cotθ=3, cos<0. Find the exact value of sin θ/2 using half-angle formula. Please help!
if cot > 0 and cos < 0, then you are in QIII
cotθ = 3
cosθ = -3/√10
sin θ/2 = √((1-cosθ)/2)
= √((1+3/√10)/2)
Note that since θ is in QIII, θ/2 is in QII, and so sinθ is positive.
WOW!! Thanks so much! Looks like I knew how to solve it the whole time, just needed to see the proof I guess
To find the exact value of sin(θ/2) using the half-angle formula, we first need to find the exact value of cos(θ/2) using the given information.
Since cot(θ) = 3 and cos(θ) < 0, we can use the trigonometric identity:
cot^2(θ) = 1 / tan^2(θ) = 1 / (1 / cot^2(θ)) = cot^2(θ)
Substituting the given value of cot(θ) = 3:
3^2 = cot^2(θ)
9 = cot^2(θ)
Also, since cos(θ) < 0:
cos(θ) = -√(1 + cot^2(θ)) = -√(1 + 9) = -√10
Now, we can use the half-angle formula:
sin(θ/2) = ±√((1 - cos(θ)) / 2)
Substituting the value of cos(θ) = -√10:
sin(θ/2) = ±√((1 - (-√10)) / 2)
sin(θ/2) = ±√((1 + √10) / 2)
Therefore, the exact value of sin(θ/2) is ±√((1 + √10) / 2).