I am trying to prove that $$\cos A+\cos B+\cos C=4\sin\frac A2\sin\frac B2\sin\frac C2$$ for ABC is a triangle. I tried up to the stage of $$-2\sin^2 C+2\cos\frac{180-C}2 \cos\frac{A+B}2$$ but how do I proceed from here?

  • 1
    $\begingroup$ You seem to be missing a $1$ somewhere in there. $\endgroup$
    – robjohn
    Jul 9, 2013 at 18:30

3 Answers 3


$$\cos A+\cos B+\cos C$$

$$=2\cos\frac{A+B}2\cos\frac{A-B}2+1-2\sin^2\frac C2$$

$$=2\sin\frac C2\cos\frac{A-B}2+1-2\sin^2\frac C2\text{ as } \cos\frac{A+B}2=\cos\left(\frac{\pi-C}2\right)=\sin \frac C2$$

$$=1+2\sin\frac C2\left(\cos\frac{A-B}2-\cos\frac{A+B}2\right)\text{ as }\sin \frac C2=\cos\frac{A+B}2$$

$$=1+2\sin\frac C22\sin\frac A2\sin\frac B2$$

$$=1+4\sin\frac A2\sin\frac B2\sin\frac C2$$


Recall the identities $$ \begin{align} 2\sin(x)\sin(y)&=\cos(x-y)-\cos(x+y)\tag{1}\\ 2\sin(x)\cos(y)&=\sin(x-y)+\sin(x+y)\tag{2} \end{align} $$

First use $(1)$ on $2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)$, then $(2)$ on the results. Finally, use $A+B+C=\pi$. $$ \begin{align} &4\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)\\ &=\color{#C00000}{2\sin\left(\frac{C}{2}\right)\cos\left(\frac{A-B}{2}\right)} \color{#00A000}{-2\sin\left(\frac{C}{2}\right)\cos\left(\frac{A+B}{2}\right)}\\ &=\color{#C00000}{\sin\left(\frac{C+B-A}{2}\right)+\sin\left(\frac{C+A-B}{2}\right)} \color{#00A000}{+\sin\left(\frac{A+B-C}{2}\right)-\sin\left(\frac{A+B+C}{2}\right)}\\[6pt] &=\cos(A)+\cos(B)+\cos(C)-1 \end{align} $$


PROOF: There is something wrong in the question
TO PROVE :$\cos A+\cos B+\cos C -1=4\sin\frac A2\sin\frac B2\sin\frac C2$
PROOF: You can proceed from the right hand side too
$4\sin\frac A2\sin\frac B2\sin\frac C2 = 2\sin\frac A2\sin\frac B2 2\sin\frac C2$
NOW, $2\sin\frac A2\sin\frac B2 = cos\frac{A-B}2 - \cos\frac{A+B}2 = cos\frac{A-B}2 - \sin\frac C2$
The expression becomes,
$2\sin\frac C2\cos\frac{A-B}2 - 2\sin^2\frac C2 = \sin\frac{A-B+C}2 - \sin\frac{A-B-C}2 + cosC - 1$ = $cosB + cosA+ cosC- 1$


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