MHB Is Your Trigonometry Strong Enough to Solve This Challenge?

[anemone]

Evaluate $$\cos (A-C)+4\cos B$$ if $$b=\frac{a+c}{2}$$ 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:

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

Hence:

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

By solving:

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

And we get:

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

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

 

[anemone]

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

We get:

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

Hence:

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

By solving:

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

And we get:

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

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

You're not only correct...you're brilliant!(Clapping):)