. Elements of plane and spherical trigonometry . tan a . (1) sin2 d . tan 0 (2) sec a cos 0 cot (3) cot2 x +tan2 a; (4) sin y cos y tan y cot y. 18. Reduce the following to expressions containing only thetangent of the angle : (1) tan x -f-1. cot x + 1 (3) (1 + cos 0) (1 — cos 0) (4) sin (f> cos ^ + cos 0 tan (2) cot /? + sec |8 - f tan 0 (cot2 0 — 1)A + cot 0 sin 0. esc (3; 28. Functions of 90° -f 0, 180° + 0, 270° + 0, in termsof d. All mathematical tables give the trigonometric func-tions of angles only up to 90°, but in practice we are con-stantly meeting with angles greater than 90°. I
. Elements of plane and spherical trigonometry . tan a . (1) sin2 d . tan 0 (2) sec a cos 0 cot (3) cot2 x +tan2 a; (4) sin y cos y tan y cot y. 18. Reduce the following to expressions containing only thetangent of the angle : (1) tan x -f-1. cot x + 1 (3) (1 + cos 0) (1 — cos 0) (4) sin (f> cos ^ + cos 0 tan (2) cot /? + sec |8 - f tan 0 (cot2 0 — 1)A + cot 0 sin 0. esc (3; 28. Functions of 90° -f 0, 180° + 0, 270° + 0, in termsof d. All mathematical tables give the trigonometric func-tions of angles only up to 90°, but in practice we are con-stantly meeting with angles greater than 90°. It thereforebecomes necessary to investigate the relations between func-tions of angles greater than 90° and those of angles less than90°, so that, knowing the latter,we may be able to find theformer. Let ACPV Fig. 15, be anyangle less than 90°, and let itbe represented by 6. Draw CPrCPV and CPV making with GP1angles of 90°, 180°, and 270°respectively. We shall then have ACP2= 90°+ 0, (a)ACP3=18O° + 0, (/5) ACP, = 270° -f 0. GO. It is readily seen that the triangles CPXMV CP2M2, GPZMVCP±M± are all equal. We have, therefore, sin AGPn CPn CM,OR cos ACP, and cos A CP2 = ——*0P„ -MA CP -sin ACPV THE TRIGONOMETRIC FUNCTIONS. 37 Hence, by (a), sin (90° + 0) = cos 0,cos (90° -j- 0) == —sin 0, from which we derive, by (23), (14), and (15),tan (90° -f 0) = —cot 0,cot (90° + 0) = — tan 0,sec (90° + 0) = —esc 0,esc (90° + 0) = sec 0. In the same way we have in the third quadrant (39) 3 CPS OP, -sin AC1 inn /^r»Q ,4 (!P 3 1 ^^ cos ACPj 3 CP3 GPX Hence, by (/5), sin (1S0° -f 0) = —sin 0/ cos (180° + 0) = —cos 0, and by (23), (14), and (15), tan (180° -f 0) = tan 0, ? cot (180° -f 0) = cot 0, sec (180° -f 0) = —sec 0, esc (180° + 0) = —esc 0., The proof of the following is left to the student: sin (270° -f 0) = —cos 0; cos (270° -f 0) = sin 0. tan (270° + 0) = —cot 0, cot (270° + 0) = —tan 0, sec (270° + 0) = esc 0, esc (270° -
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