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14. (a sin 0 cos + r cos 0 cos ) (6 sin o sin • + r sin 6 cos o) - (b sin 0 cos
r sin o sin o) (a sine sin 0 + r cos e sin o)
= r sin o (r cos 0 + a sin o). 15. If x = r sin 0 cos , Y = r sin o sin $, % = r cos 0, show that " + y2 + m 2
16. If a = b cos C+ c cos B, b = a cos C + c cos A, C = ( cos B + b cos A, show that a2 + 72 - C 2 ab cos C. 17. Given sino A + 3 sin A 1, find sin A. 18. Given cos” A
sin A =
25, find cos A.
1 19. Solve sin A
for sin A.
V2 20. Find tan A, when tan A + 1 . sec A. 21. Given a cos A b sin A + a, find cot A. 22. Given tan’ A 7 tan A + 6 O, find tan A. 23. Show that VI + 2 sin A cos A + VI – 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 cosA,
m cos' A + n cos' B b sino A, find sin A and sin B.
CONTRARIETY OF SIGNS.- CHANGES OF MAGNITUDE AND SIGN OF
RIGONOMETRICAL RATIOS THROUGH THE FOUR QUADRANTS. 9. We have explained at some length the meaning and
use of the signs + and in algebra. They have a similar interpretation in trigonometry.
1. Lines.—Draw the horizontal A
line A 'A, and draw BB' at right angles, meeting it in 0. Then considering O as origin,
(1.) All lines drawn to the right B'
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 revolving line by which angles may be conceived to be generated.
2. Angles.--A similar convention is made forangles. Let
A OA be an initial line, and let a revolving line about the centre O take up the positions
P 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.
A A ZAOP A, then ZAOP'= - A.
10. We will now examine the trigonometrical ratios for
P4 angles greater than a right angle, and for negative angles.
Let OP, OP, OP, OP, represent the position of the revolving line at any period of revolution in the several quadrants respectively,
And let PLN, P.Ñ,, P2N3, P.N4 be the respective perpendiculars from the end of the revolving line upon the initial line.
Then P N, P,N, P,N, PLN, are respectively the perpendiculars corresponding to the angles generated.
Also, ON, ON, ON, ON, are respectively the bases of the right-angled triangles with respect to the angles in question. We have then in the second quadrantPN
It is therefore evident that the relations between the trigonometrical ratios, which were proved to exist in Art. 7, also hold for angles in the second quadrant—that is, angles between 90° and 180°.
And in the same way we may show that they hold for angles in the third, fourth, or any quadrant.
And again, if we suppose the line to revolve in a negative direction, and take the position OP', we shall have PN the perpendicular corresponding to the negative angle AOP', and ON' the base.
ONS And the relations proved in Art. 7 may be also similarly proved to exist here.
Hence the relations proved in Art. 7 hold for any angles whatever.
Changes of Magnitude and Sign of the Trigonometrical
Ratios. 11. Let OP, OP,, OP3, OP, be positions of the revolving line in the several quadrants respectively; P N, P,N,
* Sin 0°
P..N3, P, N4, the respective perpendiculars; and ON, ON2, ON, ON the bases of the corresponding right-angled triangles.
OP At the commencement of the motion of the revolving line, the angle AOP, = 0°;
Also, the perpendicular P N, = 0,
0 tan 0°
OP, As the revolving line moves from OA towards OB, PAN, increases and ON, diminishes; and when it arrives at OB, we have P,N, OP ,and ON, = 0. But the angle generated is now a right angle. Hence we have—
OP, tan 90°
0 Hence, as the angle increases from 0° to 90°The sine changes in magnitude from 0 to 1 and is +. The cosine changes in magnitude from 1 to 0 and is +. The tangent changes in magnitude from 0 to co and is +.
(2.) In the second quadrantHere the perpendicular P,N, is +
and the base ON, is * The student ought properly to look upon the values 0, 1, 0 here obtained as the limiting values of the sine, cosine, and tangent respectively, when the angle is indefinitely diminished.
Hence the sine during the second quadrant is +, the cosine is and the tangent is
Again, as the revolving line moves from OB to OA', the perpendicular P,N, diminishes until it becomes zero. Also, the base ON, increases in magnitude, until it finally coincides with OA', and .:: – OP, But the angle now described is 180°. Hence we have
0 Sin 180°
0, cos 180o = OP
OP Hence in the second quadrantThe sine changes in magnitude from 1 to 0, and is positive. The cosine changes in magnitude from 0 to 1, and is negative. The tangent changes in magnitude from a to 0, and is negative.
And in the same way may we trace the changes of magnitude and sign in the third and fourth quadrants.
Thus we shall find
(3.) In the third quadrantThe sine changes in magnitude from 0 to 1, and is negative. The cosine changes in magnitude from 1 to 0, and is negative. The tangent changes in magnitude from 0 to co, and is positive.
(4.) In the fourth quadrantThe sine changes in magnitude from 1 to 0, and is negative. The cosine changes in magnitude from 0 to 1, and is positive. The tangent changes in magnitude from 0 to 0,and is negative.
Moreover, as the cosecant, secant, and cotangent are respectively the reciprocals of the sine, cosine, and tangent, it follows that their signs will follow respectively the latter, and that their magnitudes will be their reciprocals.
TRIGONOMETRICAL RATIOS CONTINUED. ARITHMETICAL VALUES
OF THE TRIGONOMETRICAL RATIOS OF 30°, 45°, 60°, &c. 12. To prove that sin A = cos (90° A), and that