Question

Suppose that cscθ=4
and that π/2<θ<π.
Find the values of sinθ,
cosθ
, tanθ
, secθ
, and cotθ.

Answer 1

Given that \(\csc \theta = 4\), we can find \(\sin \theta\) using the reciprocal relationship:

\[
\sin \theta = \frac{1}{\csc \theta} = \frac{1}{4}.
\]

Since \( \frac{\pi}{2} < \theta < \pi \), we know that \(\theta\) is in the second quadrant, where sine is positive.

Next, we use the Pythagorean identity to find \(\cos \theta\):

\[
\sin^2 \theta + \cos^2 \theta = 1.
\]

Substituting \(\sin \theta\):

\[
\left(\frac{1}{4}\right)^2 + \cos^2 \theta = 1,
\]
\[
\frac{1}{16} + \cos^2 \theta = 1,
\]
\[
\cos^2 \theta = 1 - \frac{1}{16} = \frac{16}{16} - \frac{1}{16} = \frac{15}{16}.
\]

Thus,

\[
\cos \theta = -\sqrt{\frac{15}{16}} = -\frac{\sqrt{15}}{4},
\]

since cosine is negative in the second quadrant.

Now we can find \(\tan \theta\):

\[
\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{4}}{-\frac{\sqrt{15}}{4}} = -\frac{1}{\sqrt{15}} = -\frac{\sqrt{15}}{15}.
\]

Next, we find \(\sec \theta\), which is the reciprocal of \(\cos \theta\):

\[
\sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{\sqrt{15}}{4}} = -\frac{4}{\sqrt{15}} = -\frac{4\sqrt{15}}{15} \quad \text{(after rationalizing the denominator)}.
\]

Finally, we find \(\cot \theta\), which is the reciprocal of \(\tan \theta\):

\[
\cot \theta = \frac{1}{\tan \theta} = -\frac{\sqrt{15}}{1} = -\sqrt{15}.
\]

Thus, we have:

\[
\sin \theta = \frac{1}{4},
\]
\[
\cos \theta = -\frac{\sqrt{15}}{4},
\]
\[
\tan \theta = -\frac{\sqrt{15}}{15},
\]
\[
\sec \theta = -\frac{4\sqrt{15}}{15},
\]
\[
\cot \theta = -\sqrt{15}.
\]