Question

Suppose that secθ=(√34)/5
and that (3π)/2<θ<2π
. Find the values of sinθ
, cosθ
, tanθ
, cscθ
, and cotθ
.

Answer 1

To find the trigonometric values given that \(\sec \theta = \frac{\sqrt{34}}{5}\) and \(\frac{3\pi}{2} < \theta < 2\pi\), we first need to calculate \(\cos \theta\).

Recall that:

\[
\sec \theta = \frac{1}{\cos \theta}
\]

Thus, we have:

\[
\cos \theta = \frac{1}{\sec \theta} = \frac{5}{\sqrt{34}}
\]

Since \(\theta\) is in the fourth quadrant, \(\cos \theta\) is positive, which is consistent with our calculation.

Next, we can find \(\sin \theta\) using the Pythagorean identity:

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

Substituting for \(\cos \theta\):

\[
\sin^2 \theta + \left(\frac{5}{\sqrt{34}}\right)^2 = 1
\]

Calculating \(\left(\frac{5}{\sqrt{34}}\right)^2\):

\[
\left(\frac{5}{\sqrt{34}}\right)^2 = \frac{25}{34}
\]

Now substituting this value back into the identity:

\[
\sin^2 \theta + \frac{25}{34} = 1
\]

Subtracting \(\frac{25}{34}\) from both sides:

\[
\sin^2 \theta = 1 - \frac{25}{34} = \frac{34}{34} - \frac{25}{34} = \frac{9}{34}
\]

Taking the square root, and since \(\theta\) is in the fourth quadrant where sine is negative:

\[
\sin \theta = -\sqrt{\frac{9}{34}} = -\frac{3}{\sqrt{34}} = -\frac{3\sqrt{34}}{34}
\]

Now we can find the other trigonometric values:

1. **Cosecant** (\(\csc \theta\)):

\[
\csc \theta = \frac{1}{\sin \theta} = \frac{1}{-\frac{3}{\sqrt{34}}} = -\frac{\sqrt{34}}{3}
\]

2. **Tangent** (\(\tan \theta\)):

\[
\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{3}{\sqrt{34}}}{\frac{5}{\sqrt{34}}} = -\frac{3}{5}
\]

3. **Cotangent** (\(\cot \theta\)):

\[
\cot \theta = \frac{1}{\tan \theta} = \frac{1}{-\frac{3}{5}} = -\frac{5}{3}
\]

So, the values are:

- \(\sin \theta = -\frac{3\sqrt{34}}{34}\)
- \(\cos \theta = \frac{5}{\sqrt{34}}\)
- \(\tan \theta = -\frac{3}{5}\)
- \(\csc \theta = -\frac{\sqrt{34}}{3}\)
- \(\cot \theta = -\frac{5}{3}\)