Suppose 3π/2 < θ < 2π and cos θ = a. Find sin θ, tan θ, sec θ, and csc θ in terms of a.

Answer 1

cosx = a

= 1/secx = a

where a>0 in quadrant IV

secx = 1/cosx = 1/a

>0 in quadrant IV

sinx = -sqr(1-a^2)

<0 in quadrant IV

cscx = 1/sinx = -1/sqr(1-a^2)

<0 in quadrant IV

tanx = sinx/cosx = -sqr(1-a^2)/a

<0 in quadrant IV

Answer 2

View cos(θ) = a as the fraction a/1 and label a right triangle with adjacent side a and hypotenuse 1. Then the opposite side is sqrt(1 - a²) by the Pythagorean theorem. Since θ is in quadrant 4, sin(θ) < 0 and cos(θ) = a > 0. From the right triangle, sin(θ) = -sqrt(1 - a²)/1 = -sqrt(1 - a²). Then

tan(θ) = sin(θ)/cos(θ) = -sqrt(1 - a²)/a

sec(θ) = 1/cos(θ) = 1/a

csc(θ) = 1/sin(θ) = -1/sqrt(1 - a²).