please look at the image and help me, there is a few parts but for finding the other functions I can do that as long as I know how to find them based on the first one we do. Thank you!

please look at the image and help me there is a few parts but for finding the other functions I can do that as long as I know how to find them based on the firs class=

Respuesta :

Answer:

To answer this question, we will use the following diagram as reference:

Now, recall that in a right triangle the trigonometric functions are defined as follows:

[tex]\begin{gathered} \sin \theta=\frac{\text{opposite leg}}{hypotenuse}, \\ \cos \theta=\frac{adjacent\text{ leg}}{hypotenuse}, \\ \tan \theta=\frac{\text{opposite leg}}{adjacent\text{ leg}}. \end{gathered}[/tex]

Also, recall that:

[tex]\begin{gathered} \csc \theta=\frac{1}{\sin \theta}, \\ \sec \theta=\frac{1}{\cos \theta}, \\ \cot =\frac{1}{\tan \theta}\text{.} \end{gathered}[/tex]

From the diagram, we get that:

[tex]\sin \theta=\frac{y}{r},[/tex]

substituting r=√x²+y² we get:

[tex]\sin \theta=\frac{y}{\sqrt[]{x^2+y^2}}.[/tex]

Analogously, we get that:

[tex]\begin{gathered} \cos \theta=\frac{x}{\sqrt[]{x^2+y^2}}, \\ \text{tan}\theta=\frac{y}{x}, \\ \csc \theta=\frac{\sqrt[]{x^2+y^2}}{y}, \\ \sec \theta=\frac{\sqrt[]{x^2+y^2}}{x}, \\ \cot \theta=\frac{x}{y}\text{.} \end{gathered}[/tex]

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