= tan A. P'M' PM tan(180°- A)= tan AOP'= OM OM Similarly the other relations may be obtained. 31. To prove sin (90°+A)= cos A, cos(90° +A)=– sin A, and tan (90° + A)=-cot A. B P' Let OP and OP' be the positions of the radius for any angle AOP= A, and AOP'= 90° + A. A А Since OP = OP', and AOP = P'OB M' M] = OP'M', the triangles POM and POM' are equal in all respects. B' P'M' OM .: sin (90°+ A)=sin AOP'= = cos A, OP OP P'M' tan (180° + A)=tan AOP' = OM' - PM - OM tan A. B 33. To prove sin (-A)=- sin A, COS (-A)=cos A, - cos tan(- A)=– tan A. Let OP and OP' be the positions of the radius for any equal angles AOP and AOP' measured from the initial A line AO in opposite directions. Then A if the angle AOP be denoted by A, the numerically equal angle AOP' will be denoted by – A (Art. 4). B' The triangles POM and P'OM are geometrically equal. A M p' B 34. To prove sin (270° + A)=sin (270° – A)=-cos A, and cos (270° + A)=- cos (270° — A) = sin A. Let the angle AOP : = A; then the angles AOQ and AOR, measured in Ν ΟΙ the positive direction, = (270° – A) A , A M and (270°+ A) respectively. The triangles POM, QON, and ROL are geometrically equal. B' = R .: sin (270°+ A)=sin (270°— A)=-cos A. .. cos (270° + A)=-cos (270° — A)=sin A. 35. Table giving the Values of the Functions of Any Angle in Terms of the Functions of an Angle less than 90°. — The foregoing results, and other similar ones, which may be proved in the same manner, are here collected for reference. QUADRANT II. sin (180° — A)= sin A. sin (90° + A)= cos (180° - A)= -cos A. cos (90° + A)=– sin A. tan (180°– A)= -tan A. tan (90° + A)=-- cot A. cot (180°– A)=– cot A. cot (90° + A)=– tan A. sec (180° — A)=– sec A. sec (90° + A)=– cosec A. cosec (180° — A)= cosec A. cosec (90° + A)= sec A. = = QUADRANT III. sin (180° + A)=- sin A. sin (270° — A)=– cos A. cos (180° + A)= -cos A. cos (270o - A)= -sin A. tan (180° + A)= tan A. tan (270°— A)= cot A. cot (180° + A)= cot A. cot (270° — A)= tan A. sec (180° + A)=– sec A. sec (270° — A)=– cosec A. cosec (180° + A)=– cosec A. cosec (270° — A)=– sec A. QUADRANT IV. sin (360° – A)=- sin A. sin (270° + A)= -cos A. cos (360° — A)= cos (270° + A)= sin A. ( tan (360° – A)=–tan A. tan (270° + A)=-cot A. cot (360° – A)=-cot A. cot (270° + A)= -tan A. sec (360° — A) = sec (270° + A)= cosec A. cosec (360°— A)=– cosec A. cosec (270°+ A)=– sec A. sec A. NOTE.- These relations* may be remembered by noting the following rules : When A is associated with an even multiple of 90°, any function of the angle is numerically equal to the same function of A. When A is associated with an odd multiple of 90°, any function of the angle is numerically equal to the corresponding cofunction of the angle A. The sign to be prefixed will depend upon the quadrant to which the angle belongs (Art. 5), regarding A as an acute angle. * Although these relations have been proved only in case of A, an acute angle, they are true whatever A may be. Thus, cos (270° – A)=-- sin A; tbe angle 270° — A being in the 3d quadrant, and its cosine negative in consequence. For an angle in the First quadrant all the functions are positive. Fourth quadrant all are negative except the cosine and secant. 36. Periodicity of the Trigonometric Functions. — Let AOP be an angle of any magnitude, as in the figure of Art. 18; then if OP revolve in the positive or the negative direction through an angle of 360°, it will return to the position from which it started. Hence it is clear from the definitions that the trigonometric functions remain unchanged when the angle is increased or diminished by 360°, or any multiple of 360°. Thus the functions of the angle 400° are the same both in numerical value and in algebraic sign as the functions of the angle of 400°— 360°, i.e., of the angle of 40°. Also the functions of 360° + A are the same in numerical value and in sign as those of A. In general, if n denote any integer, either positive or negative, the functions of n x 360° + A are the same as those of A. Thus the functions of 1470°= the functions of 30°. If 0 denotes any angle in circular measure, the functions of (2 nT + ) are the same as those of 0. Thus sin (2n7 + )=sino, cos (2 n1 +6)= cos 0, etc. By this proposition we can reduce an angle of any magnitude to an angle less than 360° without changing the values of the functions. It is therefore unnecessary to consider the functions of angles greater than 360°; the formulæ already established are true for angles of any magnitude whatever. EXAMPLES. Express sin 700° in terms of the functions of an acute angle. sin 700°= sin (360° + 340°)= sin 340°= sin (180° + 160°) = sin 160° sin 20°. – tan 40°, Express the following functions in terms of the functions of acute angles : 1. sin 204', sin 510°. ° Ans. — sin 24°, sin 30°. 2. cos (-8000), cos 359o. cos 80°, cos 1o. 3. tan 500°, tan 300°. - cot 30°. Find the value of the sine, cosine, and tangent of the following angles : 1 4. 150°. 1 . - *, 2 V3 5. - 240°. 1 V3, -V3. 2' 6. 330°. 1 1 { V3 1 7. 225°. 1 1. V2 V2 Find the values of the following functions: 8. sin 810°, sin(- 2409), cos 210°. Ans. 1, V3, -V3. 1 9. tan (- 120°), cot 420°, cot 510°. V3, 1. V3 10. sin 930°, tan 6420°. 1 1 Ans. 3 – V3, -- V3, 2 V3 11. cot1035°, cosec 570°. -1, -2. 37. Angles corresponding to Given Functions. - When an angle is given, we can find its trigonometric functions, as in Arts. 26 and 27; and to each value of the angle there is but one value of each of the functions. But in the converse proposition - being given the value of the trigonometric functions, to find the corresponding angles — we have seen (Art. 36) that there are many angles of different magnitude which have the same functions. If two such angles are in the same quadrant, they are represented geometrically by the same position of OP, so that they differ by some multiple of four right angles. |