Hampton Ball
04/16/2024 · High School
A point on the terminal side of an angle \( \theta \) in standard position is \( (3,-5) \). Find the exact value of each of the six trigonometric functions of \( \theta \).
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Step-by-step Solution
To find the six trigonometric functions of the angle \( \theta \) whose terminal side passes through the point \( (3, -5) \), we first need to determine the radius \( r \), which is the distance from the origin to the point \( (3, -5) \). This can be calculated using the distance formula:
\[
r = \sqrt{x^2 + y^2}
\]
where \( x = 3 \) and \( y = -5 \). Plugging in these values, we get:
\[
r = \sqrt{3^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}
\]
Now that we have \( r \), we can find the six trigonometric functions:
1. **Sine**:
\[
\sin(\theta) = \frac{y}{r} = \frac{-5}{\sqrt{34}}
\]
2. **Cosine**:
\[
\cos(\theta) = \frac{x}{r} = \frac{3}{\sqrt{34}}
\]
3. **Tangent**:
\[
\tan(\theta) = \frac{y}{x} = \frac{-5}{3}
\]
4. **Cosecant** (the reciprocal of sine):
\[
\csc(\theta) = \frac{r}{y} = \frac{\sqrt{34}}{-5} = -\frac{\sqrt{34}}{5}
\]
5. **Secant** (the reciprocal of cosine):
\[
\sec(\theta) = \frac{r}{x} = \frac{\sqrt{34}}{3}
\]
6. **Cotangent** (the reciprocal of tangent):
\[
\cot(\theta) = \frac{x}{y} = \frac{3}{-5} = -\frac{3}{5}
\]
Now, summarizing all the values, we have:
\[
\sin(\theta) = -\frac{5}{\sqrt{34}}, \quad \cos(\theta) = \frac{3}{\sqrt{34}}, \quad \tan(\theta) = -\frac{5}{3}
\]
\[
\csc(\theta) = -\frac{\sqrt{34}}{5}, \quad \sec(\theta) = \frac{\sqrt{34}}{3}, \quad \cot(\theta) = -\frac{3}{5}
\]
Thus, the exact values of the six trigonometric functions of \( \theta \) are:
\[
\sin(\theta) = -\frac{5}{\sqrt{34}}, \quad \cos(\theta) = \frac{3}{\sqrt{34}}, \quad \tan(\theta) = -\frac{5}{3}
\]
\[
\csc(\theta) = -\frac{\sqrt{34}}{5}, \quad \sec(\theta) = \frac{\sqrt{34}}{3}, \quad \cot(\theta) = -\frac{3}{5}
\]
Quick Answer
\[
\sin(\theta) = -\frac{5}{\sqrt{34}}, \quad \cos(\theta) = \frac{3}{\sqrt{34}}, \quad \tan(\theta) = -\frac{5}{3}
\]
\[
\csc(\theta) = -\frac{\sqrt{34}}{5}, \quad \sec(\theta) = \frac{\sqrt{34}}{3}, \quad \cot(\theta) = -\frac{3}{5}
\]
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