How do you evaluate $arcsin - \frac{1}{2}$ $arcsin -1/2$ ?

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How do you evaluate $arcsin - \frac{1}{2}$ $arcsin -1/2$ ?

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Answer 1 · 2015-05-29T22:58:08

Consider an equilateral triangle with sides of length 2.
Each of the internal angles will be $\frac{π}{3}$ $pi/3$ (i.e. $60^{o}$ $60^o$ ).

Now split the triangle into two mirror image right angled triangles.
The shortest side of each will have length $1$ $1$ , and the smallest angle opposite it will be $\frac{π}{6}$ $pi/6$ (i.e. $30^{o}$ $30^o$ ).

Then by definition $sin (\frac{π}{6}) = \frac{1}{2}$ $sin(pi/6) = 1/2$ - the length of the shortest side divided by the length of the hypotenuse.

Now $sin (- θ) = - sin (θ)$ $sin(-theta) = -sin(theta)$ , so

$sin (- \frac{π}{6}) = - sin (\frac{π}{6}) = - \frac{1}{2}$ $sin(-pi/6) = -sin(pi/6) = -1/2$

The range of $arcsin$ $arcsin$ is $- \frac{π}{2} \leq θ \leq \frac{π}{2}$ $-pi/2 <= theta <= pi/2$ .

$- \frac{π}{6}$ $-pi/6$ lies in this range so $arcsin (- \frac{1}{2}) = - \frac{π}{6}$ $arcsin(-1/2) = -pi/6$

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How do you evaluate $arcsin - \frac{1}{2}$ $arcsin -1/2$ ?

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