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Answered 2013-03-05 17:01:15

Assume the angle u takes place in Quadrant IV.

Let u = arctan(-12). Then, tan(u) = -12.

By the Pythagorean identity, we obtain:

sec(u) = √(1 + tan²(u))

= √(1 + (-12)²)

= √145

Since secant is the inverse of cosine, we have:

cos(u) = 1/√145

Therefore:

sin(u) = -√(1 - cos²(u))

= -√(1 - 1/145)

= -12/√145

Otherwise, if the angle takes place in Quadrant II, then sin(u) = 12/√145

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