# Evaluate cos(A-C)+4cosB

#### anemone

##### MHB POTW Director
Staff member
Evaluate $$\displaystyle \cos (A-C)+4\cos B$$ if $$\displaystyle b=\frac{a+c}{2}$$ in the triangle ABC.

#### Prove It

##### Well-known member
MHB Math Helper
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

##### MHB POTW Director
Staff member
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

##### Active member
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?

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#### anemone

##### MHB POTW Director
Staff member
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!