If OP starting from OR stop in the position OP1, the angle described is one of those included in the expression

If OP starting from OR stop in the position OP2, the angle described is one of those included in the expression (2r+1) π + α.

Both of these expressions are included in

ηπια*.

Thus the solution of the equation tan ◊ = tan a is

(1) Write down the complete Algebraical Solution of each of the following equations :

(i) cos 0=1.

(ii) tan 0=1.

(iii) tan0 - 1.

4π

Зп 4

(iv) tan 0= −√3. (v) cos 0=cos (vi) tan 0=tan 5

(2) Show that each of the following angles has the same cosine :

-120°, 240°, 480o, - 480o.

(3) The angles 60° and 120° have one of the Trigonometrical Ratios the same for both; which of the ratios is it?

(4) Can the following angles have any one of their Trigonometrical Ratios the same for all?

-230, -1570 and 157o.

(5) Find four angles which satisfy each of the equations in (1).

*For if n be even, this is the first formula; if n be odd it is the second.

150. We can now point out the use of the ambiguous sign in the formula cos

= ±√√1 − sin2 0.

If we know the numerical value of the sine of an angle 0, without knowing the magnitude of the angle, we cannot from the identity, cos 0=1-sin2 0, completely de

termine cos 0, for we get cose = ±√√1 − sin3 0.

This is a general formula, and we shall find that it represents an important Geometrical truth.

151. Given sin 0 = a, we can say that is one of the angles represented by one or the other of the positions OP,, OP, of the revolving line in Fig. I. on page 110.

2

If we attempt to find the cosine of these angles we get

although equal in magnitude, are opposite in sign. Hence, if a be the least angle whose sine is equal to a, we have

cose = cos a = √1-sin3 a.

152. The same result may be obtained from the formula 0 = nπ+(−1)"a. For cos {n+(-1)"a} is of opposite sign according as n is even or odd.

and of tan 0.

(1) If be found from the equation cosa, show geometrically that there are two values of sin (2) If be found from the equation tan 0=a, show geometrically that there are two values of sin ✪ and of cos 0.

(3) If A be the least angle without regard to sign such that sin A=a, show that cos =+√1 – sin2 A.

(4) If A be the least positive angle such that cos A=a, prove that sin A+√1-cos2 4,

=

CHAPTER XI.

ON THE TRIGONOMETRICAL RATIOS OF Two ANGLES.

formulæ :

WE proceed to establish the following fundamental

sin (A + B) = sin A. cos B + cos A. sin B
cos (A + B) = cos A. cos B- sin A. sin B

sin (A – B) = sin A. cos B - cos A. sin B

cos (4-B) = cos A. cos B+ sin A. sin B

Here, A and B are angles; so that (A + B) and (A – B) are also angles.

Hence, sin (A + B) is the sine of an angle, and must not be confounded with sin A + sin B.

Sin (A + B) is a single fraction.

Sin A+ sin B is the sum of two fractions.

154. The proofs given in the next two pages are perfectly general, as will be explained below (cf. Art. 169); but the figures are drawn for the simplest case in each.

The student should notice that the words of the two proofs are very nearly the same.

To prove that sin (A+B)=sin A. cos B+ cos A. sin B,

Let ROE be the angle A, and EOF the angle B. Then in the figure, ROF is the angle (A+B).

In OF, the line which bounds the compound angle (A+B), take any point P, and from P draw PM, PN at right angles to OR and OE respectively. Draw NH, NK at right angles to MP and OR respectively. Then the angle

NPH 90° - HNP=HNO=ROE=A*.

=

KN. ON HP.NP KN ON HP NP

+

=

+

ON.OP NP.OP ON OP NP' OP

=sin ROE.cos EOF+cos HPN. sin EOF

=cos ROE.cos EOF- sin HPN. sin EOF

=cos A. cos B sin A. sin B.

Or thus. A circle goes round OMNP, because the angles OMP and ONP are right angles; therefore MPN and MON are angles in the same segment; so that MPN=MON=A.

To prove that sin (A – B)=sin A. cos B-cos A. sin B, cos (A-B)=cos A. cos B+sin A. sin B.

and that

Let ROE be the angle A, and FOE the angle B. Then in the figure, ROF is the angle (A – B).

In OF, the line which bounds the compound angle (A – B), take any point P, and from P draw PM, PN at right angles to OR and OE respectively. Draw NH, NK at right angles to MP and OR respectively. Then the angle

NPH 90° - HNP=HNE=ROE=A*.

Now sin (A-B)=sin ROF=

MP
OP

KN.ON

=

=

ON.OP

MH-PH KN PH

=

OP

OP OP PH.NP KN ON PH NP

NP.OP ON OP NP OP

=sin ROE. cos FOE-cos HPN. sin FOE

=sin A. cos B-cos A. sin B.

Also cos (A-B)=cos ROF=

+

+

OK.ON NH.NP OK ON NH NP

=

ON.OP NP.OP ON OP NP OP

=cos ROE. cos FOE+ sin HPN. sin FOE

=cos A. cos B+sin A. sin B.

*Or thus. A circle goes round OMPN, because the angles OMP and ONP are right angles; therefore MPN and MON together make up two right angles; so that HPN=MON=A.