The same formulas are of frequent occurrence in a different form. Evidently

Again, we can easily derive from the formulas for the sines and cosines of A+ B and A B, expressions for the tangents of A + B and A. B. Thus, sin. (A+B) sin. A cos. B + sin. B cos. A Tan. (A + B) = cos. (A+B) cos. A cos. B sin. A sin. B Divide both numerator of this fraction by cos. A cos. B.

=

25. Expressions in which the sum of three angles occur.

We can easily derive from the above expressions for the sines, cosines. . . of the sum of three or more angles. Thus,

Sin. (A+B+C): = sin. (A+B) cos. C + cos. (A + B) sin. C = (sin. A cos. B sin. B cos. A) cos. C+ (cos. A cos. B sin. A sin. B,) sin. C sin. A cos. B cos. Csin. B cos. C cos. A + sin. C cos. A cos. B sin. A sin. B sin. C.

In the same manner,

Cos. (A+B+ C) cos. A cos. B cos. C cos. A sin. B sin. C.

There are many similar combinations of the trigonometric ratios besides those above given. These are of very frequent occurrence, and the student who has thoroughly mastered the above will be at no loss in investigating other combinations that may occur in his subsequent reading.

27. The Sines, Cosines.... of Multiples of given Angles.

We have already seen that

Sin. (A+B)

sin. A, cos. B+ sin. B, cos. A.

This being true of all values of A and B is true when AB, and.. when A + B = 2A. 2 sin. A, cos. A. . . . (22).

Similarly, since

.. Sin. 2A

Similarly,

And

2 tan. A

Tan. 2A =

- tan.2A

....

(24).

Sin. 3A sin. (2A + A).

sin. 2A, cos. A+ cos. 2A, sin. A.

=2 sin A, cos. A, cos. A+ (Cos3A — sin.2A) sin. A.
+3 sin. A, cos2A — sin.3A.

and

3 tan. A

Tan. 3 A=

tan3A 1-3 tan A

(27).

28. Determination of Sine, &c., of an Angle in terms of the Sine, &c., of the Sub-mulples of that Angle.

From these expressions we may derive others expressing the sines, &c., of an angle in terms of the sines, &c. of the submultiples of that angle. Thus, writing we have

for A,

29. On the "Ambiguities” resulting from the use of the above Formulas. These formulas enable us to solve the following question:-Having given the sine... of an angle, we can find from it the sine. . . . of double that angle; and conversely having given the sine . . . . of an angle, we can find the sine . . . . of half that angle. (a.) Thus, having given sin. A = p, to find cos. 2A, we have

Cos. 2A1 — 2 sin.2A

and so, having given sin. A = p to find sin. 2A.

= 1-2p2.

Since sin. A p cos. A = √1 — p2 .. sin. 2A= +2p √ 1 − p2.

It will be seen, from the above formulas, that for one given value (p) of sin. A, there is one value of cos. 2A, while there are two of sine 2A equal in magnitude, but of different signs. This is sometimes spoken of as an ambiguity. It will be observed, however, that the ambiguity in the determination of sin. 2A arises necessarily from the data, since it appears by considering the values of A which satisfy the equation

Sin. AP,

that there will be one value of cos. 2A and two values of sin. 2A resulting from the

data. Thus, if A1 be one angle which satisfies the equation, then all the valu are included in the two formulas

sine 2A has the two values + sine 2A1 and 1 sine 2A1.

If, however, we know A, or even the limits between which A lies, as well as that sin. Ap, then all indeterminateness vanishes from the expression for sin. 2A. Thus, if A is less than 90°, then 2A is 180 and sin° 2A must be positive. And, again, if A 790' 180°, then 2A 7180' 360°, and the sin. 2A is negative in the former case; therefore,

Hence it appears that for a given value of cos. @ there are two values of sin.

0 2

likewise two values of cos.- equal in magnitude, but with different signs. This is,

as before, necessarily the case, if we only know the value of cos. e. For if we have a given value p of cos. e, so that

Cos. 0 = P,

and if 1 is a value of which satisfies this equation, then all the values of 0, which satisfy this equation, are expressed by,

under any circumstances; therefore there must be two values of sin. equal in magni

tude, but with different signs; the same result as that we obtained from the equations

for sin.

and cos.

02

Extracting the square root of each of these equations, and we have—

+ cos.2

=1+sin. 0

2

÷ {√ 1 + sin. 0 +√1—sin. e}

=

} {√ 1 + sin. 0 V 1

sin. es

And since each square root has two sines, it follows that if we have given merely the

value of sin. 0, i.e. sin. 0

p. we have four different values of sin.

- ·

2

viz. (√ 1+p. + √ 1 — p.) i (v 1+p. —√ 1—p.)
(−√1+p. + √ 1-p.) and (−√1+p. -v1-p.)
¦ (—

We may prove, as before, that this amount of undeterminateness is involved in the data: for if

sin. = P,

and if e′ be a value of 0. which satisfies this equation, then all the values of ✪ which satisfy the equation, are given by the formulas.

02 m. 180° + 0
(2 m. + 1) 180°

and 0 =