Question

draw the angle in standard position and state the quadrant in which the terminal side of the angle lies or the axis on which the terminal side of the angle lies.
$-\frac{7pi}{2}$

the terminal side of the angle lies in which quadrant or on which axis?

choose the correct drawing of the given angle below.
a.

b.

c.

d.

e.

f.

g.

h.

Explanation:

Step1: Convert negative angle to positive

To find the coterminal angle of $-\frac{7\pi}{2}$, we add multiples of $2\pi$ (since the period of the angle in standard position is $2\pi$). Let's add $4\pi$ (which is $2\times2\pi$) to $-\frac{7\pi}{2}$:
$$-\frac{7\pi}{2}+ 4\pi=-\frac{7\pi}{2}+\frac{8\pi}{2}=\frac{\pi}{2}$$

Step2: Determine the position

The angle $\frac{\pi}{2}$ (or $90^\circ$) in standard position has its terminal side along the positive $y$-axis. Wait, no, wait. Wait, let's check again. Wait, $-\frac{7\pi}{2}$: let's do it step by step. Let's find how many full rotations (each $2\pi$) are in $-\frac{7\pi}{2}$.

The absolute value of $-\frac{7\pi}{2}$ is $\frac{7\pi}{2}= 3\pi+\frac{\pi}{2}= 2\pi + \pi+\frac{\pi}{2}$. So when we rotate clockwise (because the angle is negative) by $\frac{7\pi}{2}$, we can think of it as rotating clockwise $3$ full rotations and then an additional $\pi+\frac{\pi}{2}=\frac{3\pi}{2}$? Wait, no, maybe a better way: add $2\pi$ until we get an angle between $0$ and $2\pi$.

$-\frac{7\pi}{2}+ 2\pi=-\frac{7\pi}{2}+\frac{4\pi}{2}=-\frac{3\pi}{2}$

$-\frac{3\pi}{2}+ 2\pi=-\frac{3\pi}{2}+\frac{4\pi}{2}=\frac{\pi}{2}$

Ah, so the coterminal angle is $\frac{\pi}{2}$, which is $90^\circ$, so the terminal side is along the positive $y$-axis? Wait, no, wait $\frac{\pi}{2}$ is positive $y$-axis. But wait, let's check the direction. A negative angle means clockwise rotation. So starting from the positive $x$-axis, rotating clockwise by $\frac{7\pi}{2}$. Let's see, $\frac{7\pi}{2}= 3\pi+\frac{\pi}{2}= 2\pi+\pi+\frac{\pi}{2}$. So rotating clockwise $2\pi$ (a full circle) brings us back to the positive $x$-axis. Then rotating clockwise another $\pi+\frac{\pi}{2}=\frac{3\pi}{2}$. Rotating clockwise $\pi$ from positive $x$-axis is negative $x$-axis, then rotating clockwise $\frac{\pi}{2}$ more is negative $y$-axis? Wait, I think I made a mistake in the coterminal angle calculation.

Wait, let's calculate the coterminal angle correctly. The formula for coterminal angles is $\theta + 2\pi n$, where $n$ is an integer. We want to find $n$ such that $\theta + 2\pi n$ is between $0$ and $2\pi$.

For $\theta=-\frac{7\pi}{2}$, let's solve for $n$:

$0\leq -\frac{7\pi}{2}+ 2\pi n< 2\pi$

Add $\frac{7\pi}{2}$ to all parts:

$\frac{7\pi}{2}\leq 2\pi n< 2\pi+\frac{7\pi}{2}=\frac{4\pi + 7\pi}{2}=\frac{11\pi}{2}$

Divide all parts by $2\pi$:

$\frac{7\pi}{2}\div(2\pi)\leq n<\frac{11\pi}{2}\div(2\pi)$

$\frac{7}{4}\leq n<\frac{11}{4}$

Since $n$ must be an integer, $n = 2$ (because $\frac{7}{4}=1.75$ and $\frac{11}{4}=2.75$). So $n = 2$:

$\theta + 2\pi\times2=-\frac{7\pi}{2}+ 4\pi=-\frac{7\pi}{2}+\frac{8\pi}{2}=\frac{\pi}{2}$

So the coterminal angle is $\frac{\pi}{2}$, which is $90^\circ$, so the terminal side is along the positive $y$-axis. Wait, but let's visualize the rotation. A negative angle is clockwise. So starting at positive $x$-axis, rotating clockwise by $\frac{7\pi}{2}$. Let's see, $\frac{7\pi}{2}= 3.5\pi$. A full clockwise rotation is $-2\pi$, so $3.5\pi$ clockwise is the same as $3.5\pi - 2\pi=1.5\pi$ clockwise? No, wait, the magnitude of the angle is $\frac{7\pi}{2}$, direction is clockwise. So the terminal side after rotating clockwise $\frac{7\pi}{2}$ is the same as rotating counterclockwise $\frac{\pi}{2}$ (since $\frac{7\pi}{2}- 2\pi\times2=\frac{7\pi}{2}-4\pi=-\frac{\pi}{2}$? Wait, I'm getting confused. Let's use the unit circle.

