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PM? AM PM? + AM? (4.) Sin? A + cos? A = AP + AP = — AP? –
Hence also, transposing and taking the square root-
AP2 AM? + PM , (7.) Sec A = AM = — AM AP2 PMR + AM" = 1 + Puz
AM? (8.) Cosec* A = PM2 = - PM2 = 1 + PMI = 1 + cot? A.
PM PM (9.) Tan A = AM =
5 = sin A • COS A sin A
AM AM PM (10) Cot A = PM = AP P = cos A ; sin A
= sin A The student must make himself thoroughly master of the results in this article.
7. To express the trigonometrical ratios in terms of the sine. (1.) Cos A = V1 – sino A, by Art. 6 (6.)
sin A (2.) Tan A =
cos A' by Art. 6 (9.),
Ex. If sin A = x, find the other trigonometrical ratios. We have, cos A = V1 - (3)% = N1 - = 4.
8. To express the trigonometrical ratios in terms of the cotangent. (1.) Sin A =
=, by Art. 6 (8.) cosec A V1 + cotA'
(2.) Cos A = d = min tone mi by Art. 6 (7.)
cot A Ncot A + 1
NI + cot' A
1 (3.) Tan A = 1
Ncot? A +' oy (.) above,
Ncot? A + 1
- cot A (5.) Cosec A = V1 + cot A, by Art. 6 (8.)
And in the same way the trigonometrical ratios may be expressed in terms of any one of them.
1. Given sin A = tf, find the other trigonometrical ratios.
2. Given tan A = 44, find the remaining trigonometrical ratios. 3. If cot A = a, show that sin A =
Vī + ai 4. If vers A = b, then tan A = ~
1-b . 5. Construct by scale and compass an angle (1.) whose cosine is 4; (2.) whose tangent is & ; (3.) whose secant is W2; (4.) whose cotangent is 2 + 13.
Prove the following identities :6. (Sin A + cos A) + (sin A - cos A)2 = 2. 7. Sec? A + cos2 B. cosec B = cosec B + sino A. seco A. 8. Sec? A. cosec? A = sec? A + coseco A. 9. Sec A. cosec A = tan A + cot A. 10. Sin A - cos® A = (sin? A – cos? A) (sin A + cosé A). 1 Sec A + tan A - cosec B - cot B Cosec B + cot B sec A – tan A
sin A + sin? A cos? A + cosa A 12. 1 + sin A cos A = 5
1 - sin A cos A i 3. (x cos 0 + y sin o) (x sine + y coso) - (x cos – y sin o) (oc sin 0 - y cosu) == 2 xy.
14. (a sin o cos o + r cos 6 cos ) (sin o sin 0 + r sin 0 cos o) – (b sin coso - r sin o sin °) (a sin o sin • + r cos e sin o)
= r sin o (r cos 0 + a sin o). 15. If x = y sin o cos , y = r sin o sin p, % = r cos 0, show that ac* + y + ma = pé.
16. If a = b cos C + c cos B, b = a cos C + c cos A, c = a cos B + b cos A, show that aż + 12 – c = 2 ab cos C.
17. Given sino A + 3 sin A = }, find sin A. 18. Given cosA – sin A = 7's, find cos A. 19. Solve sin A - cos A = 75, for sin A. 20. Find tan A, when tan A + 1 = N. sec A. 21. Given a cos A = b sin A + a, find cot A. 22. Given tan' A - 7 tan A + 6 = 0, find tan A. 23. Show that V1 + 2 sin A cos A + V1 – 2 sin A cos A = 2 cos A or 2 sin A, according as A is between 0° and 45°, or between 45° and 90°. 24. Given m sin’ A + n sinB = a cos’ A,
m cos” A + n cosa B = b sino A, find sin A and sin B.
CONTRARIETY OF SIGNS.—CHANGES OF MAGNITUDE AND SIGN OF THE TRIGONOMETRICAL RATIOS THROUGH THE FOUR QUADRANTS. 9. We have explained at some length the meaning and IB.
use of the signs + and - in algebra. They have a similar interpretation in trigonometry.
1. Lines.—Draw the horizontal line A 'A, and draw BB' at right angles, meeting it in O. Then considering () as origin,
(1.) All lines drawn to the right parallel to A 'A are called positive,
and all lines drawn to the left parallel to A'A are called negative.
(2.) All lines drawn upwards parallel to B'B are called positive, and all lines drawn downwards parallel to B'B are called negative.
(3.) Lines drawn in every other direction are considered positive, as is therefore the
P revolving line by which angles may be conceived to be generated.
2. Angles.--A similar convention is made forangles. Let OA be an initial line, and let a revolving line about the centre O take up the positions OP and OP!. Then
(1.) That direction of revolution is considered positive which is contrary to that of the hands of a watch, and the angle generated is a positive angle.
(The positive direction is then upwards.) Thus, AOP is a positive angle.
(2.) The negative direction of revolution is the same as that of the hands of a watch, and the angle thus generated is a negative angle.
(The negative direction is then downwards.)
Thus, AOP'is a negative angle.
Hence, if the angles AOP and AOP' be of the same magnitude, and We have
ZAOP = A, then ZAOP'= - A.
10. We will now examine the trigonometrical ratios for
P4 P : angles greater than a right angle, and for negative angles.