If \[ \sin x=\frac{\sqrt5}{3} \] and \(x\) lies in the IInd quadrant, find the values of \[ \cos\frac{x}{2},\quad \sin\frac{x}{2},\quad \tan\frac{x}{2} \]
Solution:
Given,
\[
\sin x=\frac{\sqrt5}{3}
\]
Using
\[
\sin^2x+\cos^2x=1
\]
we get
\[
\cos^2x
=
1-\left(\frac{\sqrt5}{3}\right)^2
\]
\[
=
1-\frac59
\]
\[
=
\frac49
\]
\[
\cos x=\pm\frac23
\]
Since \(x\) lies in the IInd quadrant,
\[
\cos x<0
\]
Therefore,
\[
\cos x=-\frac23
\]
Now,
\[
\cos\frac{x}{2}
=
\pm\sqrt{\frac{1+\cos x}{2}}
\]
Substituting the value of \(\cos x\):
\[
=
\pm\sqrt{
\frac{1-\frac23}{2}
}
\]
\[
=
\pm\sqrt{
\frac{\frac13}{2}
}
\]
\[
=
\pm\sqrt{\frac16}
\]
Since \(x\) lies in the IInd quadrant,
\[
\frac{x}{2}
\]
lies in the Ist quadrant, where cosine is positive.
Hence,
\[
\boxed{
\cos\frac{x}{2}
=
\frac{1}{\sqrt6}
}
\]
Now,
\[
\sin\frac{x}{2}
=
\pm\sqrt{\frac{1-\cos x}{2}}
\]
Substituting:
\[
=
\pm\sqrt{
\frac{1+\frac23}{2}
}
\]
\[
=
\pm\sqrt{
\frac{\frac53}{2}
}
\]
\[
=
\pm\sqrt{\frac56}
\]
Since \(\frac{x}{2}\) lies in the Ist quadrant, sine is positive.
Therefore,
\[
\boxed{
\sin\frac{x}{2}
=
\sqrt{\frac56}
}
\]
Now,
\[
\tan\frac{x}{2}
=
\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}
\]
Substituting values:
\[
=
\frac{\sqrt{\frac56}}{\sqrt{\frac16}}
\]
\[
=
\sqrt5
\]
Hence,
\[
\boxed{
\tan\frac{x}{2}=\sqrt5
}
\]