Trigonometry
0 to ± 2p
Trigonometric functions of negative arcs or angles
Trigonometric functions of complementary angles
Trigonometric functions of supplementary angles
Trigonometric functions of arcs from 0 to ± 2p
Trigonometric functions of negative arcs or angles
 We say that arcs x and x′ are opposite if x + x′ = 0  or  x′ = -x. Comparing the corresponding sides of the congruent right-angled triangles in the right figure, OPxP  and  OPxP′ OSxS1  and  OSxS1′ OSyS2  and  OSyS2′ follows that we can express trigonometric functions of an negative arc (-x) by corresponding function of opposite arc x, that is
Example:   Given trigonometric functions of negative angle, arc or number should be expressed by the same functions of the positive angle, arc or number.
a)  sin (-200°),      b)  cos (-14p/3),      c)  tan (-11),      d)  cot (-750°).
Solution:  First express the angle a in decimal degrees, i.e.
a)  sin (-200°) = - sin 200°
c)  tan (-11) = - tan11 = - tan (3p + 1.575222...) = - tan 1.575222...
d)  cot (-750°) = - cot 750° = - cot (2 · 360° + 30°) = - cot 30°.
Trigonometric functions of complementary angles
Two angles, x and p/2 - x which form the right angle, are said to be complementary.
Thus, comparing the corresponding sides of the congruent right-angled triangles in the below figure,
OPxP  and  OPxP,       OSxS1  and  OSxS2    and     OSyS2  and  OSyS1
Example:   The trigonometric functions of the given angle or arc should be expressed by corresponding function of the complementary angle.
a)  sin 30°,      b)  cos (p/2 - p/3),      c)  tan 1,      d)  cot 530°.
Solution:   a)  sin 30° = cos(90° - 30°) = cos 60°
d)  cot 530° = tan (90° - 530°) = tan (- 440°) = - tan 440° = - tan (2 · 180° + 80°)
= - tan 80° = - cot (90° - 80°) = - cot 10°.
 Example:   Simplify the expression
 Solution:
Trigonometric functions of supplementary angles
Two angles, x and p - x, which when added form a straight
angle, are said to be supplementary.
Comparing the corresponding sides of the congruent right-angled
triangles in the right figure,
 Px′P′ = PxP   => sin (p - x) = sin x
 OPx′ = -OPx   => cos (p - x) = -cos x
 SxS1′ = -SxS1   => tan (p - x) = -tan x
 SyS2′ = -SyS2   => cot (p - x) = -cot x

Example:   Trigonometric functions of a given arc, angle or number should be expressed by the corresponding function of the supplementary angle.
a)  sin 5p/6,      b)  cos (-320°),      c)  tan (p - 1),      d)  cot 30°.
Solution:   a)  sin 5p/6 = sin (p - 5p/6) = sin p/6
b)  cos (-320°) = cos (180° - 500°) = - cos 500° = - cos (360° + 140°)
= cos (180° - 140°) = cos 40°
c)  tan (p - 1) =  - tan 1
d)  cot 30° =  - cot (180° - 30°) =  - cot 150°.
Example:   Calculate,  sin 3p/2 · cos(- p) + tan 5p/4.
Solution:  sin 3p/2 · cos(- p) + tan 5p/4 = - 1 · (- 1) + tan (p + p/4) = 1 + tan p/4 = 1 + 1 = 2.
 Example:   Calculate,
Solution:
Example:  Prove the identity,
cos2 p/3 · sin (p/2 - x) - cos (p - x) · cos2 p/6 = tan (p/2 + x) · sin (2p - x).
Solution:  Since  sin (p/2 - x) = cos x,   cos (p - x) = - cos x,   tan (p/2 + x) = - cot x
and  sin (2p - x) = - sin x  then,
(cos p/3)2 · cos x - (- cos x) · (cos p/6)2 = - cot x · (- sin x),
(1/2)2 · cos x + (Ö3/2)2 · cos x = (cos x/sin x) · sin x  =>  cos x = cos x.
Example:  Prove the identity,
cot2 (p + x) · cos2 (p/2 + x) + sin (- x) · sin (p + x) = tan (2p - x) · cot (- x).
Solution:    [cot (p + x)]2 · [cos (p/2 + x)]2 + (- sin x) · sin (p + x) = (- tan x) · (- cot x),
(cot x)2 · (- sin x)2 + (- sin x) · (- sin x) = (sin x/cos x) · (cos x/sin x)
cos2 x + sin2 x = (sin x/cos x) · (cos x/sin x) = 1.
Pre-calculus contents C