Trigonometry
0 to ± 2p
p/6, p/4 and p/3
30° (p/6) and 45° (p/4)
Values of trigonometric functions of arcs p/6, p/4 and p/3
The given arcs are, one twelfth, one eighth and one sixth of the circumference 2p of the unit circle so the coordinates of terminal points of the arcs are the elements of the equilateral triangle with the side a = 1 (Fig.a, c) and the sides of the square with diagonal d = 1, (Fig. b).
The values of the trigonometric functions of arcs that are multipliers of 30° (p/6) and 45° (p/4)
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.
Trigonometry contents A