3. Find sin 22° 30′, cos 22° 30′, tan 22° 30′ by means of (1), (2). Compare the values with those given in the tables. 93. Functions of three times an angle. Functions of an angle in terms of functions of one-third the angle. To express tan 3 A in terms of tan A. Let A denote any angle. 3 tan A - tan3 A ̧ (Art. 51.) (1) sin 3 A = : sin (2 A + A) = sin 2 A cos A + cos 2 A sin A [Art. 50, (1)] = 2 sin A cos2 A + (1 − 2 sin2 A) sin A = 2 sin A(1 − sin2 A) + (1 − 2 sin2 A) sin A. In a similar way, cos 3 A can be expressed in terms of cos A. (2) 1. Derive formula (3). 2. On substituting x for 3 A, write (2), (3). 3. Express formulas (1), (2), (3), and the results of Ex. 2, in words. 4. Assuming the value of sin 30°, calculate sin 90°. (3) 5. From cos 30°, derive cos 90°; from tan 30°, derive tan 90°. 6. Derive sin 180°, cos 180°, tan 180°, from sin 60°, cos 60°, tan 60°, respectively. 7. Derive sin 75°, cos 75°, tan 75°, from sin 25°, cos 25°, tan 25°, respectively, as given in the tables. 8. Derive sin 37° 30', cos 37° 30', tan 37° 30', from the ratios of 75°. = tan A+tan B+tan C-tan A tan B tan C 1-tan A tan B-tan B tan C-tan C tan A Cor. 1. If A= B=C, (1) reduces to (1), Art. 93. COR. 2. If A + B + C = 180°, then tan(A+B+C) = 0, and, accordingly, the numerator of (1) is equal to zero. Hence, if A, B, C, are the three angles of a triangle, tan Atan B + tan C = tan A tan B tan C. (2) COR. 3. If A+B+C = 90°, then tan (A+B+C)= ∞, and, accordingly, the denominator of (1) is equal to zero. Hence, tan A tan B + tan B tan C + tan C tan A = 1, when A+B+C = 90°. (3) EXERCISES. 1. Show that sin(A+B+C)= sin A cos B cos C + cos A sin B cos C + cos A cos B sin C - sin A sin B sin C. If A + B + C = 180°, the first member is zero. Division of the second member by cos A cos B cos C will give relation (2) above. 2. Show that cos(A+B+C)= cos A cos B cos C-cos A sin B sin C · sin A cos B sin C - sin A sin B cos C. What does this become when A+B+C= 180° ? Division of the second If A + B + C = 90°, the first member is zero. member by cos A cos B cos C will give relation (3) above. 6. Also, that sin(A + B) sin (B+C) = sin A sin C. 7. Also, that sin2 A + sin2 B+ sin2 C = 2 + 2 cos A cos B cos C. 8. If A + B + C = 90°, show that sin 2 A + sin 2 B − sin 2 C = 4 sin A sin B cos C. 9. Find tan 4 A, tan 5 A, tan 6 A, tan 7 A, in terms of tan A. 95. Identities. In the following exercises it is required that the first member be changed into the second member. When it is difficult to do this, help is sometimes afforded by taking some steps in changing the second member into the first. The direct steps to be taken from the first member to the second may be indicated by this means. No general directions can be given concerning the making of these transformations. The two following suggestions, however, are frequently useful: (a) Since sin2 A+ cos2 A=1, unity can be substituted for the first expression, and the first expression can be substituted for unity. (b) The change of tan x, cot x, sec x, cosec x, into their values in terms of the sine and cosine, is sometimes helpful. The examples in Art. 52 belong to this class. be expressed in terms of the sine, and the denominator in terms of the In Ex. 2, the plan of transformation is more obvious. cosine. 4. 12 sin2(45° — A) = sin 2 A. 5. cos2 A+ sin2 A cos 2 B = cos2 B + sin2 B cos 2 A. 6. 1+cot 2 0 cot 0 =cosec 2 0 cot 0. 7. 4 sin A sin (60° + A)sin(60° — A) = sin 3 A. 8. cos 50 = cos(3 0 + 2 0) = 16 cos5 0 - 20 cos3 0 + 5 cos 0. 9. sin 50 16 sin5 0 - 20 sin3 0 + 5 sin 0. = = 1 96. For an acute angle of e radians, cose >1 By Art. 50, (7), cos 0 in2; by Art. 83, sin 0 - 2 sin2 0 2 sin=2 sin cos-2 tan cos-2 tan(1-sin') 2 i.e. cose > 1 96, 97.] COMPUTATION OF TRIGONOMETRIC FUNCTIONS. 161 But by Art. 83, tan> and sin 2' < 2 2 A 97. One method of computing the trigonometric functions. method of computing the trigonometric functions of angles which are in an arithmetic progression having the common difference D', will now be shown. .. sin (n + 1) D' + sin (n − 1) D" = 2 sin nD" cos D". [Art. 52, (5).] sin (n + 1)D" = 2 sin nD" cos D" — sin (n − 1) D'. (1) Hence, if the sines of the angles D", 2 D", 3 D", up to nD" be known, then sin (n + 1) D" can be computed by formula (1). The other functions can be derived from the sine. The functions for angles from 0° to 45° will serve for the angles from 45° to 90°, since the ratio of an angle is the co-ratio of its complement. When the functions have been computed for angles up to 30°, the computations for angles greater than 30° can be made more easily. For, if A is an angle less than 30°, sin (30° + A) + sin (30° — A) = 2 sin 30° cos A = cos A. sin (30°+4)= cos A- sin (30° — A). Similarly, cos (30° + A) = cos (30° — A) — sin A. (2) (3) In formula (1) suppose that D'=10′′, and let its radian measure be denoted by 0. .. sin 10" < .00004848 ..., >[.00004848 ... - (.00004848 ...)3]. Hence, to 12 places of decimals, sin 10" = .000048481368. From this, sin 20" can be found by (1); then sin 30", then sin 40", and so on. The functions of several angles can be found independently of the method just shown. Formulas involving these angles, and |