2Ytrigquadexample


 * Using Exact Values in 4 Quadrants **


 * Example 1 **

math \text{Given that } \; \sin \big( \theta \big) = -\dfrac{5}{13} math and q is in Quadrant 3, find cos( q ) and tan( q ).


 * Method 1: **

To find sin( q ), use: . math . \qquad \sin^2 (\theta)+\cos^2(\theta)=1 math . math \\ . \qquad \left( \dfrac{5}{13} \right)^2 + \cos^2 (\theta) = 1 \\. \\ . \qquad \dfrac{25}{169} + \cos^2 (\theta) = 1 math . math \\ . \qquad \cos^2 (\theta) = 1 - \dfrac{25}{169} \\. \\ . \qquad \cos^2 (\theta) = \dfrac{144}{169} \\. \\ . \qquad \cos (\theta) = \pm \dfrac{12}{13} math

But in Q3, so cos is negative . math . \qquad \cos (\theta) = -\dfrac{12}{13} math

To find tan( q ) use: . math . \qquad \tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta)} math . math \\ . \qquad \tan(\theta)=-\dfrac{5}{13} \div -\dfrac{12}{13} \\. \\ . \qquad \tan(\theta)=-\dfrac{5}{13} \times -\dfrac{13}{12} \\. \\ . \qquad \tan(\theta)=\dfrac{5}{12} math

In Q3, so tan is positive math . \qquad \tan(\theta)=\dfrac{5}{12} math media type="custom" key="8194382"media type="custom" key="8194430"
 * Method 2 **

Ignoring negatives, draw a media type="custom" key="8194386" from sin( q ) = 5/13:

Use Pythagoras to find the media type="custom" key="8194392": {use surd form if necessary}

Now read cos( q ) and tan( q ) from the triangle . math \\ . \qquad \cos(\theta) = \dfrac{12}{13} \\ . \textit{and} \\ . \qquad \tan(\theta) = \dfrac{5}{12} math

But in Q3, so cos is negative and tan is positive, hence solution is: . math \\ . \qquad \cos(\theta) = -\dfrac{12}{13} \\ .\textit{and} \\ . \qquad \tan(\theta) = \dfrac{5}{12} math

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