 2 5 =  for natural numbers  and . Changing to exponential form gives us
2  = 5 and then raising both sides to the th power gives 2  = 5  . But 2  is even and 5  is odd. We have arrived at a
contradiction, so we conclude that our hypothesis, that log 2 5 is rational, is false. Thus, log 2 5 is irrational.
17. Let  be the distance traveled on each half of the trip. Let  1 and  2 be the times taken for the first and second halves of the trip.
For the first half of the trip we have  1 = 30 and for the second half we have  2 = 60. Thus, the average speed for the
entire trip is
total distance
total time
=
2
 1 +  2
=
2

30
+

60
·
60
60
=
120
2 + 
=
120
3
= 40. The average speed for the entire trip
is 40 mih.
18. Let () = sin, () = , and () = . Then the left­hand side of the equation is
[ ◦ ( + )]() = sin( + ) = sin2 = 2sin cos; and the right­hand side is