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College Analytic Trigonometry
- Thread starter CandiceC
- Start date
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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
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Hello, CandiceC!
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: .\(\displaystyle \csc\theta \:=\:\dfrac{hyp}{opp} \;=\;\dfrac{2}{\text{-}\sqrt{3}} \:=\:-\dfrac{2\sqrt{3}}{3}\)
\(\displaystyle \text{If }\cos\theta =\frac{1}{2}\,\text{ and }\,\frac{3\pi}{2}\,<\,\theta\,<\,2\pi,\,\text{ find }\csc\theta.\)
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: .\(\displaystyle \csc\theta \:=\:\dfrac{hyp}{opp} \;=\;\dfrac{2}{\text{-}\sqrt{3}} \:=\:-\dfrac{2\sqrt{3}}{3}\)