Question

Find the values of $\sin \left( \dfrac{x}{2} \right)$ , $\cos \left( \dfrac{x}{2} \right)$ and $\tan \left( \dfrac{x}{2} \right)$ if
$\sin \left( x \right)=\dfrac{1}{4}$ And x is in the second quadrant.

Answer 1

Hint: In this question, we are given the value of sine of x of the angle and the quadrant in which the angle x lies. Therefore, using the definition of $\sin \left( \dfrac{x}{2} \right)$ , and other trigonometric formulas, we can obtain the values of $\sin \left( \dfrac{x}{2} \right)$ , $\cos \left( \dfrac{x}{2} \right)$ and $\tan \left( \dfrac{x}{2} \right)$ by solving the corresponding trigonometric equations.

Complete step-by-step solution -
In this given question, we are asked to find the value of $\sin \left( \dfrac{x}{2} \right)$ , $\cos \left( \dfrac{x}{2} \right)$ and $\tan \left( \dfrac{x}{2} \right)$ if
$\sin \left( x \right)=\dfrac{1}{4}$ And x is in the second quadrant.
We know that for any angle a,
${{\cos }^{2}}a=1-{{\sin }^{2}}a........(1.1)$
Therefore, taking $a=x$ in equation (1.1) and using the given value of \[sin\left( x \right)\] , we obtain
$\begin{align}
  & {{\cos }^{2}}x=1-{{\sin }^{2}}x=1-{{\left( \dfrac{1}{4} \right)}^{2}}=\dfrac{15}{16} \\
 & \Rightarrow \cos x=\pm \sqrt{\dfrac{15}{16}}.................(1.2) \\
\end{align}$
However, as x lies in the second quadrant, the value of \[cos\left( x \right)\]should be negative as shown in the figure below, thus the value of \[cos\left( x \right)\] should be equal to $-\sqrt{\dfrac{15}{16}}=-\dfrac{\sqrt{15}}{4}$ ………………(1.3)

Now, we also know that for any angle $\theta $ ,
$\sin \left( \theta \right)=\sqrt{\dfrac{1-\cos \left( 2\theta \right)}{2}}.........................(1.4)$
Therefore, using $\theta =\dfrac{x}{2}$ in equation (1.4), and using the value of $\cos 2\theta =\cos x$ from (1.3), we obtain
$\sin \left( \dfrac{x}{2} \right)=\sqrt{\dfrac{1-\cos \left( 2\times \dfrac{x}{2} \right)}{2}=}\sqrt{\dfrac{1+\dfrac{\sqrt{15}}{4}}{2}}=\sqrt{\dfrac{4+\sqrt{15}}{8}}.........................(1.5)$
Again, as x lies in the second quadrant, $\dfrac{\pi }{2}Again using equation (1.1) and (1.5), we obtain
$\begin{align}
  & {{\cos }^{2}}\left( \dfrac{x}{2} \right)=1-{{\sin }^{2}}\left( \dfrac{x}{2} \right)=1-\left( \dfrac{4+\sqrt{15}}{8} \right)=\dfrac{4-\sqrt{15}}{8} \\
 & \Rightarrow \cos \left( \dfrac{x}{2} \right)=\sqrt{\dfrac{4-\sqrt{15}}{8}}........................(1.7) \\
\end{align}$
The value of $\cos \left( \dfrac{x}{2} \right)$ in equation 1.7 should be taken as positive as explained in (1.6).
Again the tangent is defined as
$\tan \left( a \right)=\dfrac{\sin a}{\cos a}$
For any angle a, using (1.5) and (1.7) as
\[\begin{align}
  & \tan \left( \dfrac{x}{2} \right)=\dfrac{\sin \left( \dfrac{x}{2} \right)}{\cos \left( \dfrac{x}{2} \right)}=\dfrac{\sqrt{\dfrac{4+\sqrt{15}}{8}}}{\sqrt{\dfrac{4-\sqrt{15}}{8}}}=\sqrt{\dfrac{4+\sqrt{15}}{4-\sqrt{15}}}= \\
 & =\sqrt{\dfrac{\left( 4+\sqrt{15} \right)\left( 4+\sqrt{15} \right)}{\left( 4-\sqrt{15} \right)\left( 4+\sqrt{15} \right)}}=\sqrt{\dfrac{{{\left( 4+\sqrt{15} \right)}^{2}}}{{{4}^{2}}-{{\left( \sqrt{15} \right)}^{2}}}}=\sqrt{\dfrac{{{\left( 4+\sqrt{15} \right)}^{2}}}{16-15}}=4+\sqrt{15}............(1.8) \\
\end{align}\]
Therefore, form equations (1.5), (1.7) and (1.8), we obtain the answer as
$\sin \left( \dfrac{x}{2} \right)=\sqrt{\dfrac{4+\sqrt{15}}{8}},\cos \left( \dfrac{x}{2} \right)=\sqrt{\dfrac{4-\sqrt{15}}{8},}\tan \left( \dfrac{x}{2} \right)=4+\sqrt{15}$
which is the required answer to this question.

Note: We could also have found tan(x) by using $\tan (x)=\dfrac{\sin x}{\cos x}$ and then found $\tan \left( \dfrac{x}{2} \right)$ by using the identity $\tan \left( x \right)=\dfrac{2\tan \left( \dfrac{x}{2} \right)}{1-{{\tan }^{2}}\left( \dfrac{x}{2} \right)}$ and from there we could have found $\sin \left( \dfrac{x}{2} \right)$ and $\cos \left( \dfrac{x}{2} \right)$ by using other trigonometric identities. However, the answer would have remained the same as found out in the solution above.