If cos 𝜃 = − (1/4) and 𝜃 is in III quadrant. Find the exact value of a) cos 𝜃/2 b) csc 2𝜃

Question

If cos 𝜃 = − (1/4) and 𝜃 is in III quadrant. Find the exact value of a) cos 𝜃/2 (2) b) csc 2𝜃

Answer 1

Answer 2

a) cosθ = -1/4

Since θ lies in Quadrant 3, 180° < θ < 270°. So, 90° < θ/2 < 135°. Therefore, cos(θ/2) < 0.

cos(θ/2) = - √ [(1 + cosθ) / 2] = - √ [ (3/4) / 2] = - √(3/8) = - √3 / (2√2) = - √6 / 4

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b) csc(2θ) = 1 / (2sinθcosθ)

Since cos²θ + sin²θ = 1, sin²θ = 1 - cos²θ.

So, sinθ = - √ (1 - cos²θ) Recall that sinθ < 0 in Quadrant 3

sinθ = - √15/4

Therefore, csc(2θ) = 1 / [2(-√15/4)(-1/4)] = 1 / (√15 / 8) = 8 / √15 = 8√15 / 15