We have, cos x = -

1

3

, x in quadrant III
since, x is in III quadrant
⇒ π < x <

3π

2

⇒

π

2

<

x

2

<

3π

4

⇒

x

2

lies in II quadrant
⇒ sin

x

2

> 0 , cos

x

2

< 0 , tan

x

2

< 0 , cos

x

2

= ± √

1+cosx

2

= ± √

1− 1 3

2

= +

1

√3

⇒ cos

x

2

= -

1

√3

Now , sin

x

2

= ± √

1−cosx

2

= ± √

1+ 1 3

2

= ± √

2

3

= √

2

3

and tan

x

2

=

sin x 2

cos x 2

= - √2
Hence, sin

x

2

= √

2

3

or

√6

3

, cos

x

2

= −

√3

3

, tan

x

2

= - √2