Find the exact values of sin(𝜃/2), cos(𝜃/2), and tan(𝜃/2) for the given condition.

Question

Find the exact values of sin(𝜃/2), cos(𝜃/2), and tan(𝜃/2) for the given condition.

csc𝜃= -5/4 ; −90° < 𝜃 < 0°

Answer 1

First remember these:

tan(θ)=sin(θ)/cos(θ); cot(θ)=cos(θ)/sin(θ); sin(θ)=1/csc (θ); etc.

If cosecθ = -5/3 then,

sinθ = -3/5

Now, -90° < θ < 0° is the same as 270° < θ < 360°

i.e. quadrant IV

So, cosθ > 0 and tanθ < 0  

Now, sin²θ + cos²θ = 1

Hence, cos²θ = 1 - sin²θ

i.e. cos²θ = 1 - (-3/5)² => 16/25

so, cosθ = 4/5

Now, tanθ = sinθ/cosθ

=> (-3/5)/(4/5) = -3/4

Now, cos2θ = 2cos²θ - 1

so, cosθ = 2cos²(θ/2) - 1

Hence, cos²(θ/2) = (cosθ + 1)/2

=> cos²(θ/2) = ((4/5) + 1)/2

i.e. cos²(θ/2) = 9/10

Hence, cos(θ/2) = 3√10/10

Now, cos2θ = 1 - 2sin²θ

so, cosθ = 1 - 2sin²(θ/2)

Hence, sin²(θ/2) = (1 - cosθ)/2

=> sin²(θ/2) = (1 - (4/5))/2

i.e. sin²(θ/2) = 1/10

Hence, sin(θ/2) = -√10/10

And, tan(θ/2) => sin(θ/2)/cos(θ/2)

so, tan(θ/2) = (-√10/10)/(3√10/10)

Hence, tan(θ/2) = -1/3

There are different ways to solve this.