Answer:
![\bold{cos(A)/(2) = -(1)/(\sqrt3)}](https://img.qammunity.org/2021/formulas/mathematics/college/fin0w3375ib90beldqlmytak4p1wdjnzmm.png)
Explanation:
Given that:
![cosA=-\frac{1}3](https://img.qammunity.org/2021/formulas/mathematics/college/9y52cnxgv5awt7f15k0qqadmyu47mdpu13.png)
and
![tanA > 0](https://img.qammunity.org/2021/formulas/mathematics/college/ioqfjto3e1x7bf82voeqagvjx4vapebokl.png)
To find:
![cos(A)/(2) = ?](https://img.qammunity.org/2021/formulas/mathematics/college/gdo4olqat8077igm0lgheouw4vnyv4iiqg.png)
Solution:
First of all,we have cos value as negative and tan value as positive.
It is possible in the 3rd quadrant only.
will lie in the 2nd quadrant so
will be negative again.
Because Cosine is positive in 1st and 4th quadrant.
Formula:
![cos2\theta =2cos^2(\theta) - 1](https://img.qammunity.org/2021/formulas/mathematics/college/htfsg0f5jir8b8iatwl0sz1tv273p74sv6.png)
Here
![\theta = (A)/(2)](https://img.qammunity.org/2021/formulas/mathematics/college/984dcjsk9r7cbwgzc5q7vpuq3n27vl9xb8.png)
![cosA =2cos^2((A)/(2)) - 1\\\Rightarrow 2cos^2((A)/(2)) =cosA+1\\\Rightarrow 2cos^2((A)/(2)) =-\frac{1}3+1\\\Rightarrow 2cos^2((A)/(2)) =\frac{2}3\\\Rightarrow cos((A)/(2)) = \pm (1)/(\sqrt3)](https://img.qammunity.org/2021/formulas/mathematics/college/gkp3n0w6k61y3s2wqhg68btbh7jplvus7r.png)
But as we have discussed,
will be negative.
So, answer is:
![\bold{cos(A)/(2) = -(1)/(\sqrt3)}](https://img.qammunity.org/2021/formulas/mathematics/college/fin0w3375ib90beldqlmytak4p1wdjnzmm.png)