Question

given \\( \sec \theta = \frac{\sqrt{37}}{6} \\) and \\( \theta \\) is in quadrant 4, find the remaining five trig functions of \\( \theta \\).

in the answer box, enter the exact, rationalized values like this (you can omit the \\( \theta \\)):
\\( \sin = \\# \\)
\\( \cos = \\# \\)
\\( \tan = \\# \\)
\\( \csc = \\# \\)
\\( \cot = \\# \\)

Explanation:

Step1: Recall reciprocal of secant

Since \(\sec\theta=\frac{1}{\cos\theta}\), we can find \(\cos\theta\) by taking the reciprocal of \(\sec\theta\). Given \(\sec\theta = \frac{\sqrt{37}}{6}\), so \(\cos\theta=\frac{6}{\sqrt{37}}\). Rationalizing the denominator, we multiply numerator and denominator by \(\sqrt{37}\), getting \(\cos\theta=\frac{6\sqrt{37}}{37}\).

Step2: Use Pythagorean identity for \(\sin\theta\)

The Pythagorean identity is \(\sin^{2}\theta+\cos^{2}\theta = 1\). We know \(\cos\theta=\frac{6}{\sqrt{37}}\), so \(\cos^{2}\theta=\frac{36}{37}\). Then \(\sin^{2}\theta=1 - \frac{36}{37}=\frac{1}{37}\), so \(\sin\theta=\pm\frac{1}{\sqrt{37}}\). Since \(\theta\) is in quadrant 4, \(\sin\theta\) is negative, so \(\sin\theta=-\frac{1}{\sqrt{37}}=-\frac{\sqrt{37}}{37}\) (after rationalizing).

Step3: Find \(\tan\theta\)

We know that \(\tan\theta=\frac{\sin\theta}{\cos\theta}\). Substituting \(\sin\theta = -\frac{1}{\sqrt{37}}\) and \(\cos\theta=\frac{6}{\sqrt{37}}\), we get \(\tan\theta=\frac{-\frac{1}{\sqrt{37}}}{\frac{6}{\sqrt{37}}}=-\frac{1}{6}\).

Step4: Find \(\csc\theta\)

Since \(\csc\theta=\frac{1}{\sin\theta}\), and \(\sin\theta = -\frac{1}{\sqrt{37}}\), then \(\csc\theta=- \sqrt{37}\) (because \(\frac{1}{-\frac{1}{\sqrt{37}}}=-\sqrt{37}\)).

Step5: Find \(\cot\theta\)

Since \(\cot\theta=\frac{1}{\tan\theta}\), and \(\tan\theta = -\frac{1}{6}\), then \(\cot\theta=- 6\) (because \(\frac{1}{-\frac{1}{6}}=-6\)).

Answer:

\(\sin = -\frac{\sqrt{37}}{37}\)
\(\cos = \frac{6\sqrt{37}}{37}\)
\(\tan = -\frac{1}{6}\)
\(\csc = -\sqrt{37}\)
\(\cot = -6\)