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The terminal side of angle [tex]\( B \)[/tex] in standard position goes through the point [tex]\((-1, 10)\)[/tex]. Find the values of the six trigonometric functions of [tex]\( B \)[/tex]. Round your answers to 1 decimal place.

[tex]\[
\begin{aligned}
\sin(\beta) &= \square \\
\cos(\beta) &= \square \\
\tan(\beta) &= \square \\
\csc(\beta) &= \square \\
\sec(\beta) &= \square \\
\cot(\beta) &= \square
\end{aligned}
\][/tex]


Sagot :

To find the values of the six trigonometric functions for angle [tex]\( B \)[/tex], where the terminal side passes through the point [tex]\((-1, 10)\)[/tex], we follow these steps:

1. Identify the coordinates:
Let [tex]\( x = -1 \)[/tex] and [tex]\( y = 10 \)[/tex].

2. Calculate the hypotenuse (r):
[tex]\[ r = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + 10^2} = \sqrt{1 + 100} = \sqrt{101} \approx 10.05 \][/tex]

3. Determine the trigonometric functions:

- Sine ([tex]\(\sin \beta\)[/tex]):
[tex]\[ \sin \beta = \frac{y}{r} = \frac{10}{10.05} \approx 1.0 \][/tex]

- Cosine ([tex]\(\cos \beta\)[/tex]):
[tex]\[ \cos \beta = \frac{x}{r} = \frac{-1}{10.05} \approx -0.1 \][/tex]

- Tangent ([tex]\(\tan \beta\)[/tex]):
[tex]\[ \tan \beta = \frac{y}{x} = \frac{10}{-1} = -10.0 \][/tex]

- Cosecant ([tex]\(\csc \beta\)[/tex]):
[tex]\[ \csc \beta = \frac{r}{y} = \frac{10.05}{10} \approx 1.0 \][/tex]

- Secant ([tex]\(\sec \beta\)[/tex]):
[tex]\[ \sec \beta = \frac{r}{x} = \frac{10.05}{-1} \approx -10.0 \][/tex]

- Cotangent ([tex]\(\cot \beta\)[/tex]):
[tex]\[ \cot \beta = \frac{x}{y} = \frac{-1}{10} \approx -0.1 \][/tex]

So, the rounded values of the six trigonometric functions of [tex]\( B \)[/tex] are:

[tex]\[ \begin{aligned} \sin (\beta) & = 1.0 \\ \cos (\beta) & = -0.1 \\ \tan (\beta) & = -10.0 \\ \csc (\beta) & = 1.0 \\ \sec (\beta) & = -10.0 \\ \cot (\beta) & = -0.1 \end{aligned} \][/tex]
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