Question

quiz 10 - solving polynomials, find trig values
calculator - 20 minutes

1. draw a diagram and find the exact

values of all 6 trig functions given
that \\( \cos \theta = \frac { 12 } { 13 } \\) and the terminal
side of \\( \theta \\) is in quadrant i.
\\( \sin \theta = \\)
\\( \csc \theta = \\)
\\( \cos \theta = \\)
\\( \sec \theta = \\)
\\( \tan \theta = \\)
\\( \cot \theta = \\)

Explanation:

Step1: Use the Pythagorean identity

We know that \(\sin^{2}\theta+\cos^{2}\theta = 1\). Given \(\cos\theta=\frac{12}{13}\), then \(\sin^{2}\theta=1-\cos^{2}\theta\). Substitute \(\cos\theta\) into the formula: \(\sin^{2}\theta=1 - (\frac{12}{13})^{2}=1-\frac{144}{169}=\frac{169 - 144}{169}=\frac{25}{169}\). Since \(\theta\) is in quadrant I (where \(\sin\theta>0\)), \(\sin\theta=\frac{5}{13}\).

Step2: Find \(\csc\theta\)

Recall that \(\csc\theta=\frac{1}{\sin\theta}\). Substitute \(\sin\theta = \frac{5}{13}\), so \(\csc\theta=\frac{13}{5}\).

Step3: Find \(\sec\theta\)

Recall that \(\sec\theta=\frac{1}{\cos\theta}\). Substitute \(\cos\theta=\frac{12}{13}\), so \(\sec\theta=\frac{13}{12}\).

Step4: Find \(\tan\theta\)

Recall that \(\tan\theta=\frac{\sin\theta}{\cos\theta}\). Substitute \(\sin\theta=\frac{5}{13}\) and \(\cos\theta=\frac{12}{13}\), so \(\tan\theta=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12}\).

Step5: Find \(\cot\theta\)

Recall that \(\cot\theta=\frac{\cos\theta}{\sin\theta}\). Substitute \(\sin\theta=\frac{5}{13}\) and \(\cos\theta=\frac{12}{13}\), so \(\cot\theta=\frac{\frac{12}{13}}{\frac{5}{13}}=\frac{12}{5}\).

Answer:

\(\sin\theta=\frac{5}{13}\), \(\csc\theta=\frac{13}{5}\), \(\cos\theta=\frac{12}{13}\), \(\sec\theta=\frac{13}{12}\), \(\tan\theta=\frac{5}{12}\), \(\cot\theta=\frac{12}{5}\)