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23. The following is the demonstration of formulas (3) and (4) of Art. 77. By Art. 69

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These forms (1) and (2) are more suitable than (1) and (2) of Art. 77, if ỏ be

2 √ab

nearly equal to a, because then tan 4, which, in Art. 77, was equal to

sinc

is very large, or is near 90°, unless c is very small, and when such is the case the increase of the tangent is not proportioned to the increase of the arc, so that the ordinary mode of calculating logarithms not exactly found in the tables would be inaccurate.

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24. Formulas not adapted to logarithmic computation may often be rendered so by the use of subsidiary angles. Specimens have been given in the last Art. and Art. 77. The following is another example.

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may be computed from (1) by logarithms, and then 1+ from (2), and from and +, becomes known.

25. To resolve a quadratic equation by the aid of Trigonometry.

The general form of such an equation is

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Logarithms may be applied to the formulas (1), (2), and (3).

If p and q be negative the following forms should be used, which may easily be

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For the resolution of a cubic equation by the aid of Trigonometry, see Alg. Art. 378.

26. To find the increment of the sine, tangent, &c. corresponding to a small increment of the angle.

Let a represent the angle, i its increment, and 8 sin a the corresponding increment of the sine. Then

When the coefficients of a quadratic are large, the trigonometric mode of solu tion is convenient.

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The increment i being supposed very small, tan i will be very small also, and unless tan a be large, the second term in the parenthesis may be omitted. Then since sin i is equal to i very nearly, i being very small, it follows that the ratio of ó sin a to i is cos a. In other words, the difference of the sines of two angles is proportional to the difference of the angles when the difference is small.

When tan a is large this principle fails, which is the case with arcs near 900.
The reasoning for d cos a is very similar.

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which if i be very small, and a not near 90°, reduces to 8 tan a = sec2 a tan i.

EXERCISES.

(1). Prove vers (1800 — a)

=

2 vers (1800+ a) vers (180o — a). (2). Find the numerical values of sin 150, vers 150, sin 90, cos 120. (3). Prove tan 500+ cot 50° 2 sec 100.

4). If a + b + c = 90°, prove

tan a tan b+ tan a tan c + tan b tan c = 1

cot a cot b + cot c = cot a cot b cot c

tan atan b+ tan ctan a tan b tan c + sec a sec b sec c

and

and

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(6). Prove the radii of the inscribed and circumscribed circles of a regular poly

1800

gon of any given number (n) of sides to be for the former r=

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length of one side of the polygon, and for the latter R =

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This is the formula used in the solution of Kepler's problem in Astronomy.

(7.) Prove the area of a regular circumscribed polygon to be

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(8). Prove the area of a regular polygon of n sides, one of which = a to bo

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(9). Prove the radii of the inscribed and circumscribed circles of a triangle to be

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