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

Question

sec θ = 5

3

;    0° < θ < 90°

Answer 1

Alex:

Since sec(θ) = 5, cos(θ) = 1/sec(θ) = 1/5

We can find formulas for half-angles using the fact that cos(2θ) = cos²(θ)-sin²(θ)

cos²(θ) = 1 - sin²(θ)

cos(2θ) = 1-sin²(θ)-sin²(θ) = 1-2*sin²(θ)

2*sin²(θ) = 1- cos(2θ)

sin²(θ) = (1-cos(2θ))/2

sin(θ) = +/- √((1-cos(2θ)/2)

sin(θ/2) = +- √(1-cos(θ))/2)

Using cos(θ) = 1/5 sin(θ/2) = √((1-1/5)/2) = √(2/5)

Using cos²(θ) = cos²(θ) - sin²(θ) = 2*cos²(θ) -1

We derive cos(θ/2) = +/- √((1+cos(θ)/2) in a similar way to how we derived sin(θ/2) formula

Using cos(θ) = 1/5 cos(θ/2) = √((1+1/5)/2) = √(3/5)

There is a half-angle formula for tan(θ/2) as well, but it is easier to compute tan(θ/2) = sin(θ/2) / cos(θ/2)

I'll let you do that. Let me know if you need help!