# Trig identity

• Nov 21st 2008, 02:55 PM
Shapeshift
Trig identity
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?
• Nov 21st 2008, 04:44 PM
TheEmptySet
Here is the idea
$-4\cos(A)\cos(B)\cos(C)=-2\cos(A)[2\cos(B)\cos(C)]$

using your hint the product identity is

$\cos(x+y)+\cos(x-y)=2\cos(x)\cos(y)$

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

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

using the product identiy again (twice) we end up with

$-\cos(A+B+C)-\cos(-A-B+C)-\cos(A-B+C)-\cos(-A+B+C)$

now we use the fact that $A+B+C=\pi$ and

notice that $-A-B=C-\pi$
$A+C=\pi -B$
and finally $B+C=\pi -A$

sub these into the above and use the sum identity for cosine and you should be done.

(Wink)
• Nov 21st 2008, 05:09 PM
Shyam
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 ???