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I humbly ask for help in the following problem.

If \begin{equation} A+B+C=180 \end{equation} Then prove \begin{equation} \cos A+\cos B+\cos C=1+4\sin(A/2)\sin(B/2)\sin(C/2) \end{equation} How would I begin the problem I mean I think $\cos C $ can be $\cos(180-A+B)$. But I am unsure what to do next.

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tried to google it? – Santosh Linkha Jul 30 '12 at 16:56
Oh I see its on yahoo answer, so would I delete this question? – Phantom Jul 30 '12 at 16:59
Underneath the tags--that is, trigonometry and trigonometric-identities--there's a delete option. That said, you don't have to delete it. – Cameron Buie Jul 30 '12 at 17:11
Hmm I will keep the question for posterity. – Phantom Jul 30 '12 at 17:25
Then for posterity I think we should link to Y! Answers. – user2468 Jul 30 '12 at 17:34

Your observation that $C=180^\circ-(A+B)$ is a good one. Recall also the following trigonometric identities: $$\sin(x\pm y)=\sin x\cos y\pm\cos x\sin y$$ $$\cos(x\pm y)=\cos x\cos y\mp\sin x\sin y$$ By LHS, I'll denote the expression on the left-hand side of the desired identity; by RHS, the expression on the right-hand side.

Now, the RHS is messier, so we'll start there. Let's first apply your observation to $\sin(C/2)$, along with the angle difference and sum formulas for sine, and the angle sum formula for cosine, to see that

$\begin{eqnarray*} \sin(C/2) & = & \sin\bigl(90^\circ-(A+B)/2\bigr)\\ & = & \sin 90^\circ\cos\bigl((A+B)/2\bigr)-\cos 90^\circ\sin\bigl((A+B)/2\bigr)\\ & = & \cos(A/2+B/2)\\ & = & \cos(A/2)\cos(B/2)-\sin(A/2)\sin(B/2), \end{eqnarray*}$

since $\sin 90^\circ=1$ and $\cos 90^\circ=0$. Thus, we see that

$$\mathrm{RHS} = 1+4\sin(A/2)\cos(A/2)\sin(B/2)\cos(B/2)-4\sin^2(A/2)\sin^2(B/2)\tag{1}$$

Recall also the double angle formulas for sine (a special case of angle sum with $x=y$): $$\sin(2x)=2\sin x\cos x.$$

Also, the Pythogorean identity and the angle sum formula for cosine (with $x=y$) gives us the following double angle formula for cosine: $$\cos(2x)=\cos^2x-\sin^2x=1-2\sin^2x,$$ from which we derive the identity $$2\sin^2x=1-\cos(2x).$$ Applying this identity, along with the double angle and angle sum formulas for sine, to $(1)$ gives us

$\begin{eqnarray*} \mathrm{RHS} & = & 1+\bigl(2\sin(A/2)\cos(A/2)\bigr)\bigl(2\sin(B/2)\cos(B/2)\bigr)-\bigl(2\sin^2(A/2)\bigr)(2\sin^2(B/2)\bigr)\\ & = & 1+\sin A\sin B-(1-\cos A)(1-\cos B)\\ & = & \cos A + \cos B - \cos A\cos B+\sin A\sin B\\ & = & \cos A + \cos B - \cos(A+B). \end{eqnarray*}$

At this point, we can apply your observation again, along with the angle difference formula for cosine, to see that

$\begin{eqnarray*} \mathrm{LHS} & = & \cos A + \cos B + \cos 180^\circ\cos(A+B)-\sin 180^\circ\sin(A+B)\\ & = & \cos A + \cos B - \cos(A+B), \end{eqnarray*}$

since $\cos 180^\circ=-1$ and $\sin 180^\circ=0$. Thus, LHS = RHS, as desired.

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What I might do is start with the right side. Since I don't remember half-angle formulas, let $a = A/2$, $b=B/2$. Note that (from the addition formulas for $\cos$) $\sin(x) \sin(y) = (\cos(x-y) - \cos(x+y))/2$, $\cos(x) \cos(y) = (\cos(x+y) + \cos(x-y))/2$, $\sin(90 - x) = \cos(x)$.

$$ \eqalign{1 &{}+ 4 \sin(a) \sin(b) \sin(90 - a - b)\cr = & 1 + 2 (\cos(a-b) - \cos(a+b)) \cos(a+b) \cr = & 1 + \cos(2a) + \cos(2b) - \cos(2a+2b) - \cos(0)\cr = & \cos(2a) + \cos(2b) + \cos(180 - 2a - 2b)\cr} $$

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

$=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}$ as $\cos2x=1-2\sin^2x$

Now $\cos\frac{A+B}{2}=\cos\frac{180^\circ - C}{2}=\cos(90^\circ-\frac{C}{2})=\sin\frac{C}{2}$

So, $\cos A+\cos B+\cos C$ becomes $2\sin\frac{C}{2}\cos\frac{A-B}{2}+1-2\sin^2\frac{C}{2}$


$=1+2\sin\frac{C}{2}(\cos\frac{A-B}{2}-\cos\frac{A+B}{2})$ replacing $\sin\frac{C}{2}$ with $\cos\frac{A+B}{2}$

$=1+2\sin\frac{C}{2}(2\sin\frac{A}{2}\sin\frac{B}{2})$ applying $\cos(x-y)-\cos(x+y)= 2 \sin x \sin y$


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