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Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
Prove trigonometry identity?
If $A$, $B$, and $C$ are to be taken as the angles of a triangle, then I beg someone to help me the proof of $$\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}.$$ Thanks!
I usually find these things easiest to do with complex exponentials, rather than remembering a zoo of formulas for trigonometric functions. With $C=\pi-A-B$, we have
$$ \begin{align} \def\tr#1#2{\mathrm e^{\mathrm i#1}#2\mathrm e^{-\mathrm i#1}} \def\co#1{\tr{#1}+} \def\si#1{\tr{#1}-} \def\cop#1{\left(\co{#1}\right)} \def\sip#1{\left(\si{#1}\right)} &4\cos\frac A2\cos\frac B2\cos\frac C2 \\ =&\frac12\cop{A/2}\cop{B/2}\cop{C/2} \\ =& \frac12\cop{A/2}\cop{B/2}\cop{(\pi-A-B)/2} \\ =& \frac12\mathrm i\cop{A/2}\cop{B/2}\sip{(-A-B)/2} \\ =& \frac12\mathrm i\left(-\sip A-\sip B-\sip{(A+B)}\right) \\ =& \frac1{2\mathrm i}\left(\sip A+\sip B+\sip{(\pi-A-B)}\right) \\ =& \frac1{2\mathrm i}\left(\sip A+\sip B+\sip C\right) \\ =& \sin A+\sin B+\sin C\;. \end{align} $$
How about a proof with a geometric flavor?
Let $a$, $b$, $c$ be the sides that oppose angles $A$, $B$, $C$, respectively. By the Law of Sines, $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = d$$ where $d$ is the circumdiameter of the triangle. If we conveniently scale the triangle so that $d=1$, then we can say simply that $$a = \sin A \qquad b = \sin B \qquad c = \sin C$$ This is a common simplification technique, as it nicely blurs the distinction between edges and angles, giving us things like this wonderfully-symmetric area formula: $$\mathrm{Area} \; = \frac{1}{2}a b c \qquad \left(\;=\frac{1}{2}ab\sin C =\frac{1}{2}ac\sin B = \frac{1}{2}bc\sin A\right)$$
(When you have this mindset, you can't look at the expression "$\sin A + \sin B + \sin C$" and not think, "That's perimeter!" ... and then you find yourself pursuing proof approaches like this one.) In what follows, I'll continue to use "$a$", "$b$", "$c$", because they're more compact than "$\sin A$", etc, but you should read them as "$\sin A$", etc.
By the Law of Cosines,
$$\cos{C} = \frac{a^2+b^2-c^2}{2ab}$$
By the half-angle formula for cosines,
$$\cos^2\frac{C}{2} = \frac{1+\cos{C}}{2}=\frac{a^2+2ab+b^2-c^2}{4ab}=\frac{(a+b)^2-c^2}{4ab}=\frac{(a+b+c)(a+b-c)}{4ab}$$
Likewise for $\cos(A/2)$ and $\cos(B/2)$, so that $$\cos^2\frac{A}{2} \cos^2\frac{B}{2} \cos^2\frac{C}{2}=\frac{(a+b+c)^2}{4a^2b^2c^2}\cdot \frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{16}$$ The conveniently-separated second factor just happens to be Heron's formula for the square of the area of the triangle; re-writing the area in wonderfully-symmetric form gives ... $$\cos^2\frac{A}{2} \cos^2\frac{B}{2} \cos^2\frac{C}{2}=\frac{(a+b+c)^2}{4a^2b^2c^2}\cdot \left(\frac{1}{2}abc\right)^2 = \frac{1}{16}\left(a+b+c\right)^2$$
We can now clear the fraction, expand the symbols "$a$", "$b$", "$c$" as the sines they represent, and take square roots (secure in the knowledge that none of the trig values is negative), so that $$4 \cos \frac{A}{2} \cos\frac{B}{2} \cos\frac{C}{2} = \sin A + \sin B + \sin C$$ as desired.
