Question

If sin(θ)=−24/25, and θ is in Quadrant III, then what is tan(θ/2)?

Answer 1

Recall the Pythagorean identity,

`\sin^2(\theta) + \cos^2(\theta) = 1`

Since
`\theta` belongs to Q3, we know both
`\sin(\theta)` and
`\cos(\theta)` are negative. Then

`\cos(\theta) = -√(1 - \sin^2(\theta)) = -\frac7{25}`

Recall the half-angle identities for sine and cosine,

`\sin^2\left(\frac\theta2\right) = \frac{1 - \cos(\theta)}2`

`\cos^2\left(\frac\theta2\right) = \frac{1 + \cos(\theta)}2`

Then by definition of tangent,

`\tan^2\left(\frac\theta2\right) = (\sin^2\left(\frac\theta2\right))/(\cos^2\left(\frac\theta2\right)) = (1 - \cos(\theta))/(1 + \cos(\theta))`

`\theta` belonging to Q3 means
`180^\circ < \theta < 270^\circ`, or
`90^\circ < \theta < 135^\circ`, so that the half-angle belongs to Q2. Then
`\sin\left(\frac\theta2\right)` is positive and
`\cos\left(\frac\theta2\right)` is negative, so
`\tan\left(\frac\theta2\right)` is negative.

It follows that

`\tan\left(\frac\theta2\right) = -\sqrt{(1 - \cos(\theta))/(1 + \cos(\theta))} = \boxed{-\frac43}`