Proof This you can use sine rule and cos rule.

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

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- November 6th 2013, 01:53 AMsrirahulanCos Rule----- Sine Rule
Proof This you can use sine rule and cos rule.

Can You give me a hint to simplify this Problem>>><<<<<< - November 6th 2013, 02:55 AMMelody2Re: 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. - November 6th 2013, 05:08 AMsrirahulanRe: Cos Rule----- Sine Rule
,that I did not mention>>>>>>

- November 6th 2013, 05:12 AMSlipEternalRe: Cos Rule----- Sine Rule
Write out the law of cosines for each angle:

Solve for

Solve for

Solve for

Once you solve for all three, plug them back into the original expression. - November 13th 2013, 01:45 AMibduttRe: 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