Question

a point on the terminal side of angle theta is given. Find the exact value of each of the six trigonometric functions of theta. given point- (3,7)

Answer 1

Given point: (3,7)

Opposite side: O

Adjacent side: A

Hypotenuse: H

`\begin{gathered} O=7 \\ A=3 \\ \end{gathered}`

Use pythagorean theorem to find H:

`\begin{gathered} H=\sqrt[]{A^2+O^2} \\ H=\sqrt[]{3^2+7^2} \\ H=\sqrt[]{9+49} \\ H=\sqrt[]{58} \end{gathered}`

Trigonometric functions:

`\begin{gathered} \sin \theta=(O)/(H)=\frac{7}{\sqrt[]{58}} \\ \\ \text{Razionalizing the denominator;} \\ \sin \theta=\frac{7}{\sqrt[]{58}}\cdot\frac{\sqrt[]{58}}{\sqrt[]{58}}=\frac{7\sqrt[]{58}}{58} \end{gathered}`
`\begin{gathered} \cos \theta=(A)/(H)=\frac{3}{\sqrt[]{58}}_{} \\ \\ \text{Razionalizing the denominator:} \\ \cos \theta=\frac{3}{\sqrt[]{58}}_{}\cdot\frac{\sqrt[]{58}}{\sqrt[]{58}}=\frac{3\sqrt[]{58}}{58} \end{gathered}`
`\tan \theta=(O)/(A)=(7)/(3)`
`\begin{gathered} \\ \csc \theta=(H)/(O)=\frac{\sqrt[]{58}}{7} \end{gathered}`
`\sec \theta=(H)/(A)=\frac{\sqrt[]{58}}{3}`
`\cot \theta=(A)/(O)=(3)/(7)`