The standard position: initial side on positive $x$-axis, terminal side determined by rotating counterclockwise (positive angle) or clockwise (negative angle).

For $-\frac{7\pi}{2}$:

Let's compute the angle modulo $2\pi$. The formula for modulo in radians: $\theta \mod 2\pi=\theta - 2\pi\times\lfloor\frac{\theta}{2\pi}
floor$ (for positive $\theta$), but for negative $\theta$, we can add $2\pi$ until we get a positive angle.

$-\frac{7\pi}{2}+ 2\pi=-\frac{3\pi}{2}$

$-\frac{3\pi}{2}+ 2\pi=\frac{\pi}{2}$

So the angle is coterminal with $\frac{\pi}{2}$, which is $90^\circ$, so the terminal side is along the positive $y$-axis. Wait, but $\frac{\pi}{2}$ is positive $y$-axis. But let's check the direction. Rotating clockwise $\frac{7\pi}{2}$: $\frac{7\pi}{2}= 3.5\pi$. A clockwise rotation of $2\pi$ is a full circle, so $3.5\pi$ clockwise is $3.5\pi - 2\pi=1.5\pi$ clockwise? No, $3.5\pi$ clockwise is the same as $3.5\pi$ in the clockwise direction, which is equivalent to $2\pi - 3.5\pi= -1.5\pi$ counterclockwise, which is $1.5\pi$ clockwise, which is $-\frac{3\pi}{2}$ counterclockwise, which is coterminal with $\frac{\pi}{2}$? Wait, no, I think the earlier calculation is correct. The coterminal angle is $\frac{\pi}{2}$, so the terminal side is on the positive $y$-axis. Wait, but let's check with a simpler angle. For example, $-\frac{\pi}{2}$ is clockwise $\frac{\pi}{2}$, terminal side on negative $y$-axis? No, wait $-\frac{\pi}{2}$: clockwise rotation of $\frac{\pi}{2}$ from positive $x$-axis is down to negative $y$-axis? Wait, no: positive $x$-axis, rotate clockwise $90^\circ$ (which is $\frac{\pi}{2}$ radians) is towards negative $y$-axis? Wait, no, clockwise from positive $x$-axis: 90 degrees clockwise is positive $y$-axis? Wait, no! Wait, standard position: positive $x$-axis is 0, counterclockwise is positive angles. So 90 degrees counterclockwise is positive $y$-axis (angle $\frac{\pi}{2}$). 90 degrees clockwise is negative $y$-axis (angle $-\frac{\pi}{2}$). Oh! I made a mistake earlier. So my coterminal angle calculation was wrong. Let's correct that.

So, to find the coterminal angle of $-\frac{7\pi}{2}$ between $0$ and $2\pi$, we add $2\pi$ until we get a positive angle.

First, $-\frac{7\pi}{2}+ 2\pi=-\frac{7\pi}{2}+\frac{4\pi}{2}=-\frac{3\pi}{2}$ (still negative)

$-\frac{3\pi}{2}+ 2\pi=-\frac{3\pi}{2}+\frac{4\pi}{2}=\frac{\pi}{2}$ (now positive). Wait, but according to the direction, $-\frac{\pi}{2}$ is clockwise $\frac{\pi}{2}$, terminal side on negative $y$-axis. But $\frac{\pi}{2}$ is counterclockwise $\frac{\pi}{2}$, terminal side on positive $y$-axis. So there's a contradiction here. The mistake is in the direction of rotation. When we add $2\pi$ (a full counterclockwise rotation) to a negative angle (clockwise rotation), we are effectively rotating counterclockwise by $2\pi$, which cancels a clockwise rotation of $2\pi$. So let's think of the angle $-\frac{7\pi}{2}$ as a clockwise rotation of $\frac{7\pi}{2}$. Let's break down $\frac{7\pi}{2}$ into full rotations and a remainder. $\frac{7\pi}{2}= 3\times2\pi+\frac{\pi}{2}$? No, $2\pi$ is a full rotation, so $\frac{7\pi}{2}= 3\pi+\frac{\pi}{2}= 2\pi+\pi+\frac{\pi}{2}$. So a clockwise rotation of $\frac{7\pi}{2}$ is a clockwise rotation of $2\pi$ (which brings us back to the positive $x$-axis), then a clockwise rotation of $\pi+\frac{\pi}{2}=\frac{3\pi}{2}$. A clockwise rotation of $\pi$ from positive $x$-axis is negative $x$-axis, then a clockwise rotation of $\frac{\pi}{2}$ from negative $x$-axis is positive $y$-axis? Wait, no: clockwise from negative $x$-axis (180 degrees) rotating clockwise 90 degrees ( $\frac{\pi}{2}$ radians) would be towards positive $y$-axis? Wait, no, negative $x$-axis is 180 degrees. Clockwise 90 degrees from 180 degrees is 90 degrees (180 - 90 = 90 degrees), which is positive $y$-axis. Wait, that makes sense. So after rotating clockwise $2\pi$ (back to positive $x$-axis), then clockwise $\pi$ (to negative $x$-axis), then clockwise $\frac{\pi}{2}$ (to positive $y$-axis). So the terminal side is on the positive $y$-axis? But earlier I thought $-\frac{\pi}{2}$ is negative $y$-axis, which is correct. So where is the mistake?

