To find the six trigonometric functions for the angle whose terminal side passes through the point \((-4, -5)\), we first calculate the radius \(r\), which is the distance from the origin to the point \((-4, -5)\). This is done using the distance formula:
\[ r = \sqrt{x^2 + y^2} \]
Substituting \(x = -4\) and \(y = -5\):
\[ r = \sqrt{(-4)^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41} \]
Now we can find the six trigonometric functions:
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Sine (\(\sin\)): \[ \sin(\theta) = \frac{y}{r} = \frac{-5}{\sqrt{41}} \]
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Cosine (\(\cos\)): \[ \cos(\theta) = \frac{x}{r} = \frac{-4}{\sqrt{41}} \]
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Tangent (\(\tan\)): \[ \tan(\theta) = \frac{y}{x} = \frac{-5}{-4} = \frac{5}{4} \]
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Cosecant (\(\csc\)): \[ \csc(\theta) = \frac{r}{y} = \frac{\sqrt{41}}{-5} \quad (\text{undefined when } y = 0) \]
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Secant (\(\sec\)): \[ \sec(\theta) = \frac{r}{x} = \frac{\sqrt{41}}{-4} \]
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Cotangent (\(\cot\)): \[ \cot(\theta) = \frac{x}{y} = \frac{-4}{-5} = \frac{4}{5} \]
Summarizing the results:
- \(\sin(\theta) = \frac{-5}{\sqrt{41}}\)
- \(\cos(\theta) = \frac{-4}{\sqrt{41}}\)
- \(\tan(\theta) = \frac{5}{4}\)
- \(\csc(\theta) = \frac{\sqrt{41}}{-5}\)
- \(\sec(\theta) = \frac{\sqrt{41}}{-4}\)
- \(\cot(\theta) = \frac{4}{5}\)
These are the six trigonometric functions for the angle whose terminal side passes through the point \((-4, -5)\).