Question

question #5
if an angle in quadrant i had tanθ = $\frac{sqrt{3}}{3}$, what is the cosθ?
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$\frac{sqrt{2}}{2}$
$\frac{1}{2}$
$\frac{sqrt{3}}{2}$
question #6
which quadrant would the angle -$\frac{16pi}{3}$ lie in?
n
iii
ii
i

Explanation:

Step1: Recall tangent - sine - cosine relationship

We know that $\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\sqrt{3}}{3}$. Also, $\sin^{2}\theta+\cos^{2}\theta = 1$. From $\tan\theta=\frac{\sqrt{3}}{3}$, we have $\sin\theta=\frac{\sqrt{3}}{3}\cos\theta$.

Step2: Substitute into Pythagorean identity

Substitute $\sin\theta=\frac{\sqrt{3}}{3}\cos\theta$ into $\sin^{2}\theta+\cos^{2}\theta = 1$. So $(\frac{\sqrt{3}}{3}\cos\theta)^{2}+\cos^{2}\theta=1$. Expanding gives $\frac{1}{3}\cos^{2}\theta+\cos^{2}\theta = 1$, or $\frac{1 + 3}{3}\cos^{2}\theta=1$, $\frac{4}{3}\cos^{2}\theta=1$, $\cos^{2}\theta=\frac{3}{4}$, $\cos\theta=\pm\frac{\sqrt{3}}{2}$. Since the angle is in Quadrant I where cosine is positive, $\cos\theta=\frac{\sqrt{3}}{2}$.

Step3: Determine the quadrant of $-\frac{18\pi}{3}$

First, simplify $-\frac{18\pi}{3}=- 6\pi$. Angles that are integer - multiples of $2\pi$ lie on the positive x - axis. But if we consider the general rule for quadrants: $-\frac{18\pi}{3}$ is equivalent to an angle of $0$ radians (because $-6\pi+6\pi = 0$), and the positive x - axis is not strictly in any of the four quadrants. However, if we consider the non - standard way of looking at it in terms of the cycle, we can rewrite $-\frac{18\pi}{3}$ as a full - rotation multiple. In a more standard sense, we can say that if we start from the positive x - axis and go clockwise (negative angle direction), after a full rotation of $2\pi$ multiple times, we end up back at the positive x - axis. If we consider the closest non - axis quadrant boundaries, we can think of it as being related to the first quadrant. But if we follow the strict definition of quadrants, we note that angles of the form $2k\pi,k\in\mathbb{Z}$ are not in any of the four quadrants. But if we assume a more lenient view for the purpose of this multiple - choice question, we know that $-\frac{18\pi}{3}=-6\pi$ and we can consider it related to the first - quadrant position (since it's a starting/ending point of a full - rotation).

Answer:

Question #5: $\frac{\sqrt{3}}{2}$
Question #6: I