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Given that [tex]\cot (\theta) = -\frac{1}{2}[/tex] and [tex]\theta[/tex] is in Quadrant II, what is [tex]\sin (\theta)[/tex]? Write your answer in exact form. Do not round.

Provide your answer below:
[tex]\sin (\theta) =[/tex] [tex]$\square$[/tex]


Sagot :

Given that [tex]\(\cot (\theta) = -\frac{1}{2}\)[/tex] and [tex]\(\theta\)[/tex] is in Quadrant II, we want to find [tex]\(\sin (\theta)\)[/tex].

1. Express [tex]\(\cot (\theta)\)[/tex] as [tex]\(\frac{\cos (\theta)}{\sin (\theta)}\)[/tex]:
[tex]\[ \cot (\theta) = \frac{\cos (\theta)}{\sin (\theta)} = -\frac{1}{2} \][/tex]

2. Express [tex]\(\cos (\theta)\)[/tex] in terms of [tex]\(\sin (\theta)\)[/tex]:
[tex]\[ \cos (\theta) = -\frac{1}{2} \sin (\theta) \][/tex]

3. Use the Pythagorean identity:
[tex]\[ \sin^2 (\theta) + \cos^2 (\theta) = 1 \][/tex]

4. Substitute [tex]\(\cos (\theta) = -\frac{1}{2} \sin (\theta)\)[/tex] into the Pythagorean identity:
[tex]\[ \sin^2 (\theta) + \left( -\frac{1}{2} \sin (\theta) \right)^2 = 1 \][/tex]

5. Simplify the equation:
[tex]\[ \sin^2 (\theta) + \frac{1}{4} \sin^2 (\theta) = 1 \][/tex]

6. Combine like terms:
[tex]\[ \sin^2 (\theta) \left( 1 + \frac{1}{4} \right) = 1 \][/tex]
[tex]\[ \sin^2 (\theta) \left( \frac{4}{4} + \frac{1}{4} \right) = 1 \][/tex]
[tex]\[ \sin^2 (\theta) \left( \frac{5}{4} \right) = 1 \][/tex]

7. Solve for [tex]\(\sin^2 (\theta)\)[/tex]:
[tex]\[ \sin^2 (\theta) = \frac{4}{5} \][/tex]

8. Take the square root of both sides to solve for [tex]\(\sin (\theta)\)[/tex]:
[tex]\[ \sin (\theta) = \pm \sqrt{\frac{4}{5}} \][/tex]
[tex]\[ \sin (\theta) = \pm \frac{2}{\sqrt{5}} \][/tex]

9. Since [tex]\(\theta\)[/tex] is in Quadrant II where [tex]\(\sin (\theta)\)[/tex] is positive:
[tex]\[ \sin (\theta) = \frac{2}{\sqrt{5}} \][/tex]

10. Rationalize the denominator:
[tex]\[ \sin (\theta) = \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{2 \sqrt{5}}{5} \][/tex]

Therefore, the exact value is:
[tex]\[ \sin (\theta) = \frac{2 \sqrt{5}}{5} \][/tex]