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It is agreed that when the straight line 0 C, Fig. I, revolves in the direction of the outer arrow all the angles generated will be positive, and when it revolves in the direction of the inner arrow all the angles generated will be negative. Let the straight line 0 C, commencing at C, take up the positions P, P', P" and P'", then the angles P O Q, P' O Q, P" O Q and P" O Q will all be positive. All vertical lines above DOC are positive or plus, and all vertical lines below D O C are minus; all horizontal lines to the right of N O B are plus, and all horizontal lines to the left are minus; the revolving line 0 C is always plus.
In forming the ratios like signs give plus and unlike signs give minus.
It will be observed that all the lines in the first quadrant, NC, are plus, consequently all the ratios are plus. In the second quadrant, N D, the vertical lines are plus and the horizontal lines are minus; consequently all the ratios in the second quadrant are minus, except the sine and cosecant. The student should prove this for himself by forming the ratios and observing the rule that like signs give plus and unlike signs minus. It is of the utmost importance to know whether a ratio is plus or minus, because it enables you to recognise at once the magnitude of the angle with which you are dealing.
In the plane triangle P O Q, Fig. 1, let the sides be denoted by the letters p, 0, q. Let 20 = 0.
i.e., I=sin.2 4 + cos. 2 0; i.e., cosec. 2 0 =1 + cot. 20; i.e., sec. 20 =tan.? 0 +I
- sin. 20
= cos. 2 0, and I cos. 2 0 = sin. 20 Cosec. 20 cot.20, and cosec. 2 A cot.2 0 = 1 Sec. 20 tan.2 0 = I, and sec. 2 0 -I = tan.20
When one ratio multiplied by another produces unity one is said to be the reciprocal of the other. (See Fig. I for triangle.)
Þ and x = 1.. the tan. and cot. are reciprocals.
It is very frequently a source of great saving of figures to substitute the reciprocal of a ratio in an equation.
The above results may be stated thus, and should be remembered
(1) Sine squared of an angle plus cosine squared of the same angle equals 1.
(2) Cosecant squared of an angle equals i plus cotangent squared of the same angle.
(3) Secant squared of an angle equals tangent squared of the same angle, plus 1.
As the sine of an angle increases from o at oo to unity at goo; the cosine decreases from unity at oo to o at 90°.
The tangent of an angle increases from o at oo to (infinity) at 90°, and the cotangent decreases from infinity at oo to o at 90°.
The secant increases from unity at 0° to infinity at 90°, and the cosecant decreases from infinity at oo to unity at 90°.
VALUES FOR RATIOS OF 30° AND 60° Let A B C, Fig. 2, be an equilateral triangle, then by Euclid, Book I., Prop. 32, each of its angles equals 60°. Bisect the angle B by a line cutting the base A C in D, then the angle ABD is 30°. Because A B = B C and B D is common to both, and angle A B D = angle C B D, therefore BD bisects A Cat right angles.
All the other ratios of 30° and 60° are found in like manner.
VALUES FOR RATIOS OF 45° Let A B C, Fig. 3, be a right-angled triangle, then by Euclid, Book I., Prop. 32, A CB = 45°.
Any ratio of 45° can be found in a similar manner.
VALUES FOR RATIOS OF A + B Let M C N, Fig. 4, be angle A and NCO be angle B, then M CO = (A + B).
Now FDG + FGD
FDG = FGC because each is the complement of FGD
Dividing numerator and denominator by cos. A x cos. B we get
sin. A X cos. B cos. A x sin. B
cos. A X cos. B sin. A X sin. B
tan. A + tan. B
The above results are very important, and should be committed to memory, thus
The sine of the sum of two angles equals the sine of the first into the cosine of the second plus the cosine of the first into the sine of the second.
The cosine of the sum of any two angles equals the cosine of the first angle into the cosine of the second angle minus the sine of the first into the sine of the second.
The tangent of the sum of any two angles equals the tangent of the first angle plus the tangent of the second, divided by i minus the tangent of the first into the tangent of the second.
VALUES FOR RATIOS OF A - B)
In C N take a point G, and from G draw G H perpendicular to CM; from G draw G D perpendicular to CO, through D draw D E parallel to GH, and through
E Н D draw D F parallel to C M, meeting HG
Fig. 5. produced, in F. Now
FGD + FDG and
FDG + FDO therefore FGD + FDG FDG + FDO and
FGD = F D O because each is the complement of F D G. Now
FDO = DCM = A therefore F G D = A
GH HF - FG DE - FG
= cos, A X cos. B + sin. A x sin. B
sin. (A — B) Tan. (A – B) = tan. MCN
cos. (A B) sin. A X cos. B
cos, A X sin. B
cos. A X cos. B + sin. A X sin. B Dividing numerator and denominator by cos. A X cos B then
sin. A x cos. B cos. A X sin. B
cos. A X ws. B cos. A X cos. B Tan. (A — B) COS. A X cos. B sin. A X sin. B
tan. A — tan. B