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Evaluate cos(A-C)+4cosB

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anemone

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Feb 14, 2012
3,706
Evaluate \(\displaystyle \cos (A-C)+4\cos B\) if \(\displaystyle b=\frac{a+c}{2}\) in the triangle ABC.
 

Prove It

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MHB Math Helper
Jan 26, 2012
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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?
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,706
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...:eek:
 

mathworker

Active member
May 31, 2013
118
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
Feb 14, 2012
3,706
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!(Clapping):)