Answer to Question #6026 in Calculus for Samantha

Question

Evaluate the trigonometric function for each value of theta
theta=(-3pi/4), find sin(theta),cos(theta),tan(theta),cot(theta),sec(theta),csc(theta).
theta=(-pi/4), find sin(theta),cos(theta),tan(theta),cot(theta),sec(theta),csc(theta).
theta=(2pi/3), find sin(theta),cos(theta),tan(theta),cot(theta),sec(theta),csc(theta).

Answer 1

Evaluate the trigonometric function for each value of α
α=(-3pi/4), find sin(α),cos(α),tan(α),cot(α),sec(α),csc(α).
α=(-pi/4), find sin(α),cos(α),tan(α),cot(α),sec(α),csc(α).
α=(2pi/3), find sin(α),cos(α),tan(α),cot(α),sec(α),csc(α).

We'll use next expressions:
tan(α) = sin(α)/cos(α)
cot(α) = 1/tan(α)
sec(α) = 1/cos(α)
csc(α) = 1/sin(α)

α=(-3pi/4):
sin(-3pi/4) = -1/sqrt(2)
cos(-3pi/4) = -1/sqrt(2)
tan(-3pi/4) = (-1/sqrt(2))/(-1/sqrt(2)) = 1
cot(-3pi/4) = 1/1 = 1
sec(-3pi/4) = 1/(-1/sqrt(2)) = -sqrt(2)
csc(-3pi/4) = 1/(-1/sqrt(2)) = -sqrt(2)

α=(-pi/4):
sin(-pi/4) = -1/sqrt(2)
cos(-pi/4) = 1/sqrt(2)
tan(-pi/4) = (-1/sqrt(2))/(1/sqrt(2)) = -1
cot(-pi/4) = (1/sqrt(2))/(-1/sqrt(2)) = -1
sec(-pi/4) = 1/(1/sqrt(2)) = sqrt(2)
csc(-pi/4) = 1/(-1/sqrt(2)) = -sqrt(2)

α=(2pi/3):
sin(2pi/3) = sqrt(3)/2
cos(2pi/3) = -1/2
tan(2pi/3) = (sqrt(3)/2)/(-1/2) = -sqrt(3)
cot(2pi/3) = -1/sqrt(3)
sec(2pi/3) = -2
csc(2pi/3) = 2/sqrt(3)