Trigonometry, Find the exact value.

Question

Find the exact value of

sin(θ/2) given sec θ = 5/3 and −90<θ< 0

Answer 1

The sine half-angle formula says

sin²(θ/2) = (1/2) – (1/2)cos(θ).

If sec θ = 5/3 then cos θ = 3/5 so

sin²(θ/2) = (1/2) – (1/2)(3/5)

= (5/10) – (3/10) = 2/10 = 1/5.

Since sin²(θ/2) = 1/5, you have sin(θ/2) = ±1/√5.

This may be written as ±(√5)/5, by rationalizing the denominator.

Now the question is: "plus" or "minus"? Since θ is between 0 and –90º, you have θ/2 between 0 and –45°, the sine must be negative.

Answer 2

Here the half-angle formula will come in handy.

sin(∝/2) = ± SQRT { (1/2) [ 1 - cos(∝) ] }

sec(∝) = 1/cos(∝) so cos(∝) = 3/5

sin(∝/2) = - SQRT { 0.5 (1-3/5) } = -√0.2 = -√5/5 ≈ -0.4472