54.

- sinh 2a + sinh 3a - ...

cos e cosh + cos2 0 cosh 20+ cos3 0 cosh 34 +

59.

1 1 1

12 52 72 112

The sum of the reciprocals of the squares of all integers except the multiples of r is (2-1) 2/62.

60. The sum of the reciprocals of the squares of all odd r(r+1)2

integers except the multiples of 2r+ 1 is

2 (2r+ 1)"

61. The sum of the reciprocals of the squares of the products

of all pairs of integers is ; and of all pairs of odd integers

68. Hence, using the identity, cosec 0 = 1⁄2 cot 10 + 1 tan 10,

70. Prove that sin 20 = 2 sin cos 0 from the factor ex

pressions of sin 0 and cos 0.

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

73.

40

3п

40

40

sin + cos 0 = (1 +· 40) (1 - 12) (1 + 10) (1-10)...

П

33

54

5п

...

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

82.

84.

1+ 3 · x2+ 5-3x2 + 7 − 5x2 +

Calculate to 3, 4, 5, 6, 7 decimal places respectively

from the formulæ 1, 2, 3, 4, 5 of Art. 539.

85. Calculate

of Art. 533.

86. The equation

to 5 decimal places from the formulæ

= cos has one and only one root: and

this root lies between π and π.

1

70

99.

We will use the formula = 4 tan-1-tan-1+tan-1 The multiplier, which occurs in 4 tan-1, may be written 4.10-2.

1

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

Then, by Art. 485, sin lies between 0 and 103 for any acute angle.

But if measures 10", 0 < 00005, i.e. <1⁄2. 10-4,

031210-12.10-13.

Hence writing for sin 0, the error in sin 10" will be less than 1.10-13. 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 sin2 5" (2424 x 10-8)2=5876 x 10-13,

=

553. To calculate the sines of angles which are multiples of 10".

If a denotes any angle, we have

sin (n + 1) a + sin (n − 1) a =

2 sin na cos a,

.`. sin (n + 1) a − sin na = sin na - sin (n − 1) a − (2 − 2 cos a) sin na.

In this equation let a = 10". Then we have found sin a and

2 cos a.

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.