Wait, let's take a smaller negative angle: $-\frac{\pi}{2}$. Clockwise rotation of $\frac{\pi}{2}$: from positive $x$-axis, turn clockwise 90 degrees, which is towards negative $y$-axis. So terminal side on negative $y$-axis. Now, $-\frac{3\pi}{2}$: clockwise rotation of $\frac{3\pi}{2}$. From positive $x$-axis, clockwise 270 degrees: that's the same as counterclockwise 90 degrees (since 360 - 270 = 90), so terminal side on positive $y$-axis. Ah! There we go. So $-\frac{3\pi}{2}$ is equivalent to counterclockwise $\frac{\pi}{2}$, so terminal side on positive $y$-axis. Then $-\frac{7\pi}{2}=-\frac{3\pi}{2}- 2\pi$. So $-\frac{7\pi}{2}$ is coterminal with $-\frac{3\pi}{2}$ (since we subtract $2\pi$, which is a full clockwise rotation, so same terminal side). Wait, no, coterminal angles differ by multiples of $2\pi$. So $-\frac{7\pi}{2}$ and $-\frac{3\pi}{2}$ differ by $-2\pi$, so they are coterminal. And $-\frac{3\pi}{2}$ and $\frac{\pi}{2}$ differ by $2\pi$ (since $-\frac{3\pi}{2}+ 2\pi=\frac{\pi}{2}$), so they are coterminal. So $-\frac{7\pi}{2}$, $-\frac{3\pi}{2}$, and $\frac{\pi}{2}$ all have the same terminal side, which is the positive $y$-axis? Wait, but $-\frac{3\pi}{2}$: clockwise 270 degrees is the same as counterclockwise 90 degrees, so positive $y$-axis. Yes, that's correct. So the terminal side of $-\frac{7\pi}{2}$ is on the positive $y$-axis? Wait, no, wait $-\frac{3\pi}{2}$: let's calculate the angle in degrees. $-\frac{3\pi}{2}$ radians is $-270$ degrees. A negative angle in degrees: $-270$ degrees is equivalent to $360 - 270 = 90$ degrees, which is positive $y$-axis. Yes! So $-270$ degrees is the same as 90 degrees, terminal side on positive $y$-axis. So $-\frac{7\pi}{2}$ radians: let's convert to degrees. $\pi$ radians is 180 degrees, so $\frac{7\pi}{2}$ radians is $630$ degrees. So $-\frac{7\pi}{2}$ radians is $-630$ degrees. Now, $-630 + 720$ (which is $4\times180$) = 90 degrees. So $-630$ degrees is equivalent to 90 degrees, terminal side on positive $y$-axis. Ah, there we go. So the terminal side of $-\frac{7\pi}{2}$ is on the positive $y$-axis.

Now, looking at the options, the angle with terminal side on the positive $y$-axis: we need to find the diagram where the terminal side is along the positive $y$-axis. Looking at the options, option G (maybe? Wait, the options are A to H. Let's assume that the diagram with the terminal side on the positive $y$-axis is the correct one. But the first part: the terminal side lies on the positive $y$-axis (or we can say the $y$-axis, since positive and negative $y$-axis are both $y$-axis, but in this case, it's positive $y$-axis). Wait, but let's confirm again.

Wait, $-\frac{7\pi}{2}$:

Coterminal angle: add $4\pi$ (since $2\pi\times2 = 4\pi$) to $-\frac{7\pi}{2}$:

$-\frac{7\pi}{2}+ 4\pi=-\frac{7\pi}{2}+\frac{8\pi}{2}=\frac{\pi}{2}$, which is $90^\circ$, so terminal side on positive $y$-axis. So the terminal side lies on the $y$-axis (specifically positive $y$-axis, but in terms of quadrant or axis, it's on the $y$-axis).

Answer:

The terminal side of the angle $-\frac{7\pi}{2}$ lies on the positive $y$-axis (or the $y$-axis). For the drawing, the correct diagram should have the terminal side along the positive $y$-axis. (Assuming among the options, the one with terminal side on positive $y$-axis is the correct drawing, e.g., if option G or another option has the terminal side on positive $y$-axis, that's the correct one. But based on the angle calculation, the terminal side is on the $y$-axis.)