# Thread: Cos Rule----- Sine Rule

1. ## Cos Rule----- Sine Rule

Proof This $a(cosB+cosC-1)+b(cosC+cosA-1)+c(cosA+cosB-1)=0$you can use sine rule and cos rule.

Can You give me a hint to simplify this Problem>>><<<<<<

2. ## Re: Cos Rule----- Sine Rule

Proof This a(cosB+cosC-1)+b(cosC+cosA-1)+c(cosA+cosB-1)=0 you can use sine rule and cos rule.

What the question doesn't mention is that ABC is a triangle.

I only used the cosine rule. I converted CosA, cosB and cos c for its cos rule equivant.

then I expanded each bracket.

then I collected terms with a common denominator and simplified the numerators.

then it all just fell out.

3. ## Re: Cos Rule----- Sine Rule

$A+B+C=\pi$,that I did not mention>>>>>>

4. ## Re: Cos Rule----- Sine Rule

Write out the law of cosines for each angle:

$a^2 = b^2+c^2-2bc\cos A$ Solve for $\cos A$

$b^2 = a^2+c^2-2ac\cos B$ Solve for $\cos B$

$c^2 = a^2+b^2-2ab\cos C$ Solve for $\cos C$

Once you solve for all three, plug them back into the original expression.

5. ## Re: Cos Rule----- Sine Rule

Just use sine rule and you will get sinerule : a/sinA = b/sin B = c/sin C = K ( Constant )
LHS = K [ sin A ( cos B + cos C - 1 ) + sin B ( cos C + cos A - 1 ) + sin C ( cos A + cos B - 1 )]
= K [ sin A cos B + sin A cos C - sin A + sin B cos C + sin B cos A - sin B + sin C cos A + sin C cos B - sin C] Rearrange terms
= K [ sin A cos B + sin B cos A + sin A cos C + sin C cos A + sin B cos C + sin C cos B - sin A - sin B - sin C]
= K [ sin ( A + B ) + sin ( A + C ) + sin ( B + C ) - sin A - sin B - sin C]
Now we know that A + B + C = 2 pi OR 180 degree thus we will have sin ( A + B ) = sin C etc Thus we get
LHS = K [ sin C + sin B + sin A - sin A - sin B - sin C ] = 0 = RHS

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# prove that in triangle ABC :- a(cosB cosC-1) b(cosC cosA-1) c(cosA cosB-1)=0

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