Draw any line OR, and take M in it so that the measure of OM is unity. Draw MP perpendicular to OM, and take P so that the measure of MP is a. Join OP. Then the angle ROP is the Thus an angle ROP has been drawn so that tan ROP=a. 117. DEFINITION. One angle is called the complement of another, when the two angles added together make up a right angle. Example 1. The complement of A is (90°-A). Example 2. The complement of 190° is (90° - 190°)=-100o. 5п 4 Example 3. The complement of T is -5T) Зп = 4 118. To prove that the sine of an angle A is equal to the cosine of its complement (90° – A). [This is true whatever be the magnitude of A, as will be proved later on.] If A be less than 90o, let ROP be A (see last figure). Draw PM perpendicular to OR. Then since PMO = 90°, therefore POM +0PM = 90o and therefore OPM= (90° — A). EXAMPLES. XIX. (1) Draw an angle whose sine is . (3) Draw an angle whose tangent is 2. (4) Can an angle be drawn whose tangent is 431 ? (5) Can an angle be drawn whose cosine is ? (6) Can an angle be drawn whose secant is 5? (7) Find the complements of (8) Prove that sin 70°= cos 20o. (9) Prove that cos 47° 16′=sin 42° 44'. (10) Prove that tan 79°=cot 11o. ON THE SOLUTION OF TRIGONOMETRICAL EQUATIONS. 119. A TRIGONOMETRICAL equation is an equation in which there is a letter, such as 0, which stands for an angle of unknown magnitude. The solution of the equation is the process of finding an angle which, if it be substituted for 0, satisfies the equation. Example 1. Solve cos 0=4. This is a Trigonometrical equation. To solve it we must find some angle such that its cosine is §. We know that cos 60o. Therefore if 60° be put for the equation is satisfied. .. 0=60o is a solution of the equation. The usual method of solution is to express all the Trigonometrical Ratios in terms of one of them. This is an equation in which 6, and therefore sin 0 is unknown. It will be convenient if we put x for sin 6, and then solve the equation for x as an ordinary algebraical equation. Thus we get sin 6-2, or sin 0. = The value 2 is inadmissible for sin 8, for there is no angle whose sine is numerically greater than 1 (Art. 115). Therefore one angle which satisfies this equation for ✔ is 30o. 12 Example 3. To solve the equation cosec - cot2 ◊ +1=0. therefore 30o is one angle which satisfies the equation for 0. EXAMPLES. XX. Find one angle which satisfies each of the following equations. ** MISCELLANEOUS EXAMPLES. XXI. (1) Prove that 3 sin 60° - 4 sin3 60°-4 cos3 30° - 3 cos 30°. (2) Prove that tan 30° (1 + cos 30° + cos 60°)=sin 30° + sin 60°. (3) If 2 cos2 6-7 cos 0+3=0, show there is only one value of cos 0. (4) Find cos (5) Find sin and prove that one (6) Find tan (7) If 3 cos2 from the equation 8 cos2 - 8 cos +1=0. from the equation 8 sin2 0-10 sin 0+3=0, value of is from the equation 12 tan2 0 - 13 tan 6+3=0. +2.√3. cos 0=5, show that there is only one value of cos e, and that one value of is π 6° (8) Prove that the value of sin10+ cos1 +2. sin2. cos2 is always the same. (9) Simplify cos1 A +2. sin2 A. cos2 A. (10) Express sino A + cos® A in terms of sin2 A and powers of sin2 A. (11) Express 1+tan1 0 in terms of cos 0 and its powers. (14) Trace the changes in the magnitude of cosec 6 as 0 increases from 0 to π (15) Trace the changes in the magnitude of cot as e de |