MCQ
If $\cos A = \frac{3}{4}$, then $32\sin \frac{A}{2}\cos \frac{5}{2}A = $
  • A
    $\sqrt 7 $
  • $ - \sqrt 7 $
  • C
    $7$
  • D
    $-7$

Answer

Correct option: B.
$ - \sqrt 7 $
b
(b) $\cos A = \frac{3}{4} \Rightarrow \sin A = \frac{{\sqrt 7 }}{4}$

$L.H.S.$ $ = 16(\sin 3A - \sin 2A)$ 

$ = 16\sin A(3 - 4{\sin ^2}A - 2\cos A)$ 

$ = 16.\frac{{\sqrt 7 }}{4}\left( {3 - 4.\frac{7}{{16}} - 2.\frac{3}{4}} \right) = - \sqrt 7 $.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

A bag contains tickets numbered from $1$ to $20$. Two tickets are drawn. The probability that both the numbers are prime, is
The area bounded by the curves $y = -  \sqrt { - \,x}$  and $x = -\sqrt { - \,y} $  where $x, y \le 0$
$\int_{}^{} {x{{\cos }^2}} xdx = $
If $f(x) = \cos (\log x)$, then $f({x^2})f({y^2}) - \frac{1}{2}\left[ {f\,\left( {\frac{{{x^2}}}{2}} \right) + f\left( {\frac{{{x^2}}}{{{y^2}}}} \right)} \right]$ has the value
If ${1 \over {{{\log }_3}\pi }} + {1 \over {{{\log }_4}\pi }} > x,$ then $x$ be
Let $\vec a = 2\hat i - \hat j + \hat k$, $\vec b = \hat i + 2\hat j - \hat k$ and $\vec c = \hat i + \hat j - 2\hat k$ be three vectors. A vector of the type $\vec b + \lambda \vec c$ for some scalar $\lambda $,  whose projection on $\vec a$ is of magnitude $\sqrt {\frac{2}{3}} $ is
$\int_{}^{} {\frac{{x - \sin x}}{{1 - \cos x}}dx = } $
If $\cot \,\theta + \tan \theta = m$ and $\sec \theta - \cos \theta = n,$ then which of the following is correct
Statement $1$ : The only circle having radius $\sqrt {10} $ and a diameter along line $2x + y = 5$ is $x^2 + y^2 - 6x +2y = 0$.
Statement $2$ : $2x + y = 5$ is a normal to the circle $x^2 + y^2 -6x+2y = 0$.
Let $\hat{u}=u_1 \hat{i}+u_2 \hat{j}+u_3 \hat{k}$ be a unit vector in $\mathbb{R}^3$ and $\hat{v}=\frac{1}{\sqrt{6}}(\hat{i}+\hat{j}+2 \hat{k})$. Given that there exists is(are) correct?

($A$) There is exactly one choice for such $\vec{v}$

($B$) There are infinitely many choices for such $\vec{v}$

($C$) If $\hat{u}$ lies in the $x y$-plane then $\left|u_1\right|=\left|u_2\right|$

($D$) If $\hat{u}$ lies in the $x z$-plane then $2\left|u_1\right|=\left|u_3\right|$