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

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 IInd quadrant, \[ \sin x>0 \] Therefore, \[ \sin x=\frac45 \] Now, \[ \sin2x=2\sin x\cos x \] Substituting the values: \[ = 2\left(\frac45\right)\left(-\frac35\right) \] \[ = -\frac{24}{25} \] Hence, \[ \boxed{ \sin2x=-\frac{24}{25} } \] Now, \[ \sin\frac{x}{2} = \pm\sqrt{\frac{1-\cos x}{2}} \] Substituting the value of \(\cos x\): \[ = \pm\sqrt{ \frac{1+\frac35}{2} } \] \[ = \pm\sqrt{ \frac{\frac85}{2} } \] \[ = \pm\sqrt{\frac45} \] \[ = \pm\frac{2}{\sqrt5} \] Since \(x\) lies in the IInd quadrant, \[ \frac{x}{2} \] lies in the Ist quadrant, where sine is positive. Therefore, \[ \boxed{ \sin\frac{x}{2} = \frac{2}{\sqrt5} } \]