Question

Let (-√7,3) be a point on a terminal side of Фfind the exact values of cosФ, secФ and cotФcosФ=secФ=cotФ=

Answer 1

SOLUTION

Let's make a simple diagram to help us solve this

From the diagram above, h is the hypotenuse.

Now from Pythagoras theorem

$\begin{gathered} \text{hyp}^2=opp^2+adj^2 \\ h^2=3^2+(-\sqrt[]{7})^2 \\ h^2=9+7 \\ h=\sqrt[]{16} \\ h=4 \end{gathered}$

Now we will use the trig-ratio to solve the question

(a) cosФ is

$\begin{gathered} \cos \theta=\frac{adjacent}{\text{hypotenuse}} \\ \cos \theta=\frac{-\sqrt[]{7}}{4} \\ \cos \theta=-\frac{\sqrt[]{7}}{4} \end{gathered}$

(b) secФ is

$\begin{gathered} \sec \theta=\frac{1}{\cos \theta} \\ \sec \theta=\frac{1}{\frac{-\sqrt[]{7}}{4}} \\ \sec \theta=\frac{4}{-\sqrt[]{7}} \\ \sec \theta=\frac{4}{-\sqrt[]{7}}\times\frac{-\sqrt[]{7}}{-\sqrt[]{7}} \\ \sec \theta=-\frac{4\sqrt[]{7}}{7} \end{gathered}$

(c) cotФ is

$\begin{gathered} \cot \theta=\frac{1}{\tan \theta} \\ \tan \theta=\frac{opposite}{\text{adjacent}} \\ \tan \theta=\frac{3}{-\sqrt[]{7}} \\ \cot \theta=\frac{1}{\frac{3}{-\sqrt[]{7}}} \\ \cot \theta=-\frac{\sqrt[]{7}}{3} \end{gathered}$