Resultant vector of an isosceles triangle?

[atypical]

Homework Statement

what is the resultant vector of an isosceles triangle?

Homework Equations

R^2=a^2+b^2-4abcos(theta)

The Attempt at a Solution

The books answer R=2acos(theta/2)
Using the formula above, and knowing that a=b in a isosceles triangle I am getting:
R=sqrt[2a^2+2a^2cos(theta)]

 

[Stonebridge]

It depends a little.

If the two vectors are as shown in the 1st diagram, where the sum is CA + AB in the directions shown, the sum is the vector CB.
By the cosine rule its length R would be
R² = a² + a² - 2a.a cos θ
R² = 2a² - 2a² cos θ

If, on the other hand, you mean the two vectors AB + AC as in the lower diagram with directions shown, the sum is vector AC in the diagram on the right.
It looks like this is what the book's answer refers to.

 

[atypical]

Thanks for the response. Now I have one more question about that. In the attachment, the book talks about (1-cos(theta))=2sin^2(theta/2) gives the magnitude of R as R=2Asin(theta/2). I don't see how they jump from the first equation to the second. Can anyone explain?

 

[Stonebridge]

Thanks for the response. Now I have one more question about that. In the attachment, the book talks about (1-cos(theta))=2sin^2(theta/2) gives the magnitude of R as R=2Asin(theta/2). I don't see how they jump from the first equation to the second. Can anyone explain?

Yes, using the cos rule on the triangle gives
D² = 2A² - 2A² cos θ
D² = 2A² (1- cos θ)
Using the identity for (1 - cos θ) gives
D² = 2A² (2 sin² (θ/2))
D² = 4A² sin² (θ/2)

D =