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Sagot :
Let's go through Henry's steps and identify where he made errors in his process of finding the exact value of [tex]\( \cos \frac{10 \pi}{3} \)[/tex].
### Henry's Steps:
1. Subtract [tex]\( 2 \pi \)[/tex] from [tex]\( \frac{10 \pi}{3} \)[/tex] as many times as possible:
[tex]\[ \frac{10 \pi}{3} - 2 \pi = \frac{10 \pi}{3} - \frac{6 \pi}{3} = \frac{4 \pi}{3} \][/tex]
This step is correct. [tex]\( \frac{10 \pi}{3} \)[/tex] reduced by cycle of [tex]\( 2 \pi \)[/tex] lands at [tex]\( \frac{4 \pi}{3} \)[/tex].
2. Find the reference angle for [tex]\( \frac{4 \pi}{3} \)[/tex]:
Here, Henry states:
[tex]\[ \frac{3 \pi}{2} - \frac{4 \pi}{3} = \frac{\pi}{6} \][/tex]
However, this approach is incorrect. The correct way to find the reference angle for [tex]\( \frac{4 \pi}{3} \)[/tex] is to subtract [tex]\( \pi \)[/tex]:
[tex]\[ \text{Reference angle} = \frac{4 \pi}{3} - \pi = \frac{4 \pi}{3} - \frac{3 \pi}{3} = \frac{\pi}{3} \][/tex]
3. The cosine value for [tex]\( \frac{\pi}{6} \)[/tex] is [tex]\( \frac{\sqrt{3}}{2} \)[/tex]:
While this cosine value is correct for [tex]\( \frac{\pi}{6} \)[/tex], the reference angle [tex]\( \frac{\pi}{6} \)[/tex] is wrong.
4. The cosine value is positive because [tex]\( \frac{\pi}{6} \)[/tex] is in the first quadrant:
This conclusion is also incorrect because [tex]\( \frac{4 \pi}{3} \)[/tex] is located in the third quadrant where cosine values are negative.
### Correct Procedure:
Let's correct the solution step-by-step:
1. Reduce the angle:
[tex]\[ \frac{10 \pi}{3} \equiv \frac{4 \pi}{3} \, (\text{mod} \, 2 \pi) \][/tex]
This reduction is correct.
2. Determine the quadrant:
[tex]\( \frac{4 \pi}{3} \)[/tex] is in the third quadrant of the unit circle.
3. Find the reference angle:
[tex]\[ \text{Reference angle} = \frac{4 \pi}{3} - \pi = \frac{\pi}{3} \][/tex]
4. Evaluate the cosine for the reference angle in the third quadrant:
[tex]\[ \cos \frac{\pi}{3} = \frac{1}{2} \][/tex]
Since cosine is negative in the third quadrant:
[tex]\[ \cos \frac{4 \pi}{3} = -\frac{1}{2} \][/tex]
Therefore, the correct value of [tex]\( \cos \frac{4 \pi}{3} \)[/tex] is [tex]\( -\frac{1}{2} \)[/tex].
### Identifying Henry's Errors:
- Henry incorrectly calculated the reference angle. Instead of using [tex]\( \frac{4 \pi}{3} - \pi \)[/tex], he incorrectly used [tex]\( \frac{3 \pi}{2} \)[/tex].
- He misunderstood the quadrant properties for cosine. In the third quadrant, cosine values should be negative.
### Correct Answer:
- Henry’s reference angle calculation is incorrect.
- Henry’s determination of the cosine sign is incorrect.
The correct value of [tex]\( \cos \frac{10 \pi}{3} \)[/tex] is [tex]\( -\frac{1}{2} \)[/tex].
Thus, the numerical result, [tex]\( (-0.5, 0.5) \)[/tex], implies:
- The wrong result by Henry is [tex]\( \frac{1}{2} \)[/tex] ([tex]\( 0.5 \)[/tex] in decimal).
