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Let ABC be a triangle. Thus prove that $$\sin^2\left({\frac{A}{2}}\right)+\sin^2\left(\frac{B}{2}\right)+\sin^2\left(\frac{C}{2}\right)+2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)= 1$$
How do i go about solving this?

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    $\begingroup$ Usec c/2 = pi/2 - (a +B)/2 and then the sum of angle formulas $\endgroup$ – Shailesh Nov 1 '15 at 6:57
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Let $x=\frac{A}{2}$, $y=\frac{B}{2}$, $z=\frac{C}{2}$, $x+y+z=\pi/2$ $$\sin^2(x)+\sin^2(y)+\sin^2(z)+2\sin(x)\sin(y)\sin(z)\\=\sin^2(x)+\sin^2(y)+\sin^2(\pi/2-x-y)+2\sin(x)\sin(y)\sin(\pi/2-x-y)\\=\sin^2(x)+\sin^2(y)+\cos^2(x+y)+2\sin(x)\sin(y)\cos(x+y)\\=\sin^2(x)+\sin^2(y)+\cos^2(x+y)+(\cos(x-y)-\cos(x+y))\cos(x+y)\\=\sin^2(x)+\sin^2(y)+\cos(x-y)\cos(x+y)\\=\frac{1-\cos(2x)}{2}+\frac{1-\cos(2y)}{2}+\cos(x-y)\cos(x+y)\\=1-\frac{\cos(2x)+\cos(2y)}{2}+\cos(x-y)\cos(x+y)\\=1-\cos(x-y)\cos(x+y)+\cos(x-y)\cos(x+y)\\=1$$

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$$F=\sin^2\dfrac A2+\sin^2\dfrac B2+\sin^2\dfrac C2 =1-\left(\cos^2\dfrac A2-\sin^2\dfrac B2\right)+\sin^2\dfrac C2$$

Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,

$$F=1-\cos\left(\dfrac{A-B}2\right)\cos\left(\dfrac{A+B}2\right)+\sin^2\dfrac C2$$

Using $\cos\left(\dfrac{A+B}2\right)=\cos\dfrac{\pi-C}2=\sin\dfrac C2,$

$$F=1-\sin\dfrac C2\left(\cos\left(\dfrac{A-B}2\right)-\cos\left(\dfrac{A+B}2\right)\right)$$

Now $\cos\left(\dfrac{A-B}2\right)-\cos\left(\dfrac{A+B}2\right)=2\sin\dfrac A2\sin\dfrac B2$

Hope you can take it from here!

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