Consider The Following. T = – 5pi/4 (A) Find The Reference Number T For The Value Of T. T = (B) Find The Terminal Point Determined By T. (X, Y) = (_
The Correct Answer and Explanation is:
(A) Find the Reference Number for
T=−5π4T = -\frac{5\pi}{4}
Step 1: Find the coterminal angle between 00 and 2π2\pi
To do this, we add 2π2\pi until the angle is in the interval [0,2π)[0, 2\pi): −5π4+2π=−5π4+8π4=3π4-\frac{5\pi}{4} + 2\pi = -\frac{5\pi}{4} + \frac{8\pi}{4} = \frac{3\pi}{4}
So, the coterminal angle is 3π4\frac{3\pi}{4}, which lies in Quadrant II.
Step 2: Find the reference number
In Quadrant II, the reference number is: π−3π4=π4\pi – \frac{3\pi}{4} = \frac{\pi}{4}
✅ Answer (A):
Reference Number = π4\frac{\pi}{4}
(B) Find the Terminal Point Determined by
T=−5π4T = -\frac{5\pi}{4}
Since all points on the unit circle are of the form (cosT,sinT)(\cos T, \sin T), we compute the cosine and sine of −5π4-\frac{5\pi}{4}.
Alternatively, we can recognize that −5π4-\frac{5\pi}{4} is coterminal with 3π4\frac{3\pi}{4}, but since signs differ depending on quadrant, we use the original angle’s quadrant.
- T=−5π4T = -\frac{5\pi}{4} lies in Quadrant III (moving clockwise from 0).
- In Quadrant III, both cosine and sine are negative.
From the reference number π4\frac{\pi}{4}, we know: cos(π4)=sin(π4)=22\cos\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}
Thus, the coordinates for T=−5π4T = -\frac{5\pi}{4} are: (x,y)=(−22,−22)(x, y) = \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)
✅ Answer (B):
Terminal Point = (−22,−22)\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)
📘 Explanation
To solve trigonometric problems involving angles on the unit circle, two key concepts are used: reference number and terminal point. The reference number helps simplify the angle into an acute version (between 0 and π2\frac{\pi}{2}) within the first quadrant, where trigonometric values are easiest to evaluate.
In part (A), the original angle is T=−5π4T = -\frac{5\pi}{4}, a negative angle, meaning it rotates clockwise from the positive x-axis. To bring this into a more usable form, we find a coterminal angle by adding 2π2\pi (a full circle): −5π4+2π=3π4-\frac{5\pi}{4} + 2\pi = \frac{3\pi}{4}
Now in standard position, 3π4\frac{3\pi}{4} is in Quadrant II. The reference number is the acute angle between the terminal side and the x-axis. In Quadrant II, we subtract the angle from π\pi: π−3π4=π4\pi – \frac{3\pi}{4} = \frac{\pi}{4}
This gives us the reference number: π4\frac{\pi}{4}.
In part (B), we identify which quadrant −5π4-\frac{5\pi}{4} falls in. Moving clockwise past −π-\pi (or −4π/4-4\pi/4), we land in Quadrant III. Here, both sine and cosine are negative. The terminal point uses the same values as π4\frac{\pi}{4} (since that’s the reference angle), but with the appropriate signs: (cosT,sinT)=(−22,−22)(\cos T, \sin T) = \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)
This point lies on the unit circle and represents the exact position of the angle −5π4-\frac{5\pi}{4}.
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