Question

If $\cos A=\frac{3}{4}$, then $32\sin \left(\frac{A}{2}\right)\sin \left(\frac{5A}{2}\right)=$

• A. $7$
• B. $8$
• C. $11$
• D. None of these

Answer 1

The correct option is C

$11$

Explanation for the correct options:

Finding the value of $\mathbf{32}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{(}\mathit{A}\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathbf{(}\mathbf{5}\mathit{A}\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{:}$

Given,

$\begin{array}{rcl}\cos A& =& \frac{3}{4}\\ 32\sin \left(\frac{A}{2}\right)\sin \left(\frac{5A}{2}\right)& =& 16[2\sin (\frac{A}{2})\sin (\frac{5A}{2}\left)\right]\end{array}$

$\begin{array}{rcl}32\sin \left(\frac{A}{2}\right)\sin \left(\frac{5A}{2}\right)& =& 16\left[\cos \left(\frac{A}{2}–\frac{5A}{2}\right)\right]–\left[\cos \left(\frac{A}{2}+\frac{5A}{2}\right)\right]\\ & =& 16(cos2A–cos3A)\\ & =& 16[2co{s}^{2}A–1–(4co{s}^{3}A–3cosA\left)\right]\\ & =& 16[2co{s}^{2}A–1–4co{s}^{3}A+3cosA]\\ & =& 16\left[2\left(\frac{9}{16}\right)–1–4\left(\frac{27}{64}\right)+3\left(\frac{3}{4}\right)\right]\left(\because given\cos A=¾\right)\\ & =& 16\left[\left(\frac{9}{8}\right)–1–\left(\frac{27}{16}\right)+\left(\frac{9}{4}\right)\right]\\ & =& 18–16–27+36\\ & =& 11\end{array}$

Hence. the correct answer is option (C).