All six values from one.

Given one ratio (e.g. sin x = 8/17), the reference right triangle is placed in the chosen quadrant on the unit circle; all six values appear with signs read from the coordinates.

With sin x = 8/17, choose the quadrant that makes cos x = −15/17 and tan x = −8/15 (quadrant II).

Taking the sign of cos or tan from the acute triangle without checking the quadrant.

• If P(a, b) is reached by turning through x, turning through −x reaches Q(a, −b), the mirror image of P in the x-axis. So cos (−x) = cos x and sin (−x) = −sin x, and hence tan (−x) = −tan x.
• Every point of the unit circle has both coordinates between −1 and 1, so −1 ≤ sin x ≤ 1 and −1 ≤ cos x ≤ 1 for every x.
• The signs follow the quadrant of P. In the first quadrant (0 < x < π/2) both coordinates are positive; in the second (π/2 < x < π) a < 0 < b; in the third (π < x < 3π/2) both are negative; in the fourth (3π/2 < x < 2π) b < 0 < a.
• Hence sin x > 0 for 0 < x < π and sin x < 0 for π < x < 2π, while cos x > 0 in the first and fourth quadrants and cos x < 0 for π/2 < x < 3π/2.
• Positive functions by quadrant: all six in the first, only sin and cosec in the second, only tan and cot in the third, only cos and sec in the fourth. A reciprocal always shares the sign of its partner.
• To find all six values from one, fix the size with an identity and the sign with the quadrant. If sin x = 8/17 with x in the second quadrant, cos² x = 1 − 64/289 = 225/289, and cos x < 0 there, so cos x = −15/17, tan x = −8/15, cosec x = 17/8, sec x = −17/15 and cot x = −15/8.
• Second example: if cos x = 7/25 with x in the fourth quadrant, sin² x = 1 − 49/625 = 576/625 and sin x < 0 there, so sin x = −24/25 and tan x = −24/7.

Teaching this idea? Put the simulation on your own page or LMS with this code.