First slide

Multiple and sub- multiple Angles

Question

If $2 \sin \frac{A}{2} = \sqrt{1 + \sin A} + \sqrt{1 - \sin A},$ then $\frac{A}{2}$ lies between

Moderate

A

$2 n π + \frac{π}{4}$ and $2 n π + \frac{3 π}{4}, n \in Z$
B

$2 n π - \frac{π}{4}$ and $2 n π + \frac{π}{4}, n \in Z$
C

$2 n π - \frac{3 π}{4}$ and $2 n π - \frac{π}{4}, n \in Z$
D

$- \infty$ and $+ \infty$

Solution

We have,
$\begin{array}{l} 2 \sin \frac{A}{2} = \sqrt{1 + \sin A} + \sqrt{1 - \sin A} \\ \Rightarrow 2 \sin \frac{A}{2} = \sqrt{(\cos A / 2 + \sin A / 2)^{2}} \\ + \sqrt{(\cos A / 2 - \sin A / 2)^{2}} \\ \Rightarrow 2 \sin A / 2 = | \cos A / 2 + \sin A / 2 | \\ + | \cos A / 2 - \sin A / 2 | \\ \Rightarrow \cos A / 2 + \sin A / 2 \geq 0 and \cos A / 2 - \sin A / 2 \leq 0 \\ \Rightarrow π / 4 \leq A / 2 \leq 3 π / 4 and π / 4 \leq A \leq 5 π / 4 \\ \Rightarrow π / 4 \leq A / 2 \leq 3 π / 4 \\ \Rightarrow 2 n π + π / 4 \leq A / 2 \leq 2 n π + 3 π / 4, n \in Z . \end{array}$

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