Question

Suppose that cotθ=−5
and that π/2<θ<π
. Find the values of sinθ
, cosθ
, tanθ
, cscθ
, and secθ.

Answer 1

Given that \(\cot \theta = -5\) and that \(\frac{\pi}{2} < \theta < \pi\), we can start by determining the values for \(\sin \theta\), \(\cos \theta\), and \(\tan \theta\) using the relationships between these trigonometric functions.

### Step 1: Find \(\sin \theta\) and \(\cos \theta\)

Since \(\cot \theta = \frac{\cos \theta}{\sin \theta}\), we have:

\[
\cot \theta = -5 \implies \frac{\cos \theta}{\sin \theta} = -5
\]

Let \(\sin \theta = y\) and \(\cos \theta = x\). Then we can write:

\[
\frac{x}{y} = -5 \implies x = -5y
\]

### Step 2: Use the Pythagorean Identity

We also know from the Pythagorean identity that:

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

Substituting \(x = -5y\) into the Pythagorean identity gives:

\[
y^2 + (-5y)^2 = 1
\]

\[
y^2 + 25y^2 = 1
\]

\[
26y^2 = 1 \implies y^2 = \frac{1}{26} \implies y = \frac{1}{\sqrt{26}} \text{ or } y = -\frac{1}{\sqrt{26}}
\]

Since we are in the second quadrant (where \(\sin \theta > 0\)), we take:

\[
\sin \theta = \frac{1}{\sqrt{26}}
\]

Now substituting back to find \(\cos \theta\):

\[
\cos \theta = -5y = -5 \left( \frac{1}{\sqrt{26}} \right) = -\frac{5}{\sqrt{26}}
\]

### Step 3: Calculate \(\tan \theta\), \(\csc \theta\), and \(\sec \theta\)

Now we can calculate the other trigonometric functions.

1. \(\tan \theta\):

\[
\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{1}{\sqrt{26}}}{-\frac{5}{\sqrt{26}}} = -\frac{1}{5}
\]

2. \(\csc \theta\) (the reciprocal of \(\sin \theta\)):

\[
\csc \theta = \frac{1}{\sin \theta} = \sqrt{26}
\]

3. \(\sec \theta\) (the reciprocal of \(\cos \theta\)):

\[
\sec \theta = \frac{1}{\cos \theta} = -\frac{\sqrt{26}}{5}
\]

### Summary of Results

Finally, the values are:

\[
\sin \theta = \frac{1}{\sqrt{26}}, \quad \cos \theta = -\frac{5}{\sqrt{26}}, \quad \tan \theta = -\frac{1}{5}, \quad \csc \theta = \sqrt{26}, \quad \sec \theta = -\frac{\sqrt{26}}{5}
\]