give the value without a calculator: tan[arc sin(-1/2)]?

Answer 1

Set up a right triangle within the unit circle with your info.

Since sin is opposite over hypotenuse, and sin is negative, your triangle must be in the third or fourth quadrants, with opposite side being -1 and the hypotenuse being 2.

You need the tangent of this angle, so you need to find the side adjacent to the angle, so use the Pythagorean theorem to find the missing side of the right triangle.

(-1)^2 + b^2 = 2^2
1 + b^2 = 4
b =+/- sqrt (3)

So, the tangent is opposite over adjacent, so, the answer is +/-(-1)/sqrt(3) which rationalizes to +/-sqrt(3)/3.

Alternatively, most trig teachers require their students to memorize the basic trig functions for certain angles (0, pi/3, pi/4, etc...) You may recognize that the sin of both 7pi/6 and 11pi/6 is -1/2, and then the tangent of these angles would be either +/-sqrt(3)/3


P.S. if your question read: tan[Arcsin(-1/2)], then that would limit the answer to -sqrt(3)/3.

Answer 2

Let t=arcsin(-1/2). Then sin t = -1/2, so t = -pi/6

tan[arc sin(-1/2)] = tan t = tan(-pi/6)
= (-1/2) / [sqrt(3)/2]
= -1/sqrt(3).

Answer 3

arcsin(-1/2) = -30 degrees or 210 degrees
Tan (-30) = -rt(3)/3
tan (210) rt (3)/3