Question

How do you find the 6 trig. functions of 13π / 4?

Answer 1

`sin((5pi)/4)=sin(pi+pi/4)=-sin(pi/4)=-sqrt(2)/2`
`cos((5pi)/4)=cos(pi+pi/4)=-cos(pi/4)=-sqrt(2)/2`
`tan((5pi)/4)=1`
`cot((5pi)/4)=1`
`sec((5pi)/4)=-sqrt(2)`
`csc((5pi)/4) =-sqrt(2)`

Explanation:

Let's start with the fact that all trigonometric functions are periodical and `2pi` is a common period for them all.
Therefore, the values of all these functions will not change if we subtract `2pi` from a given angle of `13pi/4`.
The result will be
`(13pi)/4 - 2pi = (5pi)/4`

To properly address this problem, we would represent an angle `(5pi)/4` as a point on a unit circle. Considering `(5pi)/4 = pi+pi/4`, we see that the point representing this angle is located in the third quadrant on its angle bisector (in degree that would be `180^o + 45^o=225^o`.

Next is to recall the definitions of all trigonometric functions in terms of abscissa (X-coordinate) and ordinate (Y-coordinate) of a point on a unit circle that represents the angle.

Here they are:
`sin(phi)` is an ordinate of such a point (Y-coordinate)
`cos(phi)` is an abscissa (X-coordinate)
`tan(phi) = sin(phi)/cos(phi)`, `cot(phi) = cos(phi)/sin(phi)`
`sec(phi) = 1/cos(phi)`, `csc(phi) = 1/sin(phi)`.

From these definitions and geometrical considerations of what are the abscissa and ordinate of a point at angle `(5pi)/4` on a unit circle, we conclude:
`sin((5pi)/4)=sin(pi+pi/4)=-sin(pi/4)=-sqrt(2)/2`
`cos((5pi)/4)=cos(pi+pi/4)=-cos(pi/4)=-sqrt(2)/2`
`tan((5pi)/4)=1`
`cot((5pi)/4)=1`
`sec((5pi)/4)=-sqrt(2)`
`csc((5pi)/4) =-sqrt(2)`