Question

If cosθ = \(\frac{-3}{5}\)- and π < θ < \(\frac{3π}{2}\), then tan \(\frac{ θ}{2}\) + sin \(\frac{ θ}{2}\)+ 2cos \(\frac{ θ}{2}\) =

• A. -1
• B. 1
• C. -2
• D. 2

Answer 1

The Correct Option is C

Given $\cos \theta = -\frac{3}{5}$ with $\pi < \theta < \frac{3\pi}{2}$ (third quadrant), we need to evaluate $\tan \frac{\theta}{2} + \sin \frac{\theta}{2} + 2 \cos \frac{\theta}{2}$.

1. Determine the Quadrant for $\frac{\theta}{2}$:
Dividing the inequality $\pi < \theta < \frac{3\pi}{2}$ by 2 gives $\frac{\pi}{2} < \frac{\theta}{2} < \frac{3\pi}{4}$.
This places $\frac{\theta}{2}$ in the second quadrant where:
- $\sin \frac{\theta}{2} > 0$
- $\cos \frac{\theta}{2} < 0$
- $\tan \frac{\theta}{2} < 0$

2. Find $\cos \frac{\theta}{2}$:
Using the double-angle identity:
$\cos \theta = 2 \cos^2 \frac{\theta}{2} - 1$
$-\frac{3}{5} = 2 \cos^2 \frac{\theta}{2} - 1$
$2 \cos^2 \frac{\theta}{2} = \frac{2}{5}$
$\cos^2 \frac{\theta}{2} = \frac{1}{5}$
$\cos \frac{\theta}{2} = -\frac{1}{\sqrt{5}}$ (negative in second quadrant)

3. Find $\sin \frac{\theta}{2}$:
Using the identity $\cos \theta = 1 - 2 \sin^2 \frac{\theta}{2}$:
$-\frac{3}{5} = 1 - 2 \sin^2 \frac{\theta}{2}$
$2 \sin^2 \frac{\theta}{2} = \frac{8}{5}$
$\sin^2 \frac{\theta}{2} = \frac{4}{5}$
$\sin \frac{\theta}{2} = \frac{2}{\sqrt{5}}$ (positive in second quadrant)

4. Calculate $\tan \frac{\theta}{2}$:
$\tan \frac{\theta}{2} = \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} = \frac{\frac{2}{\sqrt{5}}}{-\frac{1}{\sqrt{5}}} = -2$

5. Evaluate the Expression:
$\tan \frac{\theta}{2} + \sin \frac{\theta}{2} + 2 \cos \frac{\theta}{2} = -2 + \frac{2}{\sqrt{5}} + 2\left(-\frac{1}{\sqrt{5}}\right)$
$= -2 + \frac{2}{\sqrt{5}} - \frac{2}{\sqrt{5}} = -2$

Final Answer:
The value of the expression is ${-2}$.