XX. Trigonometrical Transformations. 185. From the formula of the preceding Chapter others may be derived; and by the aid of all the results we can change Trigonometrical expressions into various forms. The use of such transformations becomes apparent as the student advances in Mathematics, and even at present he will find valuable exercise in them. 186. The product of sin (A+B) and sin (A– B) takes a remarkable form. = (sin A cos B+ cos A sin B) (sin A cos B-cos A sin B) =sin2A cos2B-cos2A sin2B =sin2A (1-sin3B)-(1-sin2A) sin3B =sin A-sin B. This may also be put in the form cos2B-cos A. 187. Also cos (A + B) cos (A – B) =(cos A cos B-sin A sin B) (cos A cos B+ sin A sin B) =cos2A cos2B-sin A sin B = cos2A (1 − sin2B) — (1 — cos2A) sin2B =cos2A-sin2B =cos2B-sin❜A. 188. From the four fundamental formulæ of Chapter XIX. we have sin (A+B)+ sin (A-B)=2 sin A cos B, cos (A–B) — cos (A + B) = 2 sin A sin B. Let A+B=C, and A-B=D; therefore The last four formulæ are useful in converting a sum or difference into the form of a product of factors. Thus, for example, by the first of the last four formulæ the sum of two sines is converted into twice the product of a sine and cosine. The first four formulæ are useful in converting a product into the form of a sum or difference. Thus, for example, by the last of the first four formulæ we see that the product of two sines is equal to half the cosine of the difference of the two angles diminished by half the cosine of their sum. 194. By repeated applications of the formulæ in Chapter XIX, we can obtain expressions for the Trigonometrical Ratios of a composite angle made up of any number_of simple angles connected by the signs example, and sin (A+B+C)=sin (A+B) cos C+cos (A+B) sin C =sin A cos B cos C+ sin B cos C cos A + sin Ccos A cos B- sin A sin B sin C. cos (A+B+C)=cos (A+B) cos C-sin (A+B) sin C; =cos A cos B cos C-cos C sin A sin B tan (A+B+C)= For -cos B sin A sin C-cos A sin B sin C. sin (A+B+C), cos (A+B+C)' : substitute the expressions just found for sin (A+B+C) and cos (A+B+C); and then divide both numerator and denominator by cos A cos B cos C; thus we obtain tan (A+B+C): = tan A+tan B+tan C-tan A tan B tan C 1-tan Btan C-tan C'tan A-tan AtanB' 195. The particular case of the formulæ in the preceding Article in which C=B= A should be noticed. Thus we obtain sin 34 = 3 sin A cos2 A-sin3 A = 3 sin A (1-sin2 A) — sin3 A |