Question

If \(1 + \cos x = \alpha\), \(0 \leq x \leq \frac{\pi}{2}\), then \(\sin \frac{x}{2}\) is equal to

• A. $\sqrt{\frac{2+\alpha}{2}}$
• B. $\sqrt{\frac{2-\alpha}{2}}$
• C. $\sqrt{\frac{2-\alpha}{2}}$
• D. $\sqrt{\frac{1+\alpha}{2}}$
• E. $\sqrt{\frac{1-\alpha}{2}}$

Hint:

For expressions involving $\cos x$, converting into half-angle form simplifies calculations.

Answer 1

The Correct Option is B

Concept: Use half-angle identity: \[ \cos x = 1 - 2\sin^2\frac{x}{2} \]

Step 1: Apply identity
\[ \cos x = 1 - 2\sin^2\frac{x}{2} \]

Step 2: Substitute in given equation
\[ 1 + \cos x = \alpha \] \[ 1 + (1 - 2\sin^2\frac{x}{2}) = \alpha \] \[ 2 - 2\sin^2\frac{x}{2} = \alpha \] \[ \sin^2\frac{x}{2} = \frac{2 - \alpha}{2} \]

Step 3: Take positive root
Since $0 \leq x \leq \frac{\pi}{2}$, $\sin\frac{x}{2} \geq 0$ \[ \sin\frac{x}{2} = \sqrt{\frac{2 - \alpha}{2}} \] Final Conclusion:
Option (B)