- The correct result is [tex]\( -\frac{1}{2} \)[/tex] ([tex]\(-0.5\)[/tex] in decimal).
### Henry's Steps:
1. Subtract [tex]\( 2 \pi \)[/tex] from [tex]\( \frac{10 \pi}{3} \)[/tex] as many times as possible:
[tex]\[ \frac{10 \pi}{3} - 2 \pi = \frac{10 \pi}{3} - \frac{6 \pi}{3} = \frac{4 \pi}{3} \][/tex]
This step is correct. [tex]\( \frac{10 \pi}{3} \)[/tex] reduced by cycle of [tex]\( 2 \pi \)[/tex] lands at [tex]\( \frac{4 \pi}{3} \)[/tex].
2. Find the reference angle for [tex]\( \frac{4 \pi}{3} \)[/tex]:
Here, Henry states:
[tex]\[ \frac{3 \pi}{2} - \frac{4 \pi}{3} = \frac{\pi}{6} \][/tex]
However, this approach is incorrect. The correct way to find the reference angle for [tex]\( \frac{4 \pi}{3} \)[/tex] is to subtract [tex]\( \pi \)[/tex]:
[tex]\[ \text{Reference angle} = \frac{4 \pi}{3} - \pi = \frac{4 \pi}{3} - \frac{3 \pi}{3} = \frac{\pi}{3} \][/tex]
3. The cosine value for [tex]\( \frac{\pi}{6} \)[/tex] is [tex]\( \frac{\sqrt{3}}{2} \)[/tex]:
While this cosine value is correct for [tex]\( \frac{\pi}{6} \)[/tex], the reference angle [tex]\( \frac{\pi}{6} \)[/tex] is wrong.
4. The cosine value is positive because [tex]\( \frac{\pi}{6} \)[/tex] is in the first quadrant:
This conclusion is also incorrect because [tex]\( \frac{4 \pi}{3} \)[/tex] is located in the third quadrant where cosine values are negative.
### Correct Procedure:
Let's correct the solution step-by-step:
1. Reduce the angle:
[tex]\[ \frac{10 \pi}{3} \equiv \frac{4 \pi}{3} \, (\text{mod} \, 2 \pi) \][/tex]
This reduction is correct.
2. Determine the quadrant:
[tex]\( \frac{4 \pi}{3} \)[/tex] is in the third quadrant of the unit circle.
3. Find the reference angle:
[tex]\[ \text{Reference angle} = \frac{4 \pi}{3} - \pi = \frac{\pi}{3} \][/tex]
4. Evaluate the cosine for the reference angle in the third quadrant:
[tex]\[ \cos \frac{\pi}{3} = \frac{1}{2} \][/tex]
Since cosine is negative in the third quadrant:
[tex]\[ \cos \frac{4 \pi}{3} = -\frac{1}{2} \][/tex]
Therefore, the correct value of [tex]\( \cos \frac{4 \pi}{3} \)[/tex] is [tex]\( -\frac{1}{2} \)[/tex].
### Identifying Henry's Errors:
- Henry incorrectly calculated the reference angle. Instead of using [tex]\( \frac{4 \pi}{3} - \pi \)[/tex], he incorrectly used [tex]\( \frac{3 \pi}{2} \)[/tex].
- He misunderstood the quadrant properties for cosine. In the third quadrant, cosine values should be negative.
### Correct Answer:
- Henry’s reference angle calculation is incorrect.
- Henry’s determination of the cosine sign is incorrect.
The correct value of [tex]\( \cos \frac{10 \pi}{3} \)[/tex] is [tex]\( -\frac{1}{2} \)[/tex].
Thus, the numerical result, [tex]\( (-0.5, 0.5) \)[/tex], implies:
- The wrong result by Henry is [tex]\( \frac{1}{2} \)[/tex] ([tex]\( 0.5 \)[/tex] in decimal).
- The correct result is [tex]\( -\frac{1}{2} \)[/tex] ([tex]\(-0.5\)[/tex] in decimal).
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