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tags:
Cosine Cotangent Coterminal Angles Trigonometry Trigonometry Series Unit Circle

Trig Functions with the Unit Circle Exercise 6

Author: Hannah Solomon

The following problem is solved in this video. It is recommended that you try to solve the problem before watching the video. You can click "Reveal Answer" to see the answer to the problem.

Problem: Find all \(0^\circ \leq \theta \leq 360^\circ\) satisfying \(\sec{\theta}=-\sqrt{2}\) and \(\tan{\theta}>0.\)

Answer:  \(\theta=225^\circ\)

Solution Method: In this problem, we need to find \(0^\circ\) \(\leq \theta \leq\) \(360^\circ\) satisfying two conditions. First, let’s narrow down the possible quadrants for \(\theta\).

\(\tan\theta>0\) means tangent needs to be positive. Using our acronym ASTC, we know that tangent is positive in Q1 and Q3.

We also have that secant is negative. Secant is the reciprocal of cosine, so their signs will be the same. So we need cosine to be negative, which occurs in Q2 and Q3.

Now \(\theta\) has to satisfy BOTH conditions because it’s an "and" statement. So \(\theta\) can only be in Q3.

Now that we have the quadrant, we need to determine if \(\theta\) has a reference angle of \(30^\circ,\) \(45^\circ,\) or \(60^\circ.\) I don’t know the secant values on the unit circle off the top of my head, but I do know the cosine values because I know the x-coordinates. If \(\sec{\theta}=-\sqrt{2},\) \(\cos{\theta}=-\frac{1}{\sqrt{2}}=-\frac{\sqrt{2}}{2}\). Reference angles of \(45^\circ\) have x-coordinates of \(\pm\frac{\sqrt{2}}{2}\).

So then \(\theta\) is in Q3 and has a reference angle of \(45^\circ,\) so \(\theta=225^\circ\) is the only answer.

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