Given that [tex]\(\csc(\theta) = 2\)[/tex] and [tex]\(\theta\)[/tex] is in Quadrant I, we need to find the values of the six trigonometric functions of [tex]\(\theta\)[/tex]. Here are the step-by-step calculations:
1. Finding [tex]\(\sin(\theta)\)[/tex]:
[tex]\(\csc(\theta)\)[/tex] is the reciprocal of [tex]\(\sin(\theta)\)[/tex]. Therefore:
[tex]\[
\sin(\theta) = \frac{1}{\csc(\theta)} = \frac{1}{2} = 0.5
\][/tex]
2. Finding [tex]\(\cos(\theta)\)[/tex]:
We use the Pythagorean identity [tex]\(\sin^2(\theta) + \cos^2(\theta) = 1\)[/tex]. Given [tex]\(\sin(\theta) = 0.5\)[/tex]:
[tex]\[
\cos^2(\theta) = 1 - \sin^2(\theta) = 1 - (0.5)^2 = 1 - 0.25 = 0.75
\][/tex]
Taking the positive square root (since [tex]\(\theta\)[/tex] is in Quadrant I where [tex]\(\cos(\theta)\)[/tex] is positive):
[tex]\[
\cos(\theta) = \sqrt{0.75} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \approx 0.8660254037844386
\][/tex]
3. Finding [tex]\(\tan(\theta)\)[/tex]:
[tex]\(\tan(\theta)\)[/tex] is the ratio of [tex]\(\sin(\theta)\)[/tex] to [tex]\(\cos(\theta)\)[/tex]:
[tex]\[
\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{0.5}{0.8660254037844386} \approx 0.5773502691896258
\][/tex]
4. Finding [tex]\(\csc(\theta)\)[/tex]:
This was given directly in the problem:
[tex]\[
\csc(\theta) = 2
\][/tex]
5. Finding [tex]\(\sec(\theta)\)[/tex]:
[tex]\(\sec(\theta)\)[/tex] is the reciprocal of [tex]\(\cos(\theta)\)[/tex]:
[tex]\[
\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{0.8660254037844386} \approx 1.1547005383792517
\][/tex]
6. Finding [tex]\(\cot(\theta)\)[/tex]:
[tex]\(\cot(\theta)\)[/tex] is the reciprocal of [tex]\(\tan(\theta)\)[/tex]:
[tex]\[
\cot(\theta) = \frac{1}{\tan(\theta)} = \frac{1}{0.5773502691896258} \approx 1.732050807568877
\][/tex]
Summarizing the results, we get:
[tex]\[
\begin{array}{|l|l|l|}
\hline
\sin(\theta) = 0.5 & \cos(\theta) = 0.8660254037844386 & \tan(\theta) = 0.5773502691896258 \\
\hline
\csc(\theta) = 2 & \sec(\theta) = 1.1547005383792517 & \cot(\theta) = 1.732050807568877 \\
\hline
\end{array}
\][/tex]
