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