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Sagot :
To find [tex]\(\cos \theta\)[/tex] given that [tex]\(\tan \theta = -\frac{4}{3}\)[/tex] and knowing that [tex]\(\theta\)[/tex] is in Quadrant II, follow these steps:
1. Understand the Properties in Quadrant II:
- In Quadrant II, the cosine function ([tex]\(\cos \theta\)[/tex]) is negative, while the sine function ([tex]\(\sin \theta\)[/tex]) is positive.
2. Relate [tex]\(\tan \theta\)[/tex] to [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
- By definition, [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex].
- Given [tex]\(\tan \theta = -\frac{4}{3}\)[/tex], we can write this as:
[tex]\[ \frac{\sin \theta}{\cos \theta} = -\frac{4}{3} \][/tex]
3. Use a Pythagorean Identity:
- The Pythagorean identity states that:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
- We can let [tex]\(\sin \theta = -4k\)[/tex] and [tex]\(\cos \theta = 3k\)[/tex] or [tex]\(\cos \theta = -3k\)[/tex] since [tex]\(\tan \theta = -\frac{4}{3}\)[/tex].
4. Determine [tex]\(k\)[/tex] using the Pythagorean theorem:
- Substitute [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex] in the Pythagorean identity:
[tex]\[ \left( -4k \right)^2 + \left( 3k \right)^2 = 1 \][/tex]
[tex]\[ 16k^2 + 9k^2 = 1 \][/tex]
[tex]\[ 25k^2 = 1 \][/tex]
[tex]\[ k^2 = \frac{1}{25} \][/tex]
[tex]\[ k = \frac{1}{5} \][/tex]
5. Calculate [tex]\(\cos \theta\)[/tex]:
- Using [tex]\(k = \frac{1}{5}\)[/tex] and the fact that [tex]\(\cos \theta = 3k\)[/tex] and [tex]\(\cos \theta\)[/tex] is negative in Quadrant II, we get:
[tex]\[ \cos \theta = 3 \times \frac{1}{5} = \frac{3}{5} \][/tex]
[tex]\[ \cos \theta = -\frac{3}{5} \][/tex]
So, [tex]\(\cos \theta = -\frac{3}{5}\)[/tex]
However, none of the provided answer choices in the question include the correct result [tex]\((-0.6000000000000001\)[/tex] from the methodology).
Given that:
- Normally, the answer would be [tex]\(\cos \theta = -\frac{3}{5}\)[/tex].
Since the options A, B, C, and D do not match, there seems to be an issue with the provided answer choices, or there might be a typographical error issue in the options listed. Accordingly, none of the options A, B, C, or D seems to be correct based on our calculations for [tex]\(\cos \theta\)[/tex].
1. Understand the Properties in Quadrant II:
- In Quadrant II, the cosine function ([tex]\(\cos \theta\)[/tex]) is negative, while the sine function ([tex]\(\sin \theta\)[/tex]) is positive.
2. Relate [tex]\(\tan \theta\)[/tex] to [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
- By definition, [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex].
- Given [tex]\(\tan \theta = -\frac{4}{3}\)[/tex], we can write this as:
[tex]\[ \frac{\sin \theta}{\cos \theta} = -\frac{4}{3} \][/tex]
3. Use a Pythagorean Identity:
- The Pythagorean identity states that:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
- We can let [tex]\(\sin \theta = -4k\)[/tex] and [tex]\(\cos \theta = 3k\)[/tex] or [tex]\(\cos \theta = -3k\)[/tex] since [tex]\(\tan \theta = -\frac{4}{3}\)[/tex].
4. Determine [tex]\(k\)[/tex] using the Pythagorean theorem:
- Substitute [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex] in the Pythagorean identity:
[tex]\[ \left( -4k \right)^2 + \left( 3k \right)^2 = 1 \][/tex]
[tex]\[ 16k^2 + 9k^2 = 1 \][/tex]
[tex]\[ 25k^2 = 1 \][/tex]
[tex]\[ k^2 = \frac{1}{25} \][/tex]
[tex]\[ k = \frac{1}{5} \][/tex]
5. Calculate [tex]\(\cos \theta\)[/tex]:
- Using [tex]\(k = \frac{1}{5}\)[/tex] and the fact that [tex]\(\cos \theta = 3k\)[/tex] and [tex]\(\cos \theta\)[/tex] is negative in Quadrant II, we get:
[tex]\[ \cos \theta = 3 \times \frac{1}{5} = \frac{3}{5} \][/tex]
[tex]\[ \cos \theta = -\frac{3}{5} \][/tex]
So, [tex]\(\cos \theta = -\frac{3}{5}\)[/tex]
However, none of the provided answer choices in the question include the correct result [tex]\((-0.6000000000000001\)[/tex] from the methodology).
Given that:
- Normally, the answer would be [tex]\(\cos \theta = -\frac{3}{5}\)[/tex].
Since the options A, B, C, and D do not match, there seems to be an issue with the provided answer choices, or there might be a typographical error issue in the options listed. Accordingly, none of the options A, B, C, or D seems to be correct based on our calculations for [tex]\(\cos \theta\)[/tex].
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