Question

For $A=133°$, $2\cos \frac{A}{2}$ is equal to

• A. $-\sqrt{(1+\sin A)}–\sqrt{(1–\sin A)}$
• B. $-\sqrt{(1+\sin A)}+\sqrt{(1–\sin A)}$
• C. $\sqrt{(1+\sin A)}–\sqrt{(1–\sin A)}$
• D. $\sqrt{(1+\sin A)}+\sqrt{(1–\sin A)}$

Answer 1

The correct option is C

$\sqrt{(1+\sin A)}–\sqrt{(1–\sin A)}$

Explanation for correct option:

Step 1.Given that, $A=133°$

$\begin{array}{rcl}A& =& 133°\\ \frac{A}{2}& =& 66.5°\end{array}$

Step 2. using $\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{A}\mathbf{+}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{A}\mathbf{=}\mathbf{1}\mathbf{;}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{2}\mathit{A}\mathbf{=}\mathbf{2}\mathit{s}\mathit{i}\mathit{n}\mathit{A}\mathit{c}\mathit{o}\mathit{s}\mathit{A}$

$\begin{array}{rcl}\sqrt{1+\sin A}& =& \sqrt{{(\cos \frac{A}{2}+\sin \frac{A}{2})}^2}\\ \sqrt{1+\sin A}& =& \left|\cos \frac{A}{2}+\sin \frac{A}{2}\right|\\ \sqrt{1+\sin A}& =& \cos \frac{A}{2}+\sin \frac{A}{2}.....\left(i\right)\end{array}$

Also,

$\begin{array}{rcl}\sqrt{1-\sin A}& =& \sqrt{{(\cos \frac{A}{2}-\sin \frac{A}{2})}^2}\\ \sqrt{1-\sin A}& =& \left|\cos \frac{A}{2}-\sin \frac{A}{2}\right|[\sin ce\cos \frac{A}{2}<\sin \frac{A}{2}]\\ \sqrt{1-\sin A}& =& -\cos \frac{A}{2}+\sin \frac{A}{2}...\left(ii\right)\end{array}$

Step 3. Equation $\left(i\right)-\left(ii\right)$, we get

$\begin{array}{rcl}\sqrt{1+\sin A}-\sqrt{1-\sin A}& =& \cos \frac{A}{2}+\sin \frac{A}{2}-(-\cos \frac{A}{2}+\sin \frac{A}{2})\\ \sqrt{\mathbf{1}\mathbf{+}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{A}}\mathbf{-}\sqrt{\mathbf{1}\mathbf{-}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{A}}& \mathbf{=}& \mathbf{2}\mathbf{c}\mathbf{o}\mathbf{s}\frac{\mathbf{A}}{\mathbf{2}}\end{array}$

Hence, correct option is $\mathbf{\left(}\mathit{C}\mathbf{\right)}$