If \[ \cos x=-\frac35 \] and \(x\) lies in the IIIrd quadrant, find the values of \[ \cos\frac{x}{2},\quad \sin\frac{x}{2},\quad \text{and} \quad \sin2x \]

Solution: Given, \[ \cos x=-\frac35 \] Using \[ \sin^2x+\cos^2x=1 \] we get \[ \sin^2x = 1-\left(-\frac35\right)^2 \] \[ = 1-\frac9{25} \] \[ = \frac{16}{25} \] \[ \sin x=\pm\frac45 \] Since \(x\) lies in the IIIrd quadrant, \[ \sin x<0 \] Therefore, \[ \sin x=-\frac45 \] Now, \[ \cos\frac{x}{2} = \pm\sqrt{\frac{1+\cos x}{2}} \] Substituting the value of \(\cos x\): \[ \cos\frac{x}{2} = \pm\sqrt{ \frac{1-\frac35}{2} } \] \[ = \pm\sqrt{ \frac{\frac25}{2} } \] \[ = \pm\sqrt{\frac15} \] \[ = \pm\frac{1}{\sqrt5} \] Since \(x\) is in the IIIrd quadrant, \[ \frac{x}{2} \] lies in the IInd quadrant, where cosine is negative. Hence, \[ \boxed{ \cos\frac{x}{2} = -\frac{1}{\sqrt5} } \] Now, \[ \sin\frac{x}{2} = \pm\sqrt{\frac{1-\cos x}{2}} \] Substituting: \[ = \pm\sqrt{ \frac{1+\frac35}{2} } \] \[ = \pm\sqrt{ \frac{\frac85}{2} } \] \[ = \pm\sqrt{\frac45} \] \[ = \pm\frac{2}{\sqrt5} \] Since \[ \frac{x}{2} \] lies in the IInd quadrant, sine is positive. Therefore, \[ \boxed{ \sin\frac{x}{2} = \frac{2}{\sqrt5} } \] Now, \[ \sin2x=2\sin x\cos x \] Substituting values: \[ = 2\left(-\frac45\right)\left(-\frac35\right) \] \[ = \frac{24}{25} \] Hence, \[ \boxed{ \sin2x=\frac{24}{25} } \]