1. ## 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?

2. ## Here is the idea

$\displaystyle -4\cos(A)\cos(B)\cos(C)=-2\cos(A)[2\cos(B)\cos(C)]$

using your hint the product identity is

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

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

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

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

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

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

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

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

3. 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, $\displaystyle A+B+C=\pi$

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

Now,

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

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

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

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

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

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

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

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

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

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

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

Did you get it now ???