Question

Find the value of $\sin \left( \dfrac{x}{2} \right)$ , $\cos \left( \dfrac{x}{2} \right)$ and $\tan \left( \dfrac{x}{2} \right)$ if $\tan \left( x \right)=\dfrac{-4}{3}$ And x is in the second quadrant.

Answer 1

Hint: In this question, we are given the value of tangent of x of the angle and the quadrant in which the angle x lies. Therefore, using the definition of $\tan \left( \dfrac{x}{2} \right)$ , 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
$\tan \left( x \right)=\dfrac{-4}{3}$ And x is in the second quadrant.
We know that the formula for finding the tangent of twice an angle is given by
$\tan \left( 2a \right)=\dfrac{2\tan a}{1-{{\tan }^{2}}a}................................(1.1)$
Now, x can be written as $2\times \dfrac{x}{2}$ .
So, putting $2\times \dfrac{x}{2}$ in place of a in equation 1.1, we get,
$\begin{align}
& \tan \left( 2\times \dfrac{x}{2} \right)=\dfrac{2\tan \left( \dfrac{x}{2} \right)}{1-{{\tan }^{2}}\left( \dfrac{x}{2} \right)} \\
 & \Rightarrow \tan \left( x \right)=\dfrac{2\tan \left( \dfrac{x}{2} \right)}{1-{{\tan }^{2}}\left( \dfrac{x}{2} \right)}..........(1.2) \\
\end{align}$
Now, putting $\tan \left( x \right)=\dfrac{-4}{3}$ in equation 1.2, we get,
\[\begin{align}
  & \tan \left( x \right)=\dfrac{2\tan \left( \dfrac{x}{2} \right)}{1-{{\tan }^{2}}\left( \dfrac{x}{2} \right)} \\
 & \Rightarrow \dfrac{-4}{3}=\dfrac{2\tan \left( \dfrac{x}{2} \right)}{1-{{\tan }^{2}}\left( \dfrac{x}{2} \right)} \\
 & \Rightarrow \dfrac{-4}{3}\left[ 1-{{\tan }^{2}}\left( \dfrac{x}{2} \right) \right]=2\tan \left( \dfrac{x}{2} \right) \\
 & \Rightarrow \dfrac{-4}{3}+\dfrac{4}{3}{{\tan }^{2}}\left( \dfrac{x}{2} \right)-2\tan \left( \dfrac{x}{2} \right)=0 \\
 & \Rightarrow \dfrac{4}{3}{{\tan }^{2}}\left( \dfrac{x}{2} \right)-2\tan \left( \dfrac{x}{2} \right)-\dfrac{4}{3}=0...........(1.3) \\
\end{align}\]
Multiplying both sides by 3 and then simplifying for $\tan \left( \dfrac{x}{2} \right)$ , we get,
$\begin{align}
  & \left( \dfrac{4}{3}{{\tan }^{2}}\left( \dfrac{x}{2} \right)-2\tan \left( \dfrac{x}{2} \right)-\dfrac{4}{3} \right)\times 3=0\times 3 \\
 & \Rightarrow 4{{\tan }^{2}}\left( \dfrac{x}{2} \right)-6\tan \left( \dfrac{x}{2} \right)-4=0 \\
 & \Rightarrow \left( \tan \left( \dfrac{x}{2} \right)+\dfrac{1}{2} \right)\times \left( \tan \left( \dfrac{x}{2} \right)-2 \right)=0 \\
 & \Rightarrow \tan \left( \dfrac{x}{2} \right)=-\dfrac{1}{2}\text{ or }2..........(1.4) \\
\end{align}$
As x is in the second quadrant, tan x should be negative as shown in the figure below. So, the value of \[tan\left( x \right)\] is $-\dfrac{1}{2}$ .

Now, we know that ${{\sec }^{2}}x=1+{{\tan }^{2}}x$ , so using this, we get,
${{\sec }^{2}}\left( \dfrac{x}{2} \right)=1+{{\tan }^{2}}\left( \dfrac{x}{2} \right)...........(1.5)$
Putting the value obtained in 1.4 in 1.5, and taking square roots of both the sides we get,
$\begin{align}
  & {{\sec }^{2}}\left( \dfrac{x}{2} \right)=\dfrac{5}{4} \\
 & \Rightarrow \text{sec}\left( \dfrac{x}{2} \right)=\pm \dfrac{\sqrt{5}}{2}............\text{(1}\text{.6)} \\
\end{align}$
As, x is in second quadrant, \[sec\left( x \right)\] is negative so \[sec\left( x \right)\] is equal to $-\dfrac{\sqrt{5}}{2}$
Now, \[\begin{align}
  & \sec \left( \dfrac{x}{2} \right)=\dfrac{1}{\cos \left( \dfrac{x}{2} \right)} \\
 & \Rightarrow \cos \left( \dfrac{x}{2} \right)=\dfrac{1}{\sec \left( \dfrac{x}{2} \right)} \\
\end{align}\]
So, we get the values of \[\cos \left( \dfrac{x}{2} \right)\] as \[-\dfrac{2}{\sqrt{5}}\] .
Now, \[\begin{align}
  & \tan \left( \dfrac{x}{2} \right)=\dfrac{\sin \left( \dfrac{x}{2} \right)}{\cos \left( \dfrac{x}{2} \right)} \\
 & \Rightarrow \sin \left( \dfrac{x}{2} \right)=\tan \left( \dfrac{x}{2} \right)\times \cos \left( \dfrac{x}{2} \right) \\
 & \Rightarrow \sin \left( \dfrac{x}{2} \right)=\left( -\dfrac{1}{2} \right)\times \left( -\dfrac{2}{\sqrt{5}} \right) \\
 & \Rightarrow \sin \left( \dfrac{x}{2} \right)=\dfrac{1}{\sqrt{5}}............(1.7) \\
\end{align}\]
Using equation 1.7, we get the value of $\sin \left( \dfrac{x}{2} \right)$ as $\dfrac{1}{\sqrt{5}}$ .
Hence, we have got all our answer as the values of$\sin \left( \dfrac{x}{2} \right)$ , $\cos \left( \dfrac{x}{2} \right)$ and $\tan \left( \dfrac{x}{2} \right)$ as $\dfrac{1}{\sqrt{5}}$, \[-\dfrac{2}{\sqrt{5}}\] and $-\dfrac{1}{2}$.


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