Answer:
[tex]\bold{cos\dfrac{A}{2} = -\dfrac{1}{\sqrt3}}[/tex]
Step-by-step explanation:
Given that:
[tex]cosA=-\dfrac{1}3[/tex]
and
[tex]tanA > 0[/tex]
To find:
[tex]cos\dfrac{A}{2} = ?[/tex]
Solution:
First of all,we have cos value as negative and tan value as positive.
It is possible in the 3rd quadrant only.
[tex]\dfrac{A}{2}[/tex] will lie in the 2nd quadrant so [tex]cos\dfrac{A}{2}[/tex] will be negative again.
Because Cosine is positive in 1st and 4th quadrant.
Formula:
[tex]cos2\theta =2cos^2(\theta) - 1[/tex]
Here [tex]\theta = \frac{A}{2}[/tex]
[tex]cosA =2cos^2(\dfrac{A}{2}) - 1\\\Rightarrow 2cos^2(\dfrac{A}{2}) =cosA+1\\\Rightarrow 2cos^2(\dfrac{A}{2}) =-\dfrac{1}3+1\\\Rightarrow 2cos^2(\dfrac{A}{2}) =\dfrac{2}3\\\Rightarrow cos(\dfrac{A}{2}) = \pm \dfrac{1}{\sqrt3}[/tex]
But as we have discussed, [tex]cos\dfrac{A}{2}[/tex] will be negative.
So, answer is:
[tex]\bold{cos\dfrac{A}{2} = -\dfrac{1}{\sqrt3}}[/tex]
