Question

If $\sin x = \dfrac{1}{4}$ , x is in the second quadrant. Find the value of $\sin \dfrac{x}{2}$ .

Answer 1

Hint: The value of sinx is given. Find cosx.
Given that x is in second quadrant i.e. $\dfrac{\pi }{2} < x \leqslant \pi $, ∴ cosx is negative.
So we first find cosx.
Now, note that $\cos x = 1 - 2{\sin ^2}\dfrac{x}{2}$, i.e. \[2{\sin ^2}\dfrac{x}{2} = 1 - \cos x\]
Therefore find $\sin \dfrac{x}{2}$.

Complete step-by-step answer:
Given, $\sin x = \dfrac{1}{4}$
Also, x is in the second quadrant. Therefore, $\dfrac{\pi }{2} < x \leqslant \pi $
We know,
${\sin ^2}x + {\cos ^2}x = 1$
$ \Rightarrow {\cos ^2}x = 1 - {\sin ^2}x$
On taking square root we get,
$ \Rightarrow \cos x = \pm \sqrt {1 - {{\sin }^2}x} $
On substituting the value of sinx we get,
$ \Rightarrow \cos x = \pm \sqrt {1 - {{\left( {\dfrac{1}{4}} \right)}^2}} $
On simplification we get,
$ \Rightarrow \cos x = \pm \sqrt {1 - \dfrac{1}{{16}}} = \pm \sqrt {\dfrac{{15}}{{16}}} $
Since, x lies in second quadrant, therefore cosx is negative,
$ \Rightarrow \cos x = - \dfrac{{\sqrt {15} }}{4}$
Now, we know
\[2{\sin ^2}\dfrac{x}{2} = 1 - \cos x\]
On dividing by 2 and taking square root we get,
\[ \Rightarrow \sin \dfrac{x}{2} = \pm \sqrt {\dfrac{{1 - \cos x}}{2}} \]
On substituting the value of cosx we get,
\[ \Rightarrow \sin \dfrac{x}{2} = \pm \sqrt {\dfrac{{1 - \left( { - \dfrac{{\sqrt {15} }}{4}} \right)}}{2}} \]
On simplification we get,
\[ \Rightarrow \sin \dfrac{x}{2} = \pm \sqrt {\dfrac{{1 + \dfrac{{\sqrt {15} }}{4}}}{2}} \]
As, \[\dfrac{\pi }{2}{\text{ < x}} \leqslant \pi \Rightarrow \dfrac{\pi }{4}{\text{ < }}\dfrac{x}{2} \leqslant \dfrac{\pi }{2}\] ,hence \[{\text{sin\;}}\dfrac{x}{2}\] positive as\[\dfrac{x}{2}\] is in the first quadrant
\[ \Rightarrow \sin \dfrac{x}{2} = \sqrt {\dfrac{{4 + \sqrt {15} }}{8}} \]
Therefore, the value of $\sin \dfrac{x}{2}$ is \[\sqrt {\dfrac{{4 + \sqrt {15} }}{8}} \].

Note: Note the following important formulae:
$\cos x = \dfrac{1}{{\sec x}}$ , $\sin x = \dfrac{1}{{\cos ecx}}$ , $\tan x = \dfrac{1}{{\cot x}}$
${\sin ^2}x + {\cos ^2}x = 1$
\[{\sec ^2}x - {\tan ^2}x = 1\]
\[{\operatorname{cosec} ^2}x - {\cot ^2}x = 1\]
$\sin ( - x) = - \sin x$
$\cos ( - x) = \cos x$
$\tan ( - x) = - \tan x$
$\sin \left( {2n\pi \pm x} \right) = \sin x{\text{ , period 2}}\pi {\text{ or 3}}{60^ \circ }$
$\cos \left( {2n\pi \pm x} \right) = \cos x{\text{ , period 2}}\pi {\text{ or 3}}{60^ \circ }$
$\tan \left( {n\pi \pm x} \right) = \tan x{\text{ , period }}\pi {\text{ or 18}}{0^ \circ }$
Sign convention:

$\sin 2x = 2\sin x\cos x$
$\cos 2x = {\cos ^2}x - {\sin ^2}x = 1 - 2{\sin ^2}x = 2{\cos ^2}x - 1$
$\tan 2x = \dfrac{{2\tan x}}{{1 - {{\tan }^2}x}} = \dfrac{2}{{\cot x - \tan x}}$