Question

Given costheta = 3/5 and theta is located in the fourth quadrant; find tan theta/2.

Answer 1

Answer:

$\displaystyle \tan \frac{\theta}{2} = -\frac{1}{2}$

Step-by-step explanation:

We are given that:

$\displaystyle \cos \theta = \frac{3}{5}\text{, where } \frac{3\pi}{2} < \theta < 2\pi$

(The fourth quadrant).

And we want to find the value of:

$\displaystyle \tan \frac{\theta}{2}$

We can start by letting x = θ/2. Then θ = 2x. Hence:

$\displaystyle \cos 2x = \frac{3}{5}$

From the double-angle identity, solve for cosine:

$\displaystyle \begin{aligned} \cos 2x = 2\cos^2 x - 1 & = \frac{3}{5} \\ \\ \cos^2 x & = \frac{4}{5} \\ \\ \cos x& = \pm\frac{2}{\sqrt{5}} \Rightarrow -\frac{2}{\sqrt{5}}\end{aligned}$

Because:

$\displaystyle \displaystyle \frac{3\pi}{2} < \theta < 2\pi\text{, then } \frac{3\pi}{2} < 2x< 2\pi\Rightarrow \frac{3\pi}{4} < x < \pi$

Hence, cosine is always negative, and sine is always positive.

Solve for sine using the Pythagorean Identity:

$\displaystyle \begin{aligned} \cos^2 x + \sin^2 x & = 1 \\ \\ \left(-\frac{2}{\sqrt{5}}\right)^2 + \sin^2 x & = 1 \\ \\ \sin^2 x & = \frac{1}{5} \\ \\ \sin x & = \pm\frac{1}{\sqrt{5}} \Rightarrow \frac{1}{\sqrt{5}} \end{aligned}$

Finally, by definition:

$\displaystyle \tan x = \frac{\sin x}{\cos x}$

Substitute and evaluate:

$\displaystyle \begin{aligned} \tan x & = \frac{\left(\dfrac{1}{\sqrt{5}}\right)}{\left(-\dfrac{2}{\sqrt{5}}\right)} \\ \\ & = -\frac{1}{2} \end{aligned}$

And because x = θ/2, in conclusion:

$\displaystyle \tan \frac{\theta}{2} = -\frac{1}{2}$