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If the given point is on the terminal arm of an angle A in standard position, 0...
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You should remember that sin A is y coordinate of a point on unit circle and cos A is x coordinate of the point on unit circle.
You also should know that the functions sine and cosine are both positive in quadrant 1, `(A in (0,90^o)).`
If sine is positive and cosine is negative, then the angle is in quadrant 2, `A in (90^o,180^o).`
Hence, considering the point (6,18), you should notice that this point is not on unit circle, thus, you may evaluate sin A using the formula such that:
`sin A =` opposite leg/hypotenuse = `y/(sqrt(x^2+y^2)) => sinA = 18/(sqrt(36+324))`
`sin A = 18/(6sqrt10) => sin A = 3/sqrt10 > 1` invalid
Hence, evaluating `sin A` for the point `(6,18)` yields an invalid value, hence, you cannot evaluate sin A and cos A for the given coordinates.
Posted by sciencesolve on November 4, 2012 at 3:19 PM (Answer #1)
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