Recall the definition of a function sec(ϕ):
sec(ϕ)=1cos(ϕ)
Recall the definition a function cos(ϕ):
cos(ϕ) is an abscissa (X-coordinate) of an endpoint of a radius-vector in the unit circle that forms an angle ϕ with a positive direction of the X-axis, counting counter-clockwise from the positive direction of the X-coordinate towards a radius-vector.
From this definition of a function cos(ϕ) follows that
(a) Function cos(ϕ) is periodical with a period of 2π.
(b) Function cos(ϕ) is even in terms of cos(ϕ)=cos(−ϕ).
Using these properties, we can state that
cos(13π4)=cos(13π4−4π)=cos(−3π4)=cos(3π4)
The angle 3π4 lies in the second quarter and the abscissa of an endpoint of a unit vector that corresponds to this angle is negative.
The corresponding angle with positive but equal by absolute value abscissa is, obviously, π−3π4=π4.
So, we can conclude that cos(13π4)=−cos(π4)=−√22.
Now we can calculate the value of sec(13π4):
sec(13π4)=1cos(13π4)=1−√22=−2√2=−√2
So, the answer is
sec(13π4)=−√2