$$ \begin{align} \sin(A)+\sin(B)+\sin(C) &=\sin(A)+\sin(B)+\sin(\pi-A-B)\\[9pt] &=\color{#C00000}{\sin(A)+\sin(B)}+\color{#00A000}{\sin(A+B)}\\[6pt] &=\color{#C00000}{2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}+\color{#00A000}{2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A+B}{2}\right)}\\ &=2\sin\left(\frac{A+B}{2}\right)\left(\cos\left(\frac{A-B}{2}\right)+\cos\left(\frac{A+B}{2}\right)\right)\\ &=2\sin\left(\frac{A+B}{2}\right)\;2\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\\ &=4\cos\left(\frac\pi2-\frac{A+B}{2}\right)\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right)\\ &=4\cos\left(\frac{C}{2}\right)\cos\left(\frac{A}{2}\right)\cos\left(\frac{B}{2}\right) \end{align} $$
As Salech pointed out in the comment above, we have $A=180^\circ-(B+C)$, so applying angle sum and difference formulas for cosine and sine, we have
$\begin{eqnarray*} \cos\frac A2 & = & \cos\left(90^\circ-\frac{B+C}2\right)\\ & = & \cos 90^\circ\cos\left(\frac B2+\frac C2\right)+\sin 90^\circ\sin\left(\frac B2+\frac C2\right)\\ & = & \sin\left(\frac B2+\frac C2\right)\\ & = & \sin\frac B2\cos\frac C2+\cos\frac B2\sin\frac C2, \end{eqnarray*}$
since $\cos 90^\circ=0$ and $\sin 90^\circ=1$.
Hence, $$4\cos\frac A2\cos\frac B2\cos\frac C2 = 4\sin\frac B2\cos\frac B2\cos^2\frac C2+4\cos^2\frac B2\sin\frac C2\cos\frac C2.$$
Now, we've got the double angle identities $\sin(2x)=2\sin x\cos x$ and $\cos(2x)=2\cos^2x-1$, the latter of which gives us $2\cos^2x=1+\cos(2x)$. Hence, by substitution and application of angle addition formula from sine,
$\begin{eqnarray*} 4\cos\frac A2\cos\frac B2\cos\frac C2 & = & \left(2\sin\frac B2\cos\frac B2\right)\left(2\cos^2\frac C2\right)+\left(2\cos^2\frac B2\right)\left(2\sin\frac C2\cos\frac C2\right)\\ & = & \sin B(1+\cos C)+(1+\cos B)\sin C\\ & = & \sin B\cos C+\cos B\sin C+\sin B+\sin C\\ & = & \sin(B+C)+\sin B+\sin C. \end{eqnarray*}$
Finally, observe that $\sin(180^\circ-x)=\sin x$, so since $A=180^\circ-(B+C)$, then $\sin A=\sin(B+C)$, and therefore, $$4\cos\frac A2\cos\frac B2\cos\frac C2=\sin A+\sin B+\sin C,$$ as desired.
Since $C = \pi - A - B$, what needs to be shown is $$\sin(A) + \sin(B) + \sin(\pi - A - B) = 4\cos({A \over 2})\cos({B \over 2})\cos({\pi \over 2} - {A + B \over 2})$$ Equivalently, you need that $$\sin(A) + \sin(B) + \sin( A + B) = 4\cos({A \over 2})\cos({B \over 2})\sin( {A + B \over 2})$$ Using the trig identity $\cos(x)\sin(y) = {1 \over 2}(\sin(x + y) + \sin(y - x))$, this is the same as $$\sin(A) + \sin(B) + \sin( A + B) = 2\cos({A \over 2})\sin({A \over 2} + B) + 2\cos({A \over 2})\sin({A \over 2}) $$ Using it again on the first term on the right, this becomes $$\sin(A) + \sin(B) + \sin( A + B) = \sin(A + B) + \sin(B) + 2\cos({A \over 2})\sin({A \over 2}) $$ This collapses into the sine double angle identity $$\sin(A) = 2\cos({A \over 2})\sin({A \over 2}) $$ Reversing the above steps gives the desired result.
