Question

P is a point on the terminal side of 0 in standard position. Find the exact value of the six trigonometric functions for 0.please help

Answer 1

Let's begin by listing out the given information:

`P(4,-2)`

We will solve using the formula:

`\begin{gathered} r=\sqrt[]{x^2+y^2} \\ (x,y)=(4,-2) \\ r=\sqrt[]{4^2+(-2)^2}=\sqrt[]{16+4}=\sqrt[]{20} \\ r=\sqrt[]{20} \end{gathered}`

We will proceed to calculate for the six trigonometric function:

`\begin{gathered} sin\theta=\frac{-2}{\sqrt[]{20}}=-\frac{2\sqrt[]{20}}{20}=\frac{\sqrt[]{20}}{10}=\frac{2\sqrt[]{5}}{10}=\frac{\sqrt[]{5}}{5} \\ sin\theta=\frac{\sqrt[]{5}}{5} \\ \\ cos\theta=\frac{4}{\sqrt[]{20}}=\frac{4\sqrt[]{20}}{20}=\frac{\sqrt[]{20}}{5} \\ cos\theta=\frac{\sqrt[]{20}}{5} \\ \\ tan\theta=(-2)/(4)=-(2)/(4)=-(1)/(2) \\ tan\theta=-(1)/(2) \end{gathered}`

For the remaining, we have:

`\begin{gathered} cot\theta=(1)/(tan\theta)=(1)/((-1)/(2))=-(2)/(1) \\ cot\theta=-2 \\ \\ sec\theta=(1)/(cos\theta)=\frac{1}{\frac{\sqrt[]{20}}{5}}=\frac{5}{\sqrt[]{20}} \\ sec\theta=\frac{5}{\sqrt[]{20}} \\ \\ co\sec \theta=(1)/(sin\theta)=\frac{1}{\frac{\sqrt[]{5}}{5}}=\frac{5}{\sqrt[]{5}} \\ co\sec \theta=\frac{5}{\sqrt[]{5}} \end{gathered}`