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52.

cosa a

cos' a
COS a
1-
cosh B+
cosh 2B -

cosh 3ß + ..
1
2 2

3

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53. sinh a – - sinh 2a + 3 sinh 3a -
54. cos 6 cosh 0 + cos2 0 cosh 20+ } cos3 0 cosh 30+
55. 1 +

p (P-9)
P
cos a sinh 8 +

cos^ a sinh 28
2.92
+Pp)
P (P-9) (P-29)

cos^ a sinh 38 + ...

3.98

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59. The sum of the reciprocals of the squares of all integers except the multiples of r is (72 – 1) ++/6r%.

60. The sum of the reciprocals of the squares of all odd integers except the multiples of 2r + 1 is

r (r + 1)

2 (2r + 1) 61. The sum of the reciprocals of the squares of the products of all pairs of integers is 120#*; and of all pairs of odd integers is ਭੈ. 1 1 1 1

TA
62.
14 24 34

90

84

+

+

+

+

14

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=

68. Hence, using the identity, cosec 0 = 1 cot 10 + 1 tan 20,

1 show that

cosec 0 = 3 (-1)

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=

70. Prove that sin 20 = 2 sin 0 cos 0 from the factor expressions of sin 6 and cos 0.

71. Resolve vers 0 into factors without making use of expression for sin 0.

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40

73. sin 6 + cos 0 =(1+) -)(1+) - ...

(1
( p(1 4

)

+

40
37

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Find the general form of the component in 80 and 81.

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84. Calculate a to 3, 4, 5, 6, 7 decimal places respectively from the formulæ 1, 2, 3, 4, 5 of Art. 539.

85. Calculate i to 5 decimal places from the formulæ of Art. 533.

86. The equation 0 = cos 0 has one and only one root: and this root lies between it and ..

CHAPTER XX.

THE CONSTRUCTION OF TABLES AND INTER

POLATION.

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§ 1. THE CONSTRUCTION OF TABLES. 550. To calculate the value of T. =:8

; 1= 8 X.04=.032

; 3= '0106666666666667 ::00128

• 5=

256 = .0000512

: 7

73142857143 = '000002048

9.

2275555556+ = '00000008192

•11=

74472727 - 0000000032768

•13=

2520615+ = '000000000131072

• 15=

87381= '0000000000052429

• 17=

3084+ = '0000000000002097

•19=

110= .0000000000000084 ;21=

4+ 1:702= .0002040816326531

To=

•0142857142857143 ; 70 = .0000029154518950

9718172983+ ; 70 = .0000000416493128 :70 = .0000000005949902

1189980– : 70 = .0000000000084998 • 70 0000000000001214 • T=

174+ 1:992 = '00,01,02,03,04,05,06,07 = *0101010101010101 + :-99 = '00,00,01,03,06,10,15,21

• 3=

3435367174:99 = '00,00,00,01,04,10,20,35 ;-99 = '00,00,00,00,01,05,15,35

210307 + ;-99 = '00,00,00,00,00,01,06,21 :-99 =:00,00,00,00,00,00,01,07

15 = .8103582097472824+

·0249600463498340 1o= .7853981633974484 ..o3:141592653589793

3=

ol.

• 5=

• 7=

1

99

1

1
2 59

We will use the formula 1 = 4 tan-} - tan- to + tan-,79

The multiplier which occurs in 4 tan-1}, may be written 4.10-2.

Any pair of digits in any power of os may be found by adding the preceding pair of digits to the corresponding pair in the preceding power.

See above.

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= 100 .:7; 600

T = 600 = 9

551. To find sin 10".
We have

= .03141592653589793,
= .00523598775598299,

= .00058177641733144, T = 600 - 9 - 12 = .00004848136811095. Now let 0 = circular measure of 10"= 1 T = 90 = 60 = 6 = above calculated decimal.

Then, by Art. 485, sin 0 lies between 0 and 0 - 1 43 for any acute angle. But if & measures 10", 0 < .00005, i.e. <}. 10-4,

.: 103 <3 : 10-12 < 1.10-13. Hence writing 6 for sin 0, the error in sin 10" will be less than 10-13. That is,

For 13 places of decimals, sin 10" = circular measure of 10" = .0000484813681.

552. Similarly we have sin 5" = .00002424 nearly by halving the circular measure of 10". Thus sino 5" = (2424 x 10-8)* = 5876 x 10-13,

.:: 1 (1 - cos 10") = .0000000005876,
:. 2 (1

cos 10'') = '0000000023504.
553. To calculate the sines of angles which are multiples

<
32

=

of 10".

2 cos a.

If a denotes any angle, we have
sin (n + 1) a + sin (n − 1) a =

2 sin na cos a, i. sin (n + 1) a – sin na = sin na — sin (n − 1) a -(2 – 2 cos a) sin na. In this equation let a = 10".

10". Then we have found sin a and 2

Putting n= 1 gives us sin 20” – sin 10".

Putting n 2 gives us sin 30" – sin 20". And so on.

The advantage of the above mode of working is that the labour is reduced to the mere multiplication by the small quantity 2 - 2 cos a.

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