Find    \small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}    in 

Q (8)   

\small \tan x = - \frac{4}{3}        , x in quadrant  II

Answers (1)

tan x = -\frac{4}{3}
We know that ,
    \sec^{2}x = 1 + \tan^{2}x
                  = 1 +\left ( -\frac{4}{3} \right )^{2}
                 = 1 + \frac{16}{9}  =  \frac{25}{9}
 sec x = \sqrt{\frac{25}{9}} = \pm\frac{5}{3}
x lies in II quadrant  thats why sec x is -ve 
So,

 sec x =-\frac{5}{3}
Now,  cos x = \frac{1}{\sec x}  =  -\frac{3}{5}
We know that,
                    cos x = 2\cos^{2}\frac{x}{2}- 1                                                                               (\because \cos2x = 2\cos^{2}x - 1 \Rightarrow \cos x = 2\cos^{2}\frac{x}{2} - 1  )
                    -\frac{3}{5}+ 1 = 2   \cos^{2}\frac{x}{2}
   
                    =  \frac{-3+5}{5}    =    2\cos^{2}\frac{x}{2}
                         
                               \frac{2}{5}   =    2\cos^{2}\frac{x}{2}
                             \cos^{2}\frac{x}{2} =  \frac{1}{5} 
                             \cos\frac{x}{2}   = \sqrt{\frac{1}{5}}  = \pm\frac{1}{\sqrt5} 
 x lies in II quadrant so value of           \cos\frac{x}{2}     is +ve       

 \cos\frac{x}{2}   =  \frac{1}{\sqrt5} = \frac{\sqrt5}{5}
we know that
                  cos x =1 - 2\sin^{2}\frac{x}{2}

                 2\sin^{2}\frac{x}{2}  =  1 -  (-\frac{3}{5})   =  \frac{8}{5}
                  
                 \sin^{2}\frac{x}{2} = \frac{4}{5}\\ \\=\sin\frac{x}{2} = \sqrt{ \frac{4}{5}} = \pm \frac{2}{\sqrt{5}}
x lies in II quadrant So value of sin x is +ve
  
                 \sin\frac{x}{2} = \frac{2}{\sqrt{}5} = \frac{2\sqrt5}{5} 

    \tan \frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{2\sqrt5}{5}}{\left ( \frac{\sqrt5}{5} \right )} = 2

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