180

E

M

But suppose the sine of y to be given. The angle y may then have either of two supplementary values, y and 180°-y. The halves of the general measure of these angles will bey, 90° ty, 180°y, and 270°-y. The sines and cosines of these

four angles are all different. Therefore 180+y

the algebraic expression for the sines
ought to be susceptible of four values
instead of two, as in the first case.
(21) and (22), because they show that

E Y

X

Now this is true of equations

1+ cos y
2

1+ cos y
2

1. We have already found sin 30°; from this find the sine, cosine, tangent, and cotangent of 15° and of 75°.

2. From the values of the six trigonometric functions for 45° (830) find those for 221°.

3. From the values for 18° find those for 36°.

46. Miscellaneous relations. The following equations are of occasional use in the applications of trigonometry, and can all be derived from the formulæ of the last two chapters. Their derivations are therefore presented as an interesting exercise.

2 sin2 (45° + x) — 1 = 1 - 2 sin' (45° — ‡x). sin & cot ɛ.

11.

cos α =

1+tan a tan a

cos a = 2 cos (45° + a) cos (45° — ža).

2

tan (45° + 4a) + tan (45° — a)

2 cot 10

14. tan@=

15. tan =

16. 2 cot 2a = cot a

CHAPTER V.

TRIGONOMETRIC PROBLEMS.

47. PROBLEM I. Having given two equations of the form

where a and b are given numbers, it is required to compute r

and p.

Solution. Dividing the first equation by the second, we find

itself, then the sine or

From this value of tan we find

cosine of p, and then r from either of the equations

Example. If r sin p = 332.76, and r cos p = 290.08, it is

The numbers in brackets show the order in which the numbers of the computation are written. In writing log r sin and rcos o spaces are left for inserting log sin and log cos p after p is found, so that either of the latter may be subtracted to obtain r. It is generally best to obtain r from both r sin and r cos p, because then if any mistake is made in p it will be shown by a difference in the results.

ዋ

On the other hand, a practical computer will not write down

either sin or cos p, but will subtract in his head and write down

we have supposed r sin p and r cos p to be positive, and have taken ø in the first quadrant. But either or both of these quanФ tities may be negative. Whatever their signs, there are always two values of p, differing by 180°, corresponding to any given value of tan (§ 31). Hence the problem admits of two solutions in all cases. In the one will be positive, in the other negative.

But in practice only that solution is sought which gives a positive value of r. This being the case, sin p and cos p must have ዋ the same algebraic signs as the given quantities r sin p and r cos respectively. Now consider each case in order:

I. rsin and r cos p both positive. The angle must then be taken in the first quadrant, because only in this quadrant are sin p and cos p both positive.

II. rsin p positive and rcos p negative. Sin p is positive only in quadrants (1) and (2) (§ 21), and cos p is negative only in quadrants (2) and (3). Hence the requirement of signs can be fulfilled only in the second quadrant, and

90° < p < 180°.

III. r sin and r cos p both negative. The only quadrant in which sine and cosine are both negative is the third. Hence in this case

180°< < 270°.

IV. rsin p negative and r cos p positive. The only quadrant in which sin is negative and cos p positive is the fourth. Hence

in this case

270° < p < 360°.

NOTE. In this last exercise compute the value of (p+47° 50′) as if it were one quantity, and subtract the angle 47° 50′ from the result.

49. PROBLEM II. Having given two equations of the form x cos a + y sin a = p,

y cos a = 9,
y.

it is required to find the values of x and

The elimination is conducted by the method of addition and subtraction, as follows:

Multiply the first equation by cos a and the second by sin a. We thus have

x cos ay sin a cos a = p cos a;