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- Thread starter puremath
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[MATH]\sin(\theta)=-\frac{1}{2}[/MATH]

This implies:

[MATH]\theta=\frac{3\pi}{2}\pm\frac{\pi}{3}+2\pi k=\frac{\pi}{6}(12k+9\pm2)[/MATH] where \(k\in\mathbb{Z}\)

So, for every integral value of \(k\), we will get two angles whose sine is -1/2.

I don't understand what you did here. Sorry.

First I pictured the unit circle, and the line \(y=-\dfrac{1}{2}\). The \(y\)-coordinate of any point on the unit circle represents the sine of the angle required to get to that point. Knowing that this line would intersect the circle in the 3rd and 4th quadrants makes sense because these are the two quadrants in which the sine function is negative.

Imagining where these two points of intersection are and the symmetry of the line and the circle about the \(y\)-axis, let me realize the two angles will be of the form:

[MATH]\theta=\frac{3\pi}{2}\pm\beta[/MATH]

You see, the angle that forms the border between the 3rd and 4th quadrants is [MATH]\frac{3\pi}{2}[/MATH].

Now, since [MATH]\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}[/MATH] I know I must have:

[MATH]\beta+\frac{\pi}{6}=\frac{\pi}{2}[/MATH]

That is, \(\beta\) and the reference angle [MATH]\frac{\pi}{6}[/MATH] are complementary. Solving, we obtain:

[MATH]\beta=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}[/MATH]

So, now I know:

[MATH]\theta=\frac{3\pi}{2}\pm\frac{\pi}{3}[/MATH]

All that is left is to observe that for any angle on the unit circle, we can add a multiple of \(2\pi\) and wind up exactly where we started, since \(2\pi\) radians is equal to one complete revolution around the circle. So we can thus write:

[MATH]\theta=\frac{3\pi}{2}\pm\frac{\pi}{3}+2\pi k[/MATH] where \(k\in\mathbb{Z}\)

All that's left is to factor:

[MATH]\theta=\frac{\pi}{6}(9\pm2+12k)=\frac{\pi}{6}(12k+9\pm2)[/MATH]

And so we have stated succinctly the family of angles \(\theta\) such that:

[MATH]\sin(\theta)=-\frac{1}{2}[/MATH]

Looks like I wasted my time again.

No, you did not waste your time. I just have not been working on math problems daily. Still searching to move out.