Example 4. Shew that Since we have Σ cos 3a sin (8 - y) = 4 cos (a+B+y) II sin (6 − y). 2 cos 3a sin (8-y)=sin (3a+ẞ− y) − sin (3a - ẞ+y), 22 cos 3a sin (ẞ- y) = sin (3a +ß − y) − sin (3a - ẞ+y)+sin (3ß+y− a) − sin (3ß − y + a) + sin (3y+a - ẞ) − sin (3y − a+ß). Combining the second and third terms, the fourth and fifth terms, the sixth and first terms, and dividing by 2, we have Σ cos 3a sin (8-Y) =cos (a+ß+y) {sin 2 (ẞ − a) + sin 2 (y − ß) + sin 2 (a − y)} 307. The following example is given as a specimen of a concise solution. and Example. If (y + z) tan a+ (z+x) tan ẞ+(x+y) tan y=0, x tan ẞ tany+y tan y tan a + z tan a tan ß=x+y+%, prove that and From the given equations, we have x (1 - tan ẞ tan y) +y (1 − tan y tan a) +z (1 − tan a tan ẞ)=0, If we find the values of x y z by cross multiplication, the denominator of x = (1-tan y tan a) (tan a +tan ß) − (1 − tan a tan ß) (tan y+tan a) =(tan ẞ - tan y) +tan2a (tan ß – tan y) .. x sin 2a + y sin 28 +z sin 2y=kΣ sin 2a sec a sin (ẞ − y) =2k sin a sin (8 − y) H. K. E. T. =0. 20 Allied formulæ in Algebra and Trigonometry. 308. From well-known algebraical identities we can deduce some interesting trigonometrical identities. put Example 1. In the identity then and put (x − a) (b −c) + (x − b) (c − a) + (x − c) (a - b) = 0, x= cos 20, a= cos 2a, b=cos 28, c= cos 2y; x-a= cos 20 - cos 2a = 2 sin (a + 0) sin (a − 0), b-c=cos 28 - cos 2y= -2 sin (6+y) sin (8 − y); .. Σ sin (a + 0) sin (a − 0) sin (ß+y) sin (8 − y)=0. Example 2. In the identity then Za2 (bc) II (b−c), a=sin2a, b=sin2ß, c=sin2y; b-c=sin28-sin2y=sin (B+y) sin (6 − y); .. Σ sin1a sin (8+y) sin (8 − y) = − II sin (ẞ+y). II sin (8 − y). Σ cos 3a (cos ẞ- cos y) = -4 (cos a+ cos ẞ+cos y) II (cos ẞ - cos y). Example 4. If a+b+c=0, then a3+ b3+c3=3abc. zero; this Here a, b, c may be any three quantities whose sum is condition is satisfied if we put a=cos (a+0) sin (8-y), and b and c equal to corresponding quantities. Thus cos3 (a + 0) sin3 (ẞ − y) = 31 cos (a + 0) sin (8 − y). 309. An algebraical identity may sometimes be established by the aid of Trigonometry. Example. If x+y+z=xyz, prove that x (1-y2)(1-2)+y (1-2) (1-x)+z (1 - x2) (1-y2)=4xyz. By putting x=tana, y=tan ß, z=tany, we have tana+tan ẞ+tan y=tan a tan ẞ tan y; .. a+B+y=NT; .. 2a+28+2y=2nπ. From this relation it is easy to shew that tan 2a+tan 2ẞ+tan 2y=tan 2a tan 28 tan 2y; 2x 2y 22 + 8xyz 1 − x2 + 1 − y2 * 1 − z2 − (1 − x2) (1 − y2) (1 − z2) ' : ..x (1-y2)(1 − z2)+y (1 − z2)(1 − x2) +z (1 − x2)(1 − y) = 4xyz. EXAMPLES. XXIV. b. Prove the following identities : 1. sin (a-0) sin (B-)=0. 2. 3. cos ẞ cos y sin (B-y)=Σ sin ẞ sin y sin (B− y). sin (B-) cos (8+y+6)=0. 4. cos 2 (8-y)=4II cos (B-)-1. 5. sin ẞ sin y sin (ẞ- y) = -II sin (B—y). 6. Σcot (a-B) cot (a−y)+1=0. 7. 8. 9. 10. Σ sin 3a sin (8-y)=4 sin (a+B+y) II sin (8-y). cos (0+a) cos (B+y) sin (8 − a) sin (B− y)=0. -II sin (ẞ+y). II sin (ß − y). == 14. sin3 (B+y) sin3 (8-y)=3II sin (8+y). II sin (8-y). 15. cos3 (B+y+0) sin3 (8-y) =311 cos (B+y+0). II sin (ẞ− y). 16. If x+y+2=xyz, prove that 3x-x3 3x-x3 1-3x2 1-3.x2 17. If yz+zx+xy=1, prove that Σx (1 − y2) (1 − z2) = 4xyz. 310. From a trigonometrical identity many others may be derived by various substitutions. For instance, if A, B, C are any angles, positive or negative, connected by the relation A+B+С=π, we know that Let then and Also sin 2a+sin 28+ sin 2y=4 sin a sin ẞ sin y. Again, let 4=-, B=1-2, C= then 2 П α П A В 2 2' |