Question

Given that tan(θ)=8/3 and θ is in quadrant iii find and sin(θ/2)

Answer 1

θ lies in quadrant III, which means

π < θ < 3π/2 ⇒ π/2 < θ/2 < 3π/4

which is to say that θ/2 lies in quadrant II. For quadrant III, we have sin(θ) < 0 and cos(θ) < 0. For quadrant II, we have sin(θ/2) > 0 and cos(θ/2) < 0.

Recall the half-angle identity for sine:

sin²(θ/2) = (1 - cos(θ))/2

Then in this case,

sin(θ/2) = - √((1 - cos(θ))/2)

Also recall the Pythagorean identity,

1 + tan²(θ) = sec²(θ)

It follows that

sec(θ) = - √(1 + tan²(θ)) = -√73/3

⇒ cos(θ) = -3/√73

⇒ sin(θ/2) = - √((1 + 3/√73)/2) = √(1/2 + 3/(2√73))