Question

Find $\sin \dfrac{x}{2},\;\cos \dfrac{x}{2}\;{\text{and}}\,\tan \dfrac{x}{2}$ if $\tan x = \dfrac{{ - 4}}{4}$ also given that $x \in 2{\text{nd}}$ quadrant.

Answer 1

Hint: To find the given trigonometric functions, first find value of tangent function with help of half angle formula of tangent. And then we will use trigonometric identity consisting of secant and tangent by which we will get the value of secant and eventually cosine. And at last use sine and cosine trigonometric identity to get the value of sine function.

Formula used:
Area of the trapezium =\[\dfrac{1}{2}\left( a+b \right)\times h\], where ‘a’ and ‘b’ are the two parallel sides and ‘h’ is the height is of the trapezium.

Complete step by step solution:
In order to find values of $\sin \dfrac{x}{2},\;\cos \dfrac{x}{2}\;{\text{and}}\,\tan \dfrac{x}{2}$ if $\tan x = \dfrac{{ - 4}}{4}$ and $x$ is in second quadrant which means interval of $x\;{\text{is}}\;\left[ {\dfrac{\pi }{2},\;\pi } \right]$ we first find the value of tangent function using its half angle formula as follows
Given
$
   \Rightarrow \tan x = \dfrac{{ - 4}}{4} \\
   \Rightarrow \tan x = - 1 \\
 $
We know that, $\tan x = \dfrac{{2\tan \dfrac{x}{2}}}{{{{\tan }^2}\dfrac{x}{2}}}$
$
   \Rightarrow \dfrac{{2\tan \dfrac{x}{2}}}{{{{\tan }^2}\dfrac{x}{2}}} = - 1 \\
   \Rightarrow 2\tan \dfrac{x}{2} = - {\tan ^2}\dfrac{x}{2} \\
   \Rightarrow 2\tan \dfrac{x}{2} + {\tan ^2}\dfrac{x}{2} = 0 \\
   \Rightarrow \tan \dfrac{x}{2}\left( {2 + \tan \dfrac{x}{2}} \right) = 0 \\
   \Rightarrow \tan \dfrac{x}{2} = 0\;{\text{and}}\;\tan \dfrac{x}{2} = - 2 \\
 $
But, it is given in the question that $x \in 2{\text{nd}}$ quadrant, and for $\tan \dfrac{x}{2} = 0$ value of $x$ will lie outside the second quadrant.
Therefore $\tan \dfrac{x}{2} = - 2$ is one of the required values
Now, we know that, ${\sec ^2}x = 1 + {\tan ^2}x \Rightarrow \sec x = \sqrt {1 + {{\tan }^2}x} $ using this, we will get
$
   \Rightarrow \sec \dfrac{x}{2} = \pm \sqrt {1 + {{\tan }^2}\dfrac{x}{2}} \\
   \Rightarrow \sec \dfrac{x}{2} = \pm \sqrt {1 + {{( - 2)}^2}} \\
   \Rightarrow \sec \dfrac{x}{2} = \pm \sqrt {1 + 4} \\
   \Rightarrow \sec \dfrac{x}{2} = \pm \sqrt 5 \\
 $
We are getting here two values, positive and negative but we know that cosine and secant functions are negative in the given second quadrant, but $\dfrac{x}{2}$ lies in first quadrant and in first quadrant all function are positive.

$ \Rightarrow \sec \dfrac{x}{2} = \sqrt 5 $
Also from trigonometric relations, we know that $\cos x = \dfrac{1}{{\sec x}}$
$
   \Rightarrow \cos \dfrac{x}{2} = \dfrac{1}{{\sec \dfrac{x}{2}}} \\
   \Rightarrow \cos \dfrac{x}{2} = \dfrac{1}{{\sqrt 5 }} \\
 $
Now, from the trigonometric identity between sine and cosine, we know that ${\sin ^2}x = 1 - {\cos ^2}x \Rightarrow \sin x = \sqrt {1 - {{\cos }^2}x} $
$
   \Rightarrow \sin \dfrac{x}{2} = \pm \sqrt {1 - {{\cos }^2}\dfrac{x}{2}} \\
   \Rightarrow \sin \dfrac{x}{2} = \pm \sqrt {1 - {{\left( {\dfrac{1}{{\sqrt 5 }}} \right)}^2}} \\
   \Rightarrow \sin \dfrac{x}{2} = \pm \sqrt {1 - \dfrac{1}{5}} \\
   \Rightarrow \sin \dfrac{x}{2} = \pm \sqrt {\dfrac{{5 - 1}}{5}} \\
   \Rightarrow \sin \dfrac{x}{2} = \pm \sqrt {\dfrac{4}{5}} \\
   \Rightarrow \sin \dfrac{x}{2} = \pm \dfrac{2}{{\sqrt 5 }} \\
 $
Again we are getting here two values, but we know that sine is positive in first ,second quadrant,
Therefore $\sin \dfrac{x}{2} = \dfrac{2}{{\sqrt 5 }}$ is the required answer.

Note: In trigonometric problems like this, when finding solutions for a particular given interval then make sure to check the polarity (either positive or negative) of functional values of trigonometric functions in that interval.