 Unit Circle Chart Table

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Triangles constructed on the unit circle can also be used to illustrate the periodicity of the trigonometric functions. First, construct a radius OP from the origin O to a point P(x1,y1) on the unit circle such that an angle t with 0 < t < . mw-parser-output . sfrac{white-space:nowrap}. mw-parser-output . sfrac. tion,. mw-parser-output . sfrac . tion{display:inline-block;vertical-align:-0. 5em;font-size:85%;text-align:center}. mw-parser-output . sfrac . num,. mw-parser-output . sfrac . den{display:block;line-height:1em;margin:0 0. 1em}. mw-parser-output . sfrac . den{border-top:1px solid}. mw-parser-output . sr-only{border:0;clip:rect(0,0,0,0);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px}π/2 is formed with the positive arm of the x-axis. Now consider a point Q(x1,0) and line segments PQ ⊥ OQ. The result is a right triangle △OPQ with ∠QOP = t. Because PQ has length y1, OQ length x1, and OP has length 1 as a radius on the unit circle, sin(t) = y1 and cos(t) = x1. Having established these equivalences, take another radius OR from the origin to a point R(−x1,y1) on the circle such that the same angle t is formed with the negative arm of the x-axis. Now consider a point S(−x1,0) and line segments RS ⊥ OS. The result is a right triangle △ORS with ∠SOR = t. It can hence be seen that, because ∠ROQ = π − t, R is at (cos(π − t),sin(π − t)) in the same way that P is at (cos(t),sin(t)). The conclusion is that, since (−x1,y1) is the same as (cos(π − t),sin(π − t)) and (x1,y1) is the same as (cos(t),sin(t)), it is true that sin(t) = sin(π − t) and −cos(t) = cos(π − t). It may be inferred in a similar manner that tan(π − t) = −tan(t), since tan(t) = y1/x1 and tan(π − t) = y1/−x1. A simple demonstration of the above can be seen in the equality sin(π/4) = sin(3π/4) = 1/√2.

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