Question

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!

Answer 1

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:

`\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}`

Also, recall that:

`\begin{gathered} \csc \theta=(1)/(\sin \theta), \\ \sec \theta=(1)/(\cos \theta), \\ \cot =(1)/(\tan \theta)\text{.} \end{gathered}`

From the diagram, we get that:

`\sin \theta=(y)/(r),`

substituting r=√x²+y² we get:

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

Analogously, we get that:

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