24. Use of the Preceding Formulæ. I. To express all the other functions in terms of the sine. Since sin20+ cos2 0 = 1, .. cos@= II. To express all the other functions in terms of the tan gent. Similarly, any one of the functions of an angle may be expressed in terms of any other function of that angle. The sign of the radical will in all cases depend upon the quadrant in which the angle ◊ lies. 25. Graphic Method of finding All the Functions in Terms of One of them. To express all the other functions in terms of the cosecant. Construct a right triangle ABC, having the side BC = 1. Then cosec cosec2-1 and similarly the other functions may be expressed in terms of cosec A. 26. To find the Trigonometric Functions of 45°. - Let ABC be an isosceles right triangle in which B m A m 27. To find the Trigonometric Functions of 60° and 30°. —Let AB be an equilateral triangle. Draw AD perpendicular to BC. Then AD bisects the angle BAC and the side BC. Therefore BAD = 30°, and ABD = 60°. 2 m B m Let BAC be the angle, and BC be perpendicular to AC. Represent BC by 3, AB by 5, and consequently AC by √25-9= 4. 1 5. Given tan0=√3; find sin and cos 0. 2mn m2 + n2 28. Reduction of Trigonometric Functions to the 1st Quadrant. All mathematical tables give the trigonometric functions of angles between 0° and 90° only, but in practice we constantly have to deal with angles greater than 90°. The object of the following six Articles is to show that the trigonometric functions of any angle, positive or negative, can be expressed in terms of the trigonometric functions of an angle less than 90°, so that, if a given angle is greater than 90°, we can find an angle in the 1st quadrant whose trigonometric function has the same absolute value. 29. Functions of Complemental Angles. Let AA', BB' be two diameters of a circle at right angles, and let OP and OP' be the positions of the radius for any angle AOP = A, and its comple ment AOP' 90° – A (Art. 12). = B Hence the triangles OPM and OP'M' are equal in all The other relations are obtained by inverting the above. |