Question

If $\alpha ,\beta ,\gamma ,ẟ$ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity $k$, then the value of $4\sin \frac{\alpha }{2}+3\sin \frac{\beta }{2}+2\sin \frac{\gamma }{2}+\sin \frac{ẟ}{2}$ is equal to

• A. $2\surd (1–k)$
• B. $\frac{1}{2}\surd (1+k)$
• C. $2\surd (1+k)$
• D. None of these

Answer 1

The correct option is C

$2\surd (1+k)$

Explanation for the correct option:

Step 1. Find the value of $4\sin \frac{\alpha }{2}+3\sin \frac{\beta }{2}+2\sin \frac{\gamma }{2}+\sin \frac{ẟ}{2}$:

Given, $\alpha <\beta <\gamma <\delta$ and $\left(\sin \alpha =\sin \beta =\sin \gamma =\sin \delta \right)=k$

Now,

$\sin \alpha =\sin \beta$

$\Rightarrow$ $\sin (\pi -\alpha )=\sin \beta$

$\Rightarrow$ $\pi -\alpha =\beta$ ….(i)

$\sin \alpha =\sin \gamma$

$\Rightarrow$$\sin (2\pi +\alpha )=\gamma$

$\Rightarrow$ $2\pi +\alpha =\gamma$ ….(ii)

$\sin \alpha =\sin \delta$

$\Rightarrow$$\sin (3\pi -\alpha )=\delta$

$\Rightarrow$ $3\pi -\alpha =\delta$ ….(iii)

Step 2. Put the values of $\alpha ,\beta ,\gamma ,ẟ$ in given expression, we get

$4\sin \frac{\alpha }{2}+3\sin \frac{\beta }{2}+2\sin \frac{\gamma }{2}+\sin \frac{ẟ}{2}=4\sin \frac{\alpha }{2}+3\sin \frac{(–\alpha )}{2}+2\sin \frac{(2+\alpha )}{2}+\sin \frac{(3–\alpha )}{2}$

$=4\sin \frac{\alpha }{2}+3\cos \frac{\alpha }{2}–2\sin \frac{\alpha }{2}–\cos \frac{\alpha }{2}$

$=2\left[\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2}\right]$

$=2\left[\sqrt{{\left(\sin \frac{\alpha }{2}+\cos \frac{\alpha }{2}\right)}^2}\right]$

$=2\sqrt{\left[{\sin }^2\left(\frac{\alpha }{2}\right)+{\cos }^2\left(\frac{\alpha }{2}\right)+2\sin \frac{\alpha }{2}\cos \frac{\alpha }{2}\right]}$

$=2\sqrt{(1+\sin \alpha )}$$\left[\mathbf{\because }\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{2}\mathit{A}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{A}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathit{A}\mathbf{,}\mathbf{}\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{A}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

$\mathbf{\therefore }\mathbf{4}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\frac{\mathbf{\alpha }}{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\frac{\mathbf{\beta }}{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\frac{\mathbf{\gamma }}{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\frac{\mathbf{ẟ}}{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{\surd }\mathbf{(}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathit{k}\mathbf{)}$

Hence, Option ‘C’ is Correct.