Hence, the projection of one side of a triangle is equal to the sum of the projections of the other two sides taken in order. Thus projection of OQ=projection of OP+projection of PQ; General Proof of the Addition Formulæ. 285. In Fig. 1, let a line starting from OX revolve until it has traced the angle A, taking up the position OM, and then let it further revolve until it has traced the angle B, taking up the final position ON. Thus XON is the angle A+B. In ON take any point P, and draw PQ perpendicular to OM; also draw OR equal and parallel to QP. Projecting upon OX, we have projection of OP=projection of OQ+projection of QP =projection of OQ+ projection of OR. .. OP cos XOP=OQ cos XOQ+OR cos XOR ......... that is, .(1) =OP cos B cos XOQ+OP sin B cos XOR; .. cos XOP=cos B cos XOQ+sin B cos XOR; cos (A+B)=cos B cos A+ sin B cos (90° +A) Projecting upon OY, we have only to write Y for X in (1); cos (A+B 90°) = cos B cos (A − 90°) + sin B cos A; .. sin (A+B)=sin A cos B+cos A sin B. In Fig. 2, let a line starting from OX revolve until it has traced the angle A, taking up the position OM, and then let it revolve back again until it has traced the angle B, taking up the final position ON. Thus XON is the angle Ă - B. In ON take any point P, and draw PQ perpendicular to MO produced; also draw OR equal and parallel to QP. Projecting upon OX, we have as in the previous case cos (A - B) = cos B cos (4 – 180°) + sin B cos (4 – 90°) = cos B(-cos A) +sin B sin A =cos A cos B+sin A sin B. Projecting upon OY, we have that is, OP cos YOP=OQ cos YOQ+OR cos YOR; =OP cos (180° – B) cos YOQ cos (A - B - 90°)= −cos B cos (A − 270°)+sin B cos (4 – 180°); .'. sin(A−B)= — cos B ( − sin A)+sin B ( − cos A) =sin A cos B-cos A sin B. 286. The above method of proof is applicable to every case, and therefore the Addition Formulæ are universally established. The universal truth of the Addition Formulæ may also be deduced from the special geometrical investigations of Arts. 110 and 111 by analysis, as in the next article. 287. When each of the angles A, B, A+B is less than 90°, we have shewn that also and cos (A+B)=cos A cos B- sin A sin B............(1). But cos (A+B)=sin (A+B+90°)=sin (A +90° + B) ; cos Asin (4+90°), - sin A = cos (4+90°). [Art. 98.] Hence by substitution in (1), we have sin (A +90° + B)=sin (A+90°) cos B+cos (4 +90°) sin B. In like manner, it may be proved that )=cos (A+90°) cos B-sin (A+90°) sin B. cos (A+90° + B)= Thus the formulæ for the sine and cosine of A+B hold when A is increased by 90°. Similarly we may shew that they hold when B is increased by 90°. By repeated applications of the same process it may be proved that the formulæ are true when either or both of the angles A and B is increased by any multiple of 90°. also and Again, cos (A+B)=cos A cos B-sin A sin B............(1). But cos (A+B)= − sin (A+B− 90°) — — sin (A – 90° +B) ; sin A =cos (A - 90°). [Arts. 99 and 102.] Hence by substitution in (1), we have sin (A − 90° +B)=sin (A – 90°) cos B+cos (4 −90°) sin B. Similarly we may shew that cos (A – 90°+B)=cos (A – 90°) cos B − sin (A – 90°) sin B. Thus the formulæ for the sine and cosine of A+B hold when A is diminished by 90°. In like manner we may prove that they are true when B is diminished by 90°. By repeated applications of the same process it may be shewn that the formula hold when either or both of the angles A and B is diminished by any multiple of 90°. Further, it will be seen that the formulæ are true if either of the angles A or B is increased by a multiple of 90° and the other is diminished by a multiple of 90°. and Thus where sin (P+Q)=sin P cos Q+cos Psin Q, cos (P+Q)=cos P cos Q-sin P sin Q, m and n being any positive integers, and A and B any acute angles. Thus the Addition Formulæ are true for the algebraical sum of any two angles. MISCELLANEOUS EXAMPLES. H. 1. If the sides of a right-angled triangle are cos 2a+cos 28+ 2 cos (a+B) and sin 2a+sin 28+ sin 2 (a+B), shew that the hypotenuse is 4 cos2 a-B 2. If the in-centre and circum-centre be at equal distances from BC, prove that cos B+cos C=1. 3. The shadow of a tower is observed to be half the known height of the tower, and some time afterwards to be equal to the height how much will the sun have gone down in the interval? Given log 2, L tan 63° 26′ =10.3009994, diff. for 1'=3159. 4. If (1+sin a) (1 + sin ẞ) (1+sin y) = (1 − sin a) (1 − sin ß) (1 − sin y), shew that each expression is equal to ±cos a cos B cos y. 5. Two parallel chords of a circle lying on the same side of the centre subtend 72° and 144° at the centre: prove that the distance between them is one-half of the radius. Also shew that the sum of the squares of the chords is equal to five times the square of the radius. 6. Two straight railways are inclined at an angle of 60°. From their point of intersection two trains A and B start at the same time, one along each line. A travels at the rate of 48 miles per hour, at what rate must B travel so that after one hour they shall be 43 miles apart? 8. If p, q, r denote the sides of the ex-central triangle, prove that a2 be c2 2abc 1. pqr 9. A tower is situated within the angle formed by two straight roads OA and OB, and subtends angles a and B at the points A and B where the roads are nearest to it. If OA=α, and OB=b, shew that the height of the tower is Va2-b2 sin a sin ẞ/ √sin (a+ẞ) sin (a – B). 10. In a triangle, shew that r2+r12+r22+r32=16R2 — a2 — b2 — c2. 11. If AD be a median of the triangle ABC, shew that (1) cot BAD=2 cot A +cot B ; (2) 2 cot ADC=cot B-cot C. 12. If p, q, r are the distances of the orthocentre from the sides, prove that Graphical Representation of the Circular Functions. 288. DEFINITION. Let f (x) be a function of x which has a single value for all values of x, and let the values of x be represented by lines measured from O along OX or OX', and the values of f(x) by lines drawn perpendicular to XX. Then with the figure of the next article, if OM represent any value of x, and MP the corresponding value of f(x), the curve traced out by the point P is called the Graph of ƒ (x). |