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Find sinx2 cosx2 and tanx2 in cos x is equal to 13 x in quadrant three

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


Q (9)

\small \cos x = -\frac{1}{3}   , x in quadrant III

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\pi < x < \frac{3\pi}{2}\\ \\ \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}

We know that
     cos x  = 2\cos^{2}\frac{x}{2} - 1
              2\cos^{2}\frac{x}{2} =  cos x + 1
                                =  \left ( -\frac{1}{3} \right )   + 1   =  \left ( \frac{-1+3}{3} \right )   =   \frac{2}{3}

             \cos\frac{x}{2} = \sqrt{ \frac{1}{3}} = \pm \frac{1}{\sqrt3}  
          
          \cos\frac{x}{2} = - \frac{1}{\sqrt3} = - \frac{\sqrt3}{3}
Now,
      we know that 
 cos x = 1 - 2\sin^{2}\frac{x}{2}
          2\sin^{2}\frac{x}{2} = 1 - \cos x
                            =  1 - \left ( -\frac{1}{3} \right )   =  \frac{3+1}{3}  = \frac{4}{3}
              
               2\sin^{2}\frac{x}{2} = \frac{4}{3} \\ \\ \sin^{2}\frac{x}{2} = \frac{2}{3}\\ \sin\frac{x}{2} = \sqrt{ \frac{2}{3}} = \pm \sqrt{ \frac{2}{3}} = \frac{\sqrt6}{3}       
Because  \sin\frac{x}{2}  is +ve in given quadrant

\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt6}{3}}{\frac{-\sqrt3}{3}} = - \sqrt2
                                     

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