College Analytic Trigonometry

[CandiceC]

if cos(s)=1/2 and 3pi/2<s<2pi, find csc(s)

I'm not even sure how to get started on this.. if I know how to get started I should be able to finish the rest

please help, thanks

 

[CandiceC]

I figured it out

cos1/2=pi/3 which implies s=5pi/3 which implies csc5pi/3=1/sin5pi/3= -2root3/3

 

[Deleted member 4993]

cos1/2=pi/3 which implies s=5pi/3 which implies csc5pi/3=1/sin5pi/3= -2root3/3

Final answer is correct but somehow the work above does not make sense to me!

if cos(s)=1/2 and 3pi/2<s<2pi, find csc(s)

rewriting: The work should be:

if cos(Θ)=1/2 and 3π/2 < Θ < 2π,

find csc(Θ)

cos(Θ)=1/2 → Θ = π/3 or (2π - π/3 = ) 5π/3

here Θ = 5π/3

sin(Θ) = ± √[1- cos²(Θ)] → sin(5π/3) = - √[1- cos²(5π/3)] = -(√3)/2

csc(Θ) = 1/sin(Θ) = 1/[-(√3)/2] = -2/√3 = -2(√3)/3

 

[soroban]

Hello, CandiceC!

$\displaystyle \text{If }\cos\theta =\frac{1}{2}\,\text{ and }\,\frac{3\pi}{2}\,<\,\theta\,<\,2\pi,\,\text{ find }\csc\theta.$

We are given: .$\displaystyle \cos\theta \,=\,\dfrac{1}{2} \,=\,\dfrac{adj}{hyp}$

$\displaystyle \theta$ is in a right triangle with: $\displaystyle adj = 1,\;hyp = 2.$
Pythagorus gives us: .$\displaystyle opp = \pm\sqrt{3}$

Since $\displaystyle \theta$ is in Quadrant 4, $\displaystyle opp = \text{-}\sqrt{3}$

Therefore: .$\displaystyle \csc\theta \:=\:\dfrac{hyp}{opp} \;=\;\dfrac{2}{\text{-}\sqrt{3}} \:=\:-\dfrac{2\sqrt{3}}{3}$