Evaluate cos(A-C)+4cosB

[anemone]

Evaluate \(\displaystyle \cos (A-C)+4\cos B\) if \(\displaystyle b=\frac{a+c}{2}\) in the triangle ABC.

 

[Prove It]

Am I correct in assuming that you're using a, b, c to represent the lengths of the triangle and A, B, C to represent the angles opposite their corresponding letter side?

 

[anemone]

Am I correct in assuming that you're using a, b, c to represent the lengths of the triangle and A, B, C to represent the angles opposite their corresponding letter side?

Yes, Prove It. You're right and I'm sorry for not being clear...

 

[mathworker]

Sorry I don't know how to use $\LaTeX$...

We get:

\(\displaystyle 2\sin(B)=\sin(A)+\sin(C)\)

Hence:

\(\displaystyle 2\left(2\sin\left(\frac{B}{2} \right)\cos\left(\frac{B}{2} \right) \right)=2\sin\left(\frac{A+C}{2} \right)\cos\left(\frac{A-C}{2} \right)
\)

By solving:

\(\displaystyle 2\cos\left(\frac{B}{2} \right)=\cos\left(\frac{A-C}{2} \right)\)

And we get:

\(\displaystyle \cos(A-C)=3-4\cos(B)\)

And the final result is 3...am I correct?

 

[anemone]

Sorry I don't know how to use $\LaTeX$...

We get:

\(\displaystyle 2\sin(B)=\sin(A)+\sin(C)\)

Hence:

\(\displaystyle 2\left(2\sin\left(\frac{B}{2} \right)\cos\left(\frac{B}{2} \right) \right)=2\sin\left(\frac{A+C}{2} \right)\cos\left(\frac{A-C}{2} \right)
\)

By solving:

\(\displaystyle 2\cos\left(\frac{B}{2} \right)=\cos\left(\frac{A-C}{2} \right)\)

And we get:

\(\displaystyle \cos(A-C)=3-4\cos(B)\)

And the final result is 3...am I correct?

You're not only correct...you're brilliant!