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Francesca drew the point [tex]$(-2, -10)$[/tex] on the terminal ray of angle [tex]$\theta$[/tex], which is in standard position. She found values for the six trigonometric functions using the steps below.

\begin{tabular}{|l|l|}
\hline
Step 1 & Find the hypotenuse using [tex]$r = \sqrt{x^2 + y^2} = \sqrt{(-2)^2 + (-10)^2} = \sqrt{4 + 100} = \sqrt{104} = 2\sqrt{26}$[/tex] \\
\hline
Step 2 & [tex]$\sin \theta = \frac{y}{r} = \frac{-10}{2\sqrt{26}} = -\frac{10}{2\sqrt{26}} = -\frac{5}{\sqrt{26}} = -\frac{5\sqrt{26}}{26}$[/tex] \\
\hline
Step 3 & [tex]$\cos \theta = \frac{x}{r} = \frac{-2}{2\sqrt{26}} = -\frac{1}{\sqrt{26}} = -\frac{\sqrt{26}}{26}$[/tex] \\
\hline
\end{tabular}

(Note: Corrected steps and values should be verified by Francesca.)

Sagot :

GinnyAnsweridn
Sure, let's work through each step to determine the values for the six trigonometric functions for the point [tex]\((-2, -10)\)[/tex] on the terminal ray of angle [tex]\(\theta\)[/tex]. Here is a detailed step-by-step solution.

### Step 1: Calculate the Distance [tex]\( r \)[/tex]
First, we need to find the distance [tex]\( r \)[/tex] from the origin to the point [tex]\((-2, -10)\)[/tex]. This distance is the hypotenuse of the right triangle formed by the point and the origin. We use the Pythagorean theorem:

[tex]\[ r = \sqrt{x^2 + y^2} \][/tex]

Substituting [tex]\( x = -2 \)[/tex] and [tex]\( y = -10 \)[/tex]:

[tex]\[ r = \sqrt{(-2)^2 + (-10)^2} = \sqrt{4 + 100} = \sqrt{104} \approx 10.198 \][/tex]

So, [tex]\( r \approx 10.198 \)[/tex].

### Step 2: Calculate [tex]\(\sin \theta\)[/tex]
[tex]\(\sin \theta\)[/tex] is given by:

[tex]\[ \sin \theta = \frac{y}{r} = \frac{-10}{10.198} \approx -0.981 \][/tex]

### Step 3: Calculate [tex]\(\cos \theta\)[/tex]
[tex]\(\cos \theta\)[/tex] is given by:

[tex]\[ \cos \theta = \frac{x}{r} = \frac{-2}{10.198} \approx -0.196 \][/tex]

### Step 4: Calculate [tex]\(\tan \theta\)[/tex]
[tex]\(\tan \theta\)[/tex] is given by:

[tex]\[ \tan \theta = \frac{y}{x} = \frac{-10}{-2} = 5 \][/tex]

### Step 5: Calculate [tex]\(\csc \theta\)[/tex]
[tex]\(\csc \theta\)[/tex] (the reciprocal of [tex]\(\sin \theta\)[/tex]) is given by:

[tex]\[ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{-0.981} \approx -1.020 \][/tex]

### Step 6: Calculate [tex]\(\sec \theta\)[/tex]
[tex]\(\sec \theta\)[/tex] (the reciprocal of [tex]\(\cos \theta\)[/tex]) is given by:

[tex]\[ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{-0.196} \approx -5.099 \][/tex]

### Step 7: Calculate [tex]\(\cot \theta\)[/tex]
[tex]\(\cot \theta\)[/tex] (the reciprocal of [tex]\(\tan \theta\)[/tex]) is given by:

[tex]\[ \cot \theta = \frac{1}{\tan \theta} = \frac{1}{5} = 0.2 \][/tex]

### Summary of Results
So, summarizing all the calculated values, we have:

[tex]\[ \begin{aligned} r &\approx 10.198 \\ \sin \theta &\approx -0.981 \\ \cos \theta &\approx -0.196 \\ \tan \theta &= 5 \\ \csc \theta &\approx -1.020 \\ \sec \theta &\approx -5.099 \\ \cot \theta &= 0.2 \\ \end{aligned} \][/tex]
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