Question

An angle A standard position has a terminal side which passes through (-5,8) Determine values of the following sin Acos Atan A cot Asec Acsc A

Answer 1

If the point in the angle's terminal side is P = (x,y) then the trigonometric functions can be calculated as:

sin α = y/r

cos α = x/r

tan α = y/x

cot α = x/y

sec α = r/x

csc α = r/y

Where r is

$\sqrt{x^2+y^2}$

For the given point we have:

$r=\sqrt{(-5)^2+(8)^2}=\sqrt{25+64}=\sqrt{89}$

So the functions are:

Answer

$\begin{gathered} \sin\alpha=\frac{8}{\sqrt{89}} \\ \cos\alpha=\frac{-5}{\sqrt{89}} \\ \tan\alpha=-\frac{8}{5} \\ cot\text{ }\alpha=-\frac{5}{8} \\ sec\text{ }\alpha=-\frac{\sqrt{89}}{5} \\ csc\text{ }\alpha=\frac{\sqrt{89}}{8} \end{gathered}$