# PracticeThree Angles

#### puremath

##### New member
List three angles (in radian measure) that have a sine of -1/2.

#### MarkFL

##### La Villa Strangiato
Math Helper
[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. • puremath

#### puremath

##### New member
[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.

#### MarkFL

##### La Villa Strangiato
Math Helper
I used the fact that the quadrant 3 and quadrant 4 angles satisfying the problem are symmetric about the negative $$y$$-axis. The reference angle must be [MATH]\frac{\pi}{6}[/MATH] and so the value to be added to or subtracted from the negative $$y$$-axis must be complementary to the reference angle, which is [MATH]\frac{\pi}{3}[/MATH]. Then we observe that the period of the sine function is $$2\pi$$ so adding any multiple of this will result in the sine function having the same value.

• puremath

#### MarkFL

##### La Villa Strangiato
Math Helper
I have more time now, so I'll try to outline exactly how I approached this. Consider the following diagram: 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]

#### MarkFL

##### La Villa Strangiato
Math Helper
Looks like I wasted my time again.

#### puremath

##### New member
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.