Question

# If $\cos x = - \dfrac{4}{5}$ , where $x \in \left[ {0,\pi } \right]$ , then the value of $\cos \left( {\dfrac{x}{2}} \right)$ is equal to:$\left( a \right)\dfrac{1}{{10}} \\ \left( b \right)\dfrac{2}{5} \\ \left( c \right)\dfrac{1}{{\sqrt {10} }} \\ \left( d \right) - \dfrac{2}{5} \\ \left( e \right) - \dfrac{1}{{\sqrt {10} }} \\$

Hint: Use double angle identities of trigonometry . We know this relation $\cos \left( {2x} \right) = 2{\cos ^2}\left( x \right) - 1$ and also we can write like $\cos \left( x \right) = 2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1$.

Given, $\cos x = - \dfrac{4}{5},x \in \left[ {0,\pi } \right]$
Now, we use the double angle identity of trigonometry.
$\cos \left( x \right) = 2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1$
Use the value of $\cos x$ in above identity.
$\Rightarrow - \dfrac{4}{5} = 2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1 \\ \Rightarrow 2{\cos ^2}\left( {\dfrac{x}{2}} \right) = 1 - \dfrac{4}{5} \\ \Rightarrow 2{\cos ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{1}{5} \\ \Rightarrow {\cos ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{1}{{10}} \\$
Take square root
$\Rightarrow \cos \left( {\dfrac{x}{2}} \right) = \pm \dfrac{1}{{\sqrt {10} }}$
We can see two values of $\cos \left( {\dfrac{x}{2}} \right)$ but we have to choose only one value. So, we use $x \in \left[ {0,\pi } \right]$ .
Given, $x \in \left[ {0,\pi } \right]$
$0 \leqslant x \leqslant \pi \\ \Rightarrow 0 \leqslant \dfrac{x}{2} \leqslant \dfrac{\pi }{2} \\$
So, $\dfrac{x}{2} \in \left[ {0,\dfrac{\pi }{2}} \right]$
We know the graph of cosine is positive from 0 to $\dfrac{\pi }{2}$ .
Now, the value of $\cos \left( {\dfrac{x}{2}} \right) = \dfrac{1}{{\sqrt {10} }}$ .
So, the correct option is (c).

Note: Whenever we face such types of problems we use some important points. First we use trigonometric identities and after solving we get two answers one is positive and other is negative. So, for the correct answer we use range of x and observe whether the graph of that function is positive or negative on range of x.