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Math 116 CALCULUS II
SOLUTION FOR QUIZ – XXXI (07/18)
July 18 (Mon), 2016
Instructor:
Yasuyuki Kachi
Line #: 81300.
[I] (2pts)
The definition of tan x is
sin x
.
cos x
tan x =
[II] (10pts)
(1)
Find the following special values:
tan 0 = 0,
(6)
tan
2π
3
√
= − 3,
(2)
tan
π
6
1
= √ ,
3
(7)
tan
3π
4
= −1,
(3)
tan
π
4
= 1,
(8)
tan
5π
6
1
= −√ ,
3
(4)
tan
π
3
=
3.
(9)
tan π = 0,
(5)
tan
π
2
= ‘undefined’,
(10)
tan 2 π
= 0.
[III] (2pts)
(a)
(2)
2
d tan x
= 1 + tan x .
dx
1
d tan x
=
dx
(b)
[IV] (2pts)
√
(1)
tan
π
2
tan
−x
−x
1
cos x
2 .
=
1
.
tan x
=
− tan x.
[V] (8pts)
(1a)
Recall
sin
x+y
=
sin x
cos
x+y
=
cos x
cos y
cos y
+
−
cos x
sin y ,
sin x
sin y .
cos x
From these, you may find
tan
x+y
sin
=
cos
=
x+y
x+y
sin x
cos x
cos y
cos y
+
−
sin x
sin y
sin y
(1b)
.
In the resulting fraction in (1a) above, you divide both the numerator and the
denominator by
cos x
cos y , and obtain
tan
(2)
x+y
=
tan x
+
tan y
1 −
tan x
tan y
Substitute y with x into the outcome of (1b) above:
tan
2x
=
tan x
+
tan x
1 −
tan x
tan x
tan x
2
1 − tan x
2
=
2
.
.
(3)
Substitute
y = −
π
4
outcome of (1b) above:
tan
[VI] (6pts)
x−
and hence
π =
4
tan
2α

1 −
2
5
24
25
=
2·
tan
4α
1
5

5
12
1 −
=
2

5
.
12
5
12
=
10
12
119
144
=
3
− 1
2

120
.
119
.
+ 1
tan α =
1
5
=
=
(2)
tan x
Let α be a real number such that
2·
(1)
tan x
tan y = − 1
1
.
5
into the
(3)
tan
4α −
π
4
!
∞
X
π =
n= 0
=
1
−
7
− 1
239
119
+ 1
=
239
119
1
119
=
[VII] (10pts)
1
119
1
.
239
=
(1) Machin’s formula is as follows:

n
−1
16

−

2n+1
5
2n + 1
4
16
−
1
5
2391
!
−
1
3
4
16
−
7
5
2397
!
1
+
9
BBP formula is as follows:

!n
∞
X
1
4

π =

16
8n + 1

4
2392n+1
4
16
−
3
5
2393
!
+
4
16
−
9
5
2399
!
− ··· .
1
5


4
16
−
5
5
2395
!
(2)
−
2
8n + 4
−
1
8n + 5
−
1
8n + 6
n= 0



2
1
1
4
−
−
−
1
4
5
6
!
1
+
16
2
1
1
4
−
−
−
9
12
13
14
!
1
+
162
2
1
1
4
−
−
−
17
20
21
22
!
1
+
163
2
1
1
4
−
−
−
25
28
29
30
!
1
+
164
2
1
1
4
−
−
−
33
36
37
38
!
+ ··· .
=
4
(2b)
The sum of the first three terms of the BBP formula is
4
1
−
2
4
−
1
5
−
!
1
6
1
+
16
+
·
1
·
162
2
1
1
−
−
−
12
13
14
!
2
1
1
4
−
−
−
17
20
21
22
!
4
9
We may rely on a calculator, and find the answer as
h
Answer
(2c)
i
:
3
. 1 4 1 5 8 7 3 ...
From calculator,
π = 3.1415926535.. .
The difference between the above answer in (2) and π is
0.000005... .
It is between 10−5 and 10−6 .
5
.