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Shyam · #3 November 21st, 2008, 07:09 PM

Quote:

Originally Posted by Shapeshift

We got this question the other day and I'm not exactly sure how to do it.

Given A+B+C= pi

Prove:

cos(2A)+cos(2B)+cos(2C)+1 = -4cos(A)cos(B)cos(C)

The hint was to use product formulas.

Can anybody help me with how I would go about doing this?

Given, $A+B+C=\pi$

$\Rightarrow A+B=\pi-C$

Now,

LS $= (\cos 2A + \cos 2B) + (\cos 2C) + 1$

$= 2 \cos (A+B) \cos (A-B) + (2 \cos^2 C - 1) + 1$

$[since,\;\; \cos A + \cos B = 2\cos \left( \frac{A+B}{2}\right) \cos \left( \frac{A-B}{2}\right)\;\;\;and \;\;\; \cos 2\theta=2 \cos^2 \theta - 1 \;\;]$

$= 2 \cos (A+B) \cos (A-B) + 2 \cos^2 C$

$= 2 \cos (\pi-C) \cos (A-B) + 2 \cos^2 C$

$= -2 \cos C \cos (A-B) + 2 \cos^2 C$

$=-2 \cos C \left[ \cos (A-B) - \cos C \right]$

$=-2 \cos C \left[ \cos (A-B) - \cos (\pi-(A+B)) \right]$

$=-2 \cos C \left[ \cos (A-B) + \cos (A+B) \right]$

$=-2 \cos C (2 \cos A \cos B)$

$=-4 \cos A \cos B \cos C$

Did you get it now ???

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