Answer:
Choice A: Only one equation will be sufficient for describing [tex]\theta[/tex] as in [tex]\tan \theta = 0[/tex].
For [tex]\tan \theta[/tex] to be equal to [tex]0[/tex], [tex]\theta = 0 + k \pi[/tex] where [tex]k[/tex] is a whole number (including zero or negative values.)
There are two specific solutions for [tex]\theta[/tex] on the interval [tex][0,\, 2\pi)[/tex]:
- [tex]\theta= 0[/tex] (from [tex]k = 0[/tex],) and
- [tex]\theta = \pi[/tex] (from [tex]k = 1[/tex].)
Step-by-step explanation:
In a right triangle, the tangent of an angle (other than the right angle) is equal to [tex]\displaystyle \frac{\text{Opposite}}{\text{Adjacent}}[/tex].
In general, if [tex]\theta[/tex] corresponds to the point [tex](x,\, y)[/tex] on the unit circle, then [tex]\displaystyle \tan \theta = \frac{y}{x}[/tex].
This question asks that [tex]\tan \theta[/tex] shall be zero. In other words, [tex]\displaystyle \frac{y}{x} = 0[/tex]. The [tex]x[/tex]-coordinate is the denominator and can't be zero. The only possibility is that the [tex]y[/tex]-coordinate is zero. Graphically, there are two places on a unit circle where that could happen. (Wherefore the [tex](2\pi)\, k[/tex]? On a unit circle, each turn is [tex]2\pi[/tex] radians. The angle would appear to land at the exact same position after one or more complete turns of
- [tex]y = 0[/tex] at [tex](1,\, 0)[/tex], the intersection of the unit circle and the positive [tex]x[/tex]-axis. At that point, [tex]\theta = 0 + (2\pi)\, k[/tex].
- [tex]y = 0[/tex] at [tex](-1,\,0)[/tex],the intersection of the unit circle and the negative [tex]x[/tex]-axis. At that point, [tex]\theta = \pi + (2\pi)\, k[/tex].
Note that these two values are separated by exactly [tex]\pi[/tex] radians. In other words, there is a zero every time [tex]\theta[/tex] goes through a half circle. After [tex]0[/tex], there's
[tex]\theta = 0 + k\pi[/tex].
On the range [tex]0\le \theta < 2\pi[/tex],
[tex]0 \le 0 + k \pi <2\pi[/tex].
[tex]0 \le k\pi <2\pi[/tex].
[tex]0 \le k < 2[/tex].
In other words, [tex]k = 0[/tex] or [tex]k = 1[/tex]. That corresponds to [tex]\theta = 0[/tex] and [tex]\theta = \pi[/tex].
