From B, drop a perpendicular BC onto AD.
Draw radius AF perpendicular to DA; draw DE, FG tangent to the circle; that is, draw them at right angles to the radii; (Theorem 13)
and extend AB to meet those tangents at E, G respectively.
Then, since AG is a straight line that crosses the parallel straight lines FG, AD, the alternate angles FGA, GAD are equal. (Theorem 8)
A complete cycle of sin θ
The student with a good imagination should now try to follow the point B as it makes a complete circuit about the unit circle. In particular, let us visualize what happens to CB, the line value of sin θ.
At θ = 0°, the point B coincides with D, so that CB vanishes. sin 0° = 0.
As θ increases in the first quadrant, CB also increases until it reaches its maximum value at 90°. CB coincides there with the unit radius AF, so that CB = AF = sin 90° = 1.
When θ enters the second quadrant, CB begins to decrease in value. At 180°, B coincides with K, and CB vanishes again. sin 180° = 0.
In the third quadrant, CB increases in absolute value, but the sine in that quadrant is negative. At 270°, CB coincides with AL. sin 270° has its algebraic minimum: sin 270° = −1.
Finally, in the fourth quadrant, CB decreases in absolute value until B coincides once again with D. sin 360° = sin 0° = 0.
We see that sin θ never has a value greater than 1 (at 90°), or less than −1 (at −90°).
−1 ≤ sin θ ≤ 1.
A complete cycle of cos θ
The line value of cos θ is AC. As the point B revolves, C shuttles back and forth along the diameter DK. At 0°, AC coincides with the unit radius AD, so that cos 0° = 1.
At θ = 90°, C coincides with A, so that AC vanishes. cos 90° = 0.
At θ = 180°, C reaches K. AC = AK = cos 180° = −1.
At θ = 270°, C is back at A. cos 270° = 0.
Finally, at θ = 360°, AC coincides again with the radius AD.
cos 360° = 1.
cos θ (AC) also has a maximum value of 1 (at 0°), and minimum of −1 (at 180°).
−1 ≤ cos θ ≤ 1.
A complete cycle of tan θ
Finally, it will be extremely instructive to follow
DE and KE, the line values of tan θ.
At θ = 0° (Fig. 1), the point E coincides with D, so that the tangent line DE vanishes. tan 0° = 0.
As θ increases towards 90° (Fig. 1), DE remains tangent to the circle — its length getting longer and longer. At 90°, DE is parallel to AF -- there is no more triangle. tan 90° does not exist.
In the second quadrant (Fig. 2), the tangent line jumps to KE. (We call that a discontinuity in the function.) In the second quadrant, however, tan θ is algebraically negative, and we indicate that by the line KE'.
At θ = 180°, the line value KE is again 0.
In the third quadrant (Fig. 3), the line value is KE, but tan θ is algebraically positive (KE'). At θ = 270°, KE becomes parallel to AL, so that tan 270° does not exist.
In the fourth quadrant (Fig. 4), KE jumps over to DE (another discontinuity), and at θ = 360°, tan 360° = tan 0° = 0.
Projections
The straight line CD is called the vertical projection of the straight line AB. (The straight lines BD, AC are drawn perpendicular to CD. It is as if a light were shining on AB, and CD is the shadow.)
EF is called the horizontal projection of AB.
Now, if AB is a unit radius, then BC -- the line value of sin θ -- is the
vertical projection of the radius. And AC -- the line value of cos θ -- is the horizontal projection.
Thus in a unit circle, sin θ is the vertical projection of the radius. cos θ is the horizontal projection.