Question

An angle A in standard position has a terminal side which passes through the point (-5,8)Determine the following Sin ACos A Tan ACot A Sec ACsc A

Answer 1

The points on the terminal side of an angle α can be described as:

$(r\cos(\alpha),r\sin(\alpha))$

Where r is the distance from the origin to the point. Find r for the point (-5,8):

$r=\sqrt{(-5)^2+(8)^2}=\sqrt{89}$

Then:

$\begin{gathered} \sqrt{89}\cos(\alpha)=-5\Rightarrow\cos(\alpha)=-\frac{5}{\sqrt{89}} \\ \\ \sqrt{89}\sin(\alpha)=8\Rightarrow\sin(\alpha)=\frac{8}{\sqrt{89}} \end{gathered}$

Recall the definitions of the tangent, cotangent, secant, and cosecant of an angle in terms of its sine and its cosine:

$\begin{gathered} \tan(\alpha)=\frac{\sin(\alpha)}{\cos(\alpha)} \\ \cot(\alpha)=\frac{\cos(\alpha)}{\sin(\alpha)} \\ \sec(\alpha)=\frac{1}{\cos(\alpha)} \\ \csc(\alpha)=\frac{1}{\sin(\alpha)} \end{gathered}$

Replace the expressions for cos(α) and sin(α) to find the values of tan(α), cot(α), sec(α) and csc(α):

$\begin{gathered} \tan(\alpha)=\frac{\frac{8}{\sqrt{89}}}{-\frac{5}{\sqrt{89}}}=-\frac{8}{5} \\ \\ \cot(\alpha)=\frac{-\frac{5}{\sqrt{89}}}{\frac{8}{\sqrt{89}}}=-\frac{5}{8} \\ \\ \sec(\alpha)=\frac{1}{-\frac{5}{\sqrt{89}}}=-\frac{\sqrt{89}}{5} \\ \\ \csc(\alpha)=\frac{1}{\frac{8}{\sqrt{89}}}=\frac{\sqrt{89}}{8} \end{gathered}$