Question

If θ is an angle in standard position that terminates in Quadrant III such that tanθ = 5/12, then sinθ/2 = _____.

Answer 1

`\displaystyle \sin{(\theta)/(2)} = (5√(26))/(26)\approx 0.196`.

Explanation:

`\displaystyle \theta\in \left(\pi, (3\pi)/(2)\right)`,

such that

`\displaystyle (\theta)/(2) \in \left((\pi)/(2), (3\pi)/(4)\right)`.

As a result,

• 
  `\displaystyle 0 < \sin{(\theta)/(2)} <1`, and
• 
  `\displaystyle -1 < \cos{(\theta)/(2)} <0`.

`\displaystyle \tan{(\theta)/(2)} = \frac{\sin{\displaystyle (\theta)/(2)}}{\displaystyle \cos{(\theta)/(2)}}`,

such that

• 
  `\displaystyle \tan{(\theta)/(2)} <0`.

Let

`\displaystyle t = \tan{(\theta)/(2)}`.

`t < 0`.

By the double angle identity for tangents.

`\displaystyle \frac{\displaystyle 2\tan{(\theta)/(2)}}{1-\displaystyle \left(\tan{(\theta)/(2)}\right)^(2)} = tan(\theta)`.

`\displaystyle (2t)/(1 - t^(2)) = (5)/(12)`.

`24t = 5 - 5t^(2)`.

Solve this quadratic equation for
`t`:

• 
  `\displaystyle t_1 = (1)/(5)`, and
• 
  `t_2 = -5`.

Discard
`t_1` for it is not smaller than zero.

Let
`\displaystyle s = \sin{(\theta)/(2)}`.

`0 < s <1`.

By the definition of tangents:

`\displaystyle \tan{(\theta)/(2)} = \frac{\displaystyle \sin{(\theta)/(2)}}{\displaystyle \cos{(\theta)/(2)}}`.

Apply the Pythagorean Algorithm to express the cosine of
`\displaystyle (\theta)/(2)` in terms of
`s`. Note that
`\displaystyle \cos{(\theta)/(2)}` is expected to be smaller than zero.

`\displaystyle \cos{(\theta)/(2)} = -\sqrt{1 - \left(\sin{(\theta)/(2)}\right)^(2)}= - \sqrt{1 - s^(2)}`.

Solve for
`s`:

`\displaystyle \frac{s}{- \sqrt{1 - s^(2)}} = -5`.

`s^(2) =25(1 - s^(2))`.

`\displaystyle s = \sqrt{(25)/(26)} = (5√(26))/(26)`.

Therefore

`\displaystyle \sin{(\theta)/(2)} = (5√(26))/(26)\approx 0.196`.