Is Your Trigonometry Strong Enough to Solve This Challenge?

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Discussion Overview

The discussion revolves around evaluating the expression $$\cos (A-C)+4\cos B$$ given the condition $$b=\frac{a+c}{2}$$ in triangle ABC. Participants explore trigonometric identities and relationships within the context of triangle geometry.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant presents the expression to evaluate and sets the context with the relationship between the sides and angles of triangle ABC.
  • Another participant seeks clarification on the notation used for the sides and angles of the triangle, confirming the representation.
  • A participant derives the equation $$2\sin(B)=\sin(A)+\sin(C)$$ and manipulates it to reach $$\cos(A-C)=3-4\cos(B)$$.
  • There is a claim that the final result of the evaluation is 3, with a request for confirmation of correctness.
  • A later reply enthusiastically agrees with the correctness of the previous participant's conclusion.

Areas of Agreement / Disagreement

While there is agreement on the correctness of the final result presented by one participant, the discussion does not resolve whether the derivation leading to that result is universally accepted or if alternative interpretations exist.

Contextual Notes

Some participants express uncertainty about using LaTeX formatting, which may affect the clarity of their mathematical expressions.

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Evaluate $$\cos (A-C)+4\cos B$$ if $$b=\frac{a+c}{2}$$ in the triangle ABC.
 
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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?
 
Prove It said:
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...:o
 
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?
 
Last edited by a moderator:
mathworker said:
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):)
 

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