# If $A+B+C=\pi$, prove: $\cos(B+2C)+\cos(C+2A)+\cos(A+2B)=1-4\cos\frac {B-C}{2}\;\cos\frac {C-A}{2}\;\cos\frac {A-B}{2}$

If $A+B+C=\pi$, prove that: $$\cos(B+2C)+\cos(C+2A)+\cos(A+2B)=1-4\cos\frac {B-C}{2}\;\cos\frac {C-A}{2}\;\cos\frac {A-B}{2}$$

My Attempt:

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

Now, \begin{align} LHS &=\cos(B+2C)+\cos(C+2A)+\cos(A+2B) \\ &=\cos(B+C+C)+\cos(C+A+A)+\cos(A+B+B) \\ &=\cos(\pi-(A-C))+\cos(\pi-(B-A))+\cos(\pi-(C-B)) \\ &=-\cos(A-C)-\cos(B-A)-\cos(C-B) \end{align}

• Certainly try: $\cos (A-C)=2\cos^2\frac{A-C}{2}-1$. – Thomas Andrews Oct 3 '16 at 15:58
• Maybe try drawing a picture of a triangle with angles $A,B,C$ and seeing if you can gain some intuition from that? – Neal Oct 3 '16 at 16:22

Let $C-A=2x,B-C=2y,A-B=2z\implies2(x+y+z)=0$

$$F=\cos2x+\cos2y+\cos2z=2\cos(x+y)\cos(x-y)+2\cos^2z-1$$

Now as $\cos(x+y)=\cos(-z)=\cos z,$

$$F=2\cos z\cos(x-y)+2\cos z\cdot\cos(x+y)-1$$ $$=2\cos z\{\cos(x+y)+\cos(x-y)\}-1=2\cos z\{2\cos x\cos y\}-1=?$$

• How is $cos(x+y)=cosz$? – pi-π Oct 4 '16 at 14:10
• @user354073, $$x+y=-z\implies\cos(x+y)=\cos(-z)=?$$ – lab bhattacharjee Oct 4 '16 at 15:49
• And how is LHS=RHS? I could not get that. Please clarify. – pi-π Oct 4 '16 at 16:09
• @user354073, Replace the values of $x,y,z$ – lab bhattacharjee Oct 4 '16 at 16:10
• With that, I got: $$4cos\frac {C-A}{2}.cos\frac {B-C}{2}.cos\frac {A-B}{2} -1$$? – pi-π Oct 4 '16 at 16:13

For the RHS, you can use a known formula for the product of three cosines (see eg. here):

$\cos x \; \cos y \; \cos z = \frac{1}{4} \Big(\cos (x+y-z) + \cos (y+z-x) + \cos (z+x-y) + \cos (x+y+z)\Big)$

Considering your RHS, you set $x = \frac{B-C}{2}$, $y = \frac{C-A}{2}$, $z = \frac{A-B}{2}$, and plug it into the formula. Then, for example, $x+y-z = B-A$, likewise the other arguments.

This gives immediately the form you arrived at, $-\cos(A-C)-\cos(B-A)-\cos(C-B)$.

• Andreas, I.could not understand. Could you please elaborate? – pi-π Oct 3 '16 at 16:07
• I added more steps in the answer. – Andreas Oct 3 '16 at 